venv/lib/python3.10/site-packages/scipy/stats/_morestats.py

from __future__ import annotations
import math
import warnings
from collections import namedtuple

import numpy as np
from numpy import (isscalar, r_, log, around, unique, asarray, zeros,
                   arange, sort, amin, amax, sqrt, array, atleast_1d,  # noqa: F401
                   compress, pi, exp, ravel, count_nonzero, sin, cos,  # noqa: F401
                   arctan2, hypot)

from scipy import optimize, special, interpolate, stats
from scipy._lib._bunch import _make_tuple_bunch
from scipy._lib._util import _rename_parameter, _contains_nan, _get_nan

from ._ansari_swilk_statistics import gscale, swilk
from . import _stats_py, _wilcoxon
from ._fit import FitResult
from ._stats_py import find_repeats, _get_pvalue, SignificanceResult  # noqa: F401
from .contingency import chi2_contingency
from . import distributions
from ._distn_infrastructure import rv_generic
from ._axis_nan_policy import _axis_nan_policy_factory


__all__ = ['mvsdist',
           'bayes_mvs', 'kstat', 'kstatvar', 'probplot', 'ppcc_max', 'ppcc_plot',
           'boxcox_llf', 'boxcox', 'boxcox_normmax', 'boxcox_normplot',
           'shapiro', 'anderson', 'ansari', 'bartlett', 'levene',
           'fligner', 'mood', 'wilcoxon', 'median_test',
           'circmean', 'circvar', 'circstd', 'anderson_ksamp',
           'yeojohnson_llf', 'yeojohnson', 'yeojohnson_normmax',
           'yeojohnson_normplot', 'directional_stats',
           'false_discovery_control'
           ]


Mean = namedtuple('Mean', ('statistic', 'minmax'))
Variance = namedtuple('Variance', ('statistic', 'minmax'))
Std_dev = namedtuple('Std_dev', ('statistic', 'minmax'))


def bayes_mvs(data, alpha=0.90):
    r"""
    Bayesian confidence intervals for the mean, var, and std.

    Parameters
    ----------
    data : array_like
        Input data, if multi-dimensional it is flattened to 1-D by `bayes_mvs`.
        Requires 2 or more data points.
    alpha : float, optional
        Probability that the returned confidence interval contains
        the true parameter.

    Returns
    -------
    mean_cntr, var_cntr, std_cntr : tuple
        The three results are for the mean, variance and standard deviation,
        respectively.  Each result is a tuple of the form::

            (center, (lower, upper))

        with `center` the mean of the conditional pdf of the value given the
        data, and `(lower, upper)` a confidence interval, centered on the
        median, containing the estimate to a probability ``alpha``.

    See Also
    --------
    mvsdist

    Notes
    -----
    Each tuple of mean, variance, and standard deviation estimates represent
    the (center, (lower, upper)) with center the mean of the conditional pdf
    of the value given the data and (lower, upper) is a confidence interval
    centered on the median, containing the estimate to a probability
    ``alpha``.

    Converts data to 1-D and assumes all data has the same mean and variance.
    Uses Jeffrey's prior for variance and std.

    Equivalent to ``tuple((x.mean(), x.interval(alpha)) for x in mvsdist(dat))``

    References
    ----------
    T.E. Oliphant, "A Bayesian perspective on estimating mean, variance, and
    standard-deviation from data", https://scholarsarchive.byu.edu/facpub/278,
    2006.

    Examples
    --------
    First a basic example to demonstrate the outputs:

    >>> from scipy import stats
    >>> data = [6, 9, 12, 7, 8, 8, 13]
    >>> mean, var, std = stats.bayes_mvs(data)
    >>> mean
    Mean(statistic=9.0, minmax=(7.103650222612533, 10.896349777387467))
    >>> var
    Variance(statistic=10.0, minmax=(3.176724206..., 24.45910382...))
    >>> std
    Std_dev(statistic=2.9724954732045084,
            minmax=(1.7823367265645143, 4.945614605014631))

    Now we generate some normally distributed random data, and get estimates of
    mean and standard deviation with 95% confidence intervals for those
    estimates:

    >>> n_samples = 100000
    >>> data = stats.norm.rvs(size=n_samples)
    >>> res_mean, res_var, res_std = stats.bayes_mvs(data, alpha=0.95)

    >>> import matplotlib.pyplot as plt
    >>> fig = plt.figure()
    >>> ax = fig.add_subplot(111)
    >>> ax.hist(data, bins=100, density=True, label='Histogram of data')
    >>> ax.vlines(res_mean.statistic, 0, 0.5, colors='r', label='Estimated mean')
    >>> ax.axvspan(res_mean.minmax[0],res_mean.minmax[1], facecolor='r',
    ...            alpha=0.2, label=r'Estimated mean (95% limits)')
    >>> ax.vlines(res_std.statistic, 0, 0.5, colors='g', label='Estimated scale')
    >>> ax.axvspan(res_std.minmax[0],res_std.minmax[1], facecolor='g', alpha=0.2,
    ...            label=r'Estimated scale (95% limits)')

    >>> ax.legend(fontsize=10)
    >>> ax.set_xlim([-4, 4])
    >>> ax.set_ylim([0, 0.5])
    >>> plt.show()

    """
    m, v, s = mvsdist(data)
    if alpha >= 1 or alpha <= 0:
        raise ValueError("0 < alpha < 1 is required, but alpha=%s was given."
                         % alpha)

    m_res = Mean(m.mean(), m.interval(alpha))
    v_res = Variance(v.mean(), v.interval(alpha))
    s_res = Std_dev(s.mean(), s.interval(alpha))

    return m_res, v_res, s_res


def mvsdist(data):
    """
    'Frozen' distributions for mean, variance, and standard deviation of data.

    Parameters
    ----------
    data : array_like
        Input array. Converted to 1-D using ravel.
        Requires 2 or more data-points.

    Returns
    -------
    mdist : "frozen" distribution object
        Distribution object representing the mean of the data.
    vdist : "frozen" distribution object
        Distribution object representing the variance of the data.
    sdist : "frozen" distribution object
        Distribution object representing the standard deviation of the data.

    See Also
    --------
    bayes_mvs

    Notes
    -----
    The return values from ``bayes_mvs(data)`` is equivalent to
    ``tuple((x.mean(), x.interval(0.90)) for x in mvsdist(data))``.

    In other words, calling ``<dist>.mean()`` and ``<dist>.interval(0.90)``
    on the three distribution objects returned from this function will give
    the same results that are returned from `bayes_mvs`.

    References
    ----------
    T.E. Oliphant, "A Bayesian perspective on estimating mean, variance, and
    standard-deviation from data", https://scholarsarchive.byu.edu/facpub/278,
    2006.

    Examples
    --------
    >>> from scipy import stats
    >>> data = [6, 9, 12, 7, 8, 8, 13]
    >>> mean, var, std = stats.mvsdist(data)

    We now have frozen distribution objects "mean", "var" and "std" that we can
    examine:

    >>> mean.mean()
    9.0
    >>> mean.interval(0.95)
    (6.6120585482655692, 11.387941451734431)
    >>> mean.std()
    1.1952286093343936

    """
    x = ravel(data)
    n = len(x)
    if n < 2:
        raise ValueError("Need at least 2 data-points.")
    xbar = x.mean()
    C = x.var()
    if n > 1000:  # gaussian approximations for large n
        mdist = distributions.norm(loc=xbar, scale=math.sqrt(C / n))
        sdist = distributions.norm(loc=math.sqrt(C), scale=math.sqrt(C / (2. * n)))
        vdist = distributions.norm(loc=C, scale=math.sqrt(2.0 / n) * C)
    else:
        nm1 = n - 1
        fac = n * C / 2.
        val = nm1 / 2.
        mdist = distributions.t(nm1, loc=xbar, scale=math.sqrt(C / nm1))
        sdist = distributions.gengamma(val, -2, scale=math.sqrt(fac))
        vdist = distributions.invgamma(val, scale=fac)
    return mdist, vdist, sdist


@_axis_nan_policy_factory(
    lambda x: x, result_to_tuple=lambda x: (x,), n_outputs=1, default_axis=None
)
def kstat(data, n=2):
    r"""
    Return the nth k-statistic (1<=n<=4 so far).

    The nth k-statistic k_n is the unique symmetric unbiased estimator of the
    nth cumulant kappa_n.

    Parameters
    ----------
    data : array_like
        Input array. Note that n-D input gets flattened.
    n : int, {1, 2, 3, 4}, optional
        Default is equal to 2.

    Returns
    -------
    kstat : float
        The nth k-statistic.

    See Also
    --------
    kstatvar : Returns an unbiased estimator of the variance of the k-statistic
    moment : Returns the n-th central moment about the mean for a sample.

    Notes
    -----
    For a sample size n, the first few k-statistics are given by:

    .. math::

        k_{1} = \mu
        k_{2} = \frac{n}{n-1} m_{2}
        k_{3} = \frac{ n^{2} } {(n-1) (n-2)} m_{3}
        k_{4} = \frac{ n^{2} [(n + 1)m_{4} - 3(n - 1) m^2_{2}]} {(n-1) (n-2) (n-3)}

    where :math:`\mu` is the sample mean, :math:`m_2` is the sample
    variance, and :math:`m_i` is the i-th sample central moment.

    References
    ----------
    http://mathworld.wolfram.com/k-Statistic.html

    http://mathworld.wolfram.com/Cumulant.html

    Examples
    --------
    >>> from scipy import stats
    >>> from numpy.random import default_rng
    >>> rng = default_rng()

    As sample size increases, n-th moment and n-th k-statistic converge to the
    same number (although they aren't identical). In the case of the normal
    distribution, they converge to zero.

    >>> for n in [2, 3, 4, 5, 6, 7]:
    ...     x = rng.normal(size=10**n)
    ...     m, k = stats.moment(x, 3), stats.kstat(x, 3)
    ...     print("%.3g %.3g %.3g" % (m, k, m-k))
    -0.631 -0.651 0.0194  # random
    0.0282 0.0283 -8.49e-05
    -0.0454 -0.0454 1.36e-05
    7.53e-05 7.53e-05 -2.26e-09
    0.00166 0.00166 -4.99e-09
    -2.88e-06 -2.88e-06 8.63e-13
    """
    if n > 4 or n < 1:
        raise ValueError("k-statistics only supported for 1<=n<=4")
    n = int(n)
    S = np.zeros(n + 1, np.float64)
    data = ravel(data)
    N = data.size

    # raise ValueError on empty input
    if N == 0:
        raise ValueError("Data input must not be empty")

    # on nan input, return nan without warning
    if np.isnan(np.sum(data)):
        return np.nan

    for k in range(1, n + 1):
        S[k] = np.sum(data**k, axis=0)
    if n == 1:
        return S[1] * 1.0/N
    elif n == 2:
        return (N*S[2] - S[1]**2.0) / (N*(N - 1.0))
    elif n == 3:
        return (2*S[1]**3 - 3*N*S[1]*S[2] + N*N*S[3]) / (N*(N - 1.0)*(N - 2.0))
    elif n == 4:
        return ((-6*S[1]**4 + 12*N*S[1]**2 * S[2] - 3*N*(N-1.0)*S[2]**2 -
                 4*N*(N+1)*S[1]*S[3] + N*N*(N+1)*S[4]) /
                (N*(N-1.0)*(N-2.0)*(N-3.0)))
    else:
        raise ValueError("Should not be here.")


@_axis_nan_policy_factory(
    lambda x: x, result_to_tuple=lambda x: (x,), n_outputs=1, default_axis=None
)
def kstatvar(data, n=2):
    r"""Return an unbiased estimator of the variance of the k-statistic.

    See `kstat` for more details of the k-statistic.

    Parameters
    ----------
    data : array_like
        Input array. Note that n-D input gets flattened.
    n : int, {1, 2}, optional
        Default is equal to 2.

    Returns
    -------
    kstatvar : float
        The nth k-statistic variance.

    See Also
    --------
    kstat : Returns the n-th k-statistic.
    moment : Returns the n-th central moment about the mean for a sample.

    Notes
    -----
    The variances of the first few k-statistics are given by:

    .. math::

        var(k_{1}) = \frac{\kappa^2}{n}
        var(k_{2}) = \frac{\kappa^4}{n} + \frac{2\kappa^2_{2}}{n - 1}
        var(k_{3}) = \frac{\kappa^6}{n} + \frac{9 \kappa_2 \kappa_4}{n - 1} +
                     \frac{9 \kappa^2_{3}}{n - 1} +
                     \frac{6 n \kappa^3_{2}}{(n-1) (n-2)}
        var(k_{4}) = \frac{\kappa^8}{n} + \frac{16 \kappa_2 \kappa_6}{n - 1} +
                     \frac{48 \kappa_{3} \kappa_5}{n - 1} +
                     \frac{34 \kappa^2_{4}}{n-1} +
                     \frac{72 n \kappa^2_{2} \kappa_4}{(n - 1) (n - 2)} +
                     \frac{144 n \kappa_{2} \kappa^2_{3}}{(n - 1) (n - 2)} +
                     \frac{24 (n + 1) n \kappa^4_{2}}{(n - 1) (n - 2) (n - 3)}
    """  # noqa: E501
    data = ravel(data)
    N = len(data)
    if n == 1:
        return kstat(data, n=2) * 1.0/N
    elif n == 2:
        k2 = kstat(data, n=2)
        k4 = kstat(data, n=4)
        return (2*N*k2**2 + (N-1)*k4) / (N*(N+1))
    else:
        raise ValueError("Only n=1 or n=2 supported.")


def _calc_uniform_order_statistic_medians(n):
    """Approximations of uniform order statistic medians.

    Parameters
    ----------
    n : int
        Sample size.

    Returns
    -------
    v : 1d float array
        Approximations of the order statistic medians.

    References
    ----------
    .. [1] James J. Filliben, "The Probability Plot Correlation Coefficient
           Test for Normality", Technometrics, Vol. 17, pp. 111-117, 1975.

    Examples
    --------
    Order statistics of the uniform distribution on the unit interval
    are marginally distributed according to beta distributions.
    The expectations of these order statistic are evenly spaced across
    the interval, but the distributions are skewed in a way that
    pushes the medians slightly towards the endpoints of the unit interval:

    >>> import numpy as np
    >>> n = 4
    >>> k = np.arange(1, n+1)
    >>> from scipy.stats import beta
    >>> a = k
    >>> b = n-k+1
    >>> beta.mean(a, b)
    array([0.2, 0.4, 0.6, 0.8])
    >>> beta.median(a, b)
    array([0.15910358, 0.38572757, 0.61427243, 0.84089642])

    The Filliben approximation uses the exact medians of the smallest
    and greatest order statistics, and the remaining medians are approximated
    by points spread evenly across a sub-interval of the unit interval:

    >>> from scipy.stats._morestats import _calc_uniform_order_statistic_medians
    >>> _calc_uniform_order_statistic_medians(n)
    array([0.15910358, 0.38545246, 0.61454754, 0.84089642])

    This plot shows the skewed distributions of the order statistics
    of a sample of size four from a uniform distribution on the unit interval:

    >>> import matplotlib.pyplot as plt
    >>> x = np.linspace(0.0, 1.0, num=50, endpoint=True)
    >>> pdfs = [beta.pdf(x, a[i], b[i]) for i in range(n)]
    >>> plt.figure()
    >>> plt.plot(x, pdfs[0], x, pdfs[1], x, pdfs[2], x, pdfs[3])

    """
    v = np.empty(n, dtype=np.float64)
    v[-1] = 0.5**(1.0 / n)
    v[0] = 1 - v[-1]
    i = np.arange(2, n)
    v[1:-1] = (i - 0.3175) / (n + 0.365)
    return v


def _parse_dist_kw(dist, enforce_subclass=True):
    """Parse `dist` keyword.

    Parameters
    ----------
    dist : str or stats.distributions instance.
        Several functions take `dist` as a keyword, hence this utility
        function.
    enforce_subclass : bool, optional
        If True (default), `dist` needs to be a
        `_distn_infrastructure.rv_generic` instance.
        It can sometimes be useful to set this keyword to False, if a function
        wants to accept objects that just look somewhat like such an instance
        (for example, they have a ``ppf`` method).

    """
    if isinstance(dist, rv_generic):
        pass
    elif isinstance(dist, str):
        try:
            dist = getattr(distributions, dist)
        except AttributeError as e:
            raise ValueError("%s is not a valid distribution name" % dist) from e
    elif enforce_subclass:
        msg = ("`dist` should be a stats.distributions instance or a string "
               "with the name of such a distribution.")
        raise ValueError(msg)

    return dist


def _add_axis_labels_title(plot, xlabel, ylabel, title):
    """Helper function to add axes labels and a title to stats plots."""
    try:
        if hasattr(plot, 'set_title'):
            # Matplotlib Axes instance or something that looks like it
            plot.set_title(title)
            plot.set_xlabel(xlabel)
            plot.set_ylabel(ylabel)
        else:
            # matplotlib.pyplot module
            plot.title(title)
            plot.xlabel(xlabel)
            plot.ylabel(ylabel)
    except Exception:
        # Not an MPL object or something that looks (enough) like it.
        # Don't crash on adding labels or title
        pass


def probplot(x, sparams=(), dist='norm', fit=True, plot=None, rvalue=False):
    """
    Calculate quantiles for a probability plot, and optionally show the plot.

    Generates a probability plot of sample data against the quantiles of a
    specified theoretical distribution (the normal distribution by default).
    `probplot` optionally calculates a best-fit line for the data and plots the
    results using Matplotlib or a given plot function.

    Parameters
    ----------
    x : array_like
        Sample/response data from which `probplot` creates the plot.
    sparams : tuple, optional
        Distribution-specific shape parameters (shape parameters plus location
        and scale).
    dist : str or stats.distributions instance, optional
        Distribution or distribution function name. The default is 'norm' for a
        normal probability plot.  Objects that look enough like a
        stats.distributions instance (i.e. they have a ``ppf`` method) are also
        accepted.
    fit : bool, optional
        Fit a least-squares regression (best-fit) line to the sample data if
        True (default).
    plot : object, optional
        If given, plots the quantiles.
        If given and `fit` is True, also plots the least squares fit.
        `plot` is an object that has to have methods "plot" and "text".
        The `matplotlib.pyplot` module or a Matplotlib Axes object can be used,
        or a custom object with the same methods.
        Default is None, which means that no plot is created.
    rvalue : bool, optional
        If `plot` is provided and `fit` is True, setting `rvalue` to True
        includes the coefficient of determination on the plot.
        Default is False.

    Returns
    -------
    (osm, osr) : tuple of ndarrays
        Tuple of theoretical quantiles (osm, or order statistic medians) and
        ordered responses (osr).  `osr` is simply sorted input `x`.
        For details on how `osm` is calculated see the Notes section.
    (slope, intercept, r) : tuple of floats, optional
        Tuple  containing the result of the least-squares fit, if that is
        performed by `probplot`. `r` is the square root of the coefficient of
        determination.  If ``fit=False`` and ``plot=None``, this tuple is not
        returned.

    Notes
    -----
    Even if `plot` is given, the figure is not shown or saved by `probplot`;
    ``plt.show()`` or ``plt.savefig('figname.png')`` should be used after
    calling `probplot`.

    `probplot` generates a probability plot, which should not be confused with
    a Q-Q or a P-P plot.  Statsmodels has more extensive functionality of this
    type, see ``statsmodels.api.ProbPlot``.

    The formula used for the theoretical quantiles (horizontal axis of the
    probability plot) is Filliben's estimate::

        quantiles = dist.ppf(val), for

                0.5**(1/n),                  for i = n
          val = (i - 0.3175) / (n + 0.365),  for i = 2, ..., n-1
                1 - 0.5**(1/n),              for i = 1

    where ``i`` indicates the i-th ordered value and ``n`` is the total number
    of values.

    Examples
    --------
    >>> import numpy as np
    >>> from scipy import stats
    >>> import matplotlib.pyplot as plt
    >>> nsample = 100
    >>> rng = np.random.default_rng()

    A t distribution with small degrees of freedom:

    >>> ax1 = plt.subplot(221)
    >>> x = stats.t.rvs(3, size=nsample, random_state=rng)
    >>> res = stats.probplot(x, plot=plt)

    A t distribution with larger degrees of freedom:

    >>> ax2 = plt.subplot(222)
    >>> x = stats.t.rvs(25, size=nsample, random_state=rng)
    >>> res = stats.probplot(x, plot=plt)

    A mixture of two normal distributions with broadcasting:

    >>> ax3 = plt.subplot(223)
    >>> x = stats.norm.rvs(loc=[0,5], scale=[1,1.5],
    ...                    size=(nsample//2,2), random_state=rng).ravel()
    >>> res = stats.probplot(x, plot=plt)

    A standard normal distribution:

    >>> ax4 = plt.subplot(224)
    >>> x = stats.norm.rvs(loc=0, scale=1, size=nsample, random_state=rng)
    >>> res = stats.probplot(x, plot=plt)

    Produce a new figure with a loggamma distribution, using the ``dist`` and
    ``sparams`` keywords:

    >>> fig = plt.figure()
    >>> ax = fig.add_subplot(111)
    >>> x = stats.loggamma.rvs(c=2.5, size=500, random_state=rng)
    >>> res = stats.probplot(x, dist=stats.loggamma, sparams=(2.5,), plot=ax)
    >>> ax.set_title("Probplot for loggamma dist with shape parameter 2.5")

    Show the results with Matplotlib:

    >>> plt.show()

    """
    x = np.asarray(x)
    if x.size == 0:
        if fit:
            return (x, x), (np.nan, np.nan, 0.0)
        else:
            return x, x

    osm_uniform = _calc_uniform_order_statistic_medians(len(x))
    dist = _parse_dist_kw(dist, enforce_subclass=False)
    if sparams is None:
        sparams = ()
    if isscalar(sparams):
        sparams = (sparams,)
    if not isinstance(sparams, tuple):
        sparams = tuple(sparams)

    osm = dist.ppf(osm_uniform, *sparams)
    osr = sort(x)
    if fit:
        # perform a linear least squares fit.
        slope, intercept, r, prob, _ = _stats_py.linregress(osm, osr)

    if plot is not None:
        plot.plot(osm, osr, 'bo')
        if fit:
            plot.plot(osm, slope*osm + intercept, 'r-')
        _add_axis_labels_title(plot, xlabel='Theoretical quantiles',
                               ylabel='Ordered Values',
                               title='Probability Plot')

        # Add R^2 value to the plot as text
        if fit and rvalue:
            xmin = amin(osm)
            xmax = amax(osm)
            ymin = amin(x)
            ymax = amax(x)
            posx = xmin + 0.70 * (xmax - xmin)
            posy = ymin + 0.01 * (ymax - ymin)
            plot.text(posx, posy, "$R^2=%1.4f$" % r**2)

    if fit:
        return (osm, osr), (slope, intercept, r)
    else:
        return osm, osr


def ppcc_max(x, brack=(0.0, 1.0), dist='tukeylambda'):
    """Calculate the shape parameter that maximizes the PPCC.

    The probability plot correlation coefficient (PPCC) plot can be used
    to determine the optimal shape parameter for a one-parameter family
    of distributions. ``ppcc_max`` returns the shape parameter that would
    maximize the probability plot correlation coefficient for the given
    data to a one-parameter family of distributions.

    Parameters
    ----------
    x : array_like
        Input array.
    brack : tuple, optional
        Triple (a,b,c) where (a<b<c). If bracket consists of two numbers (a, c)
        then they are assumed to be a starting interval for a downhill bracket
        search (see `scipy.optimize.brent`).
    dist : str or stats.distributions instance, optional
        Distribution or distribution function name.  Objects that look enough
        like a stats.distributions instance (i.e. they have a ``ppf`` method)
        are also accepted.  The default is ``'tukeylambda'``.

    Returns
    -------
    shape_value : float
        The shape parameter at which the probability plot correlation
        coefficient reaches its max value.

    See Also
    --------
    ppcc_plot, probplot, boxcox

    Notes
    -----
    The brack keyword serves as a starting point which is useful in corner
    cases. One can use a plot to obtain a rough visual estimate of the location
    for the maximum to start the search near it.

    References
    ----------
    .. [1] J.J. Filliben, "The Probability Plot Correlation Coefficient Test
           for Normality", Technometrics, Vol. 17, pp. 111-117, 1975.
    .. [2] Engineering Statistics Handbook, NIST/SEMATEC,
           https://www.itl.nist.gov/div898/handbook/eda/section3/ppccplot.htm

    Examples
    --------
    First we generate some random data from a Weibull distribution
    with shape parameter 2.5:

    >>> import numpy as np
    >>> from scipy import stats
    >>> import matplotlib.pyplot as plt
    >>> rng = np.random.default_rng()
    >>> c = 2.5
    >>> x = stats.weibull_min.rvs(c, scale=4, size=2000, random_state=rng)

    Generate the PPCC plot for this data with the Weibull distribution.

    >>> fig, ax = plt.subplots(figsize=(8, 6))
    >>> res = stats.ppcc_plot(x, c/2, 2*c, dist='weibull_min', plot=ax)

    We calculate the value where the shape should reach its maximum and a
    red line is drawn there. The line should coincide with the highest
    point in the PPCC graph.

    >>> cmax = stats.ppcc_max(x, brack=(c/2, 2*c), dist='weibull_min')
    >>> ax.axvline(cmax, color='r')
    >>> plt.show()

    """
    dist = _parse_dist_kw(dist)
    osm_uniform = _calc_uniform_order_statistic_medians(len(x))
    osr = sort(x)

    # this function computes the x-axis values of the probability plot
    #  and computes a linear regression (including the correlation)
    #  and returns 1-r so that a minimization function maximizes the
    #  correlation
    def tempfunc(shape, mi, yvals, func):
        xvals = func(mi, shape)
        r, prob = _stats_py.pearsonr(xvals, yvals)
        return 1 - r

    return optimize.brent(tempfunc, brack=brack,
                          args=(osm_uniform, osr, dist.ppf))


def ppcc_plot(x, a, b, dist='tukeylambda', plot=None, N=80):
    """Calculate and optionally plot probability plot correlation coefficient.

    The probability plot correlation coefficient (PPCC) plot can be used to
    determine the optimal shape parameter for a one-parameter family of
    distributions.  It cannot be used for distributions without shape
    parameters
    (like the normal distribution) or with multiple shape parameters.

    By default a Tukey-Lambda distribution (`stats.tukeylambda`) is used. A
    Tukey-Lambda PPCC plot interpolates from long-tailed to short-tailed
    distributions via an approximately normal one, and is therefore
    particularly useful in practice.

    Parameters
    ----------
    x : array_like
        Input array.
    a, b : scalar
        Lower and upper bounds of the shape parameter to use.
    dist : str or stats.distributions instance, optional
        Distribution or distribution function name.  Objects that look enough
        like a stats.distributions instance (i.e. they have a ``ppf`` method)
        are also accepted.  The default is ``'tukeylambda'``.
    plot : object, optional
        If given, plots PPCC against the shape parameter.
        `plot` is an object that has to have methods "plot" and "text".
        The `matplotlib.pyplot` module or a Matplotlib Axes object can be used,
        or a custom object with the same methods.
        Default is None, which means that no plot is created.
    N : int, optional
        Number of points on the horizontal axis (equally distributed from
        `a` to `b`).

    Returns
    -------
    svals : ndarray
        The shape values for which `ppcc` was calculated.
    ppcc : ndarray
        The calculated probability plot correlation coefficient values.

    See Also
    --------
    ppcc_max, probplot, boxcox_normplot, tukeylambda

    References
    ----------
    J.J. Filliben, "The Probability Plot Correlation Coefficient Test for
    Normality", Technometrics, Vol. 17, pp. 111-117, 1975.

    Examples
    --------
    First we generate some random data from a Weibull distribution
    with shape parameter 2.5, and plot the histogram of the data:

    >>> import numpy as np
    >>> from scipy import stats
    >>> import matplotlib.pyplot as plt
    >>> rng = np.random.default_rng()
    >>> c = 2.5
    >>> x = stats.weibull_min.rvs(c, scale=4, size=2000, random_state=rng)

    Take a look at the histogram of the data.

    >>> fig1, ax = plt.subplots(figsize=(9, 4))
    >>> ax.hist(x, bins=50)
    >>> ax.set_title('Histogram of x')
    >>> plt.show()

    Now we explore this data with a PPCC plot as well as the related
    probability plot and Box-Cox normplot.  A red line is drawn where we
    expect the PPCC value to be maximal (at the shape parameter ``c``
    used above):

    >>> fig2 = plt.figure(figsize=(12, 4))
    >>> ax1 = fig2.add_subplot(1, 3, 1)
    >>> ax2 = fig2.add_subplot(1, 3, 2)
    >>> ax3 = fig2.add_subplot(1, 3, 3)
    >>> res = stats.probplot(x, plot=ax1)
    >>> res = stats.boxcox_normplot(x, -4, 4, plot=ax2)
    >>> res = stats.ppcc_plot(x, c/2, 2*c, dist='weibull_min', plot=ax3)
    >>> ax3.axvline(c, color='r')
    >>> plt.show()

    """
    if b <= a:
        raise ValueError("`b` has to be larger than `a`.")

    svals = np.linspace(a, b, num=N)
    ppcc = np.empty_like(svals)
    for k, sval in enumerate(svals):
        _, r2 = probplot(x, sval, dist=dist, fit=True)
        ppcc[k] = r2[-1]

    if plot is not None:
        plot.plot(svals, ppcc, 'x')
        _add_axis_labels_title(plot, xlabel='Shape Values',
                               ylabel='Prob Plot Corr. Coef.',
                               title='(%s) PPCC Plot' % dist)

    return svals, ppcc


def _log_mean(logx):
    # compute log of mean of x from log(x)
    return special.logsumexp(logx, axis=0) - np.log(len(logx))


def _log_var(logx):
    # compute log of variance of x from log(x)
    logmean = _log_mean(logx)
    pij = np.full_like(logx, np.pi * 1j, dtype=np.complex128)
    logxmu = special.logsumexp([logx, logmean + pij], axis=0)
    return np.real(special.logsumexp(2 * logxmu, axis=0)) - np.log(len(logx))


def boxcox_llf(lmb, data):
    r"""The boxcox log-likelihood function.

    Parameters
    ----------
    lmb : scalar
        Parameter for Box-Cox transformation.  See `boxcox` for details.
    data : array_like
        Data to calculate Box-Cox log-likelihood for.  If `data` is
        multi-dimensional, the log-likelihood is calculated along the first
        axis.

    Returns
    -------
    llf : float or ndarray
        Box-Cox log-likelihood of `data` given `lmb`.  A float for 1-D `data`,
        an array otherwise.

    See Also
    --------
    boxcox, probplot, boxcox_normplot, boxcox_normmax

    Notes
    -----
    The Box-Cox log-likelihood function is defined here as

    .. math::

        llf = (\lambda - 1) \sum_i(\log(x_i)) -
              N/2 \log(\sum_i (y_i - \bar{y})^2 / N),

    where ``y`` is the Box-Cox transformed input data ``x``.

    Examples
    --------
    >>> import numpy as np
    >>> from scipy import stats
    >>> import matplotlib.pyplot as plt
    >>> from mpl_toolkits.axes_grid1.inset_locator import inset_axes

    Generate some random variates and calculate Box-Cox log-likelihood values
    for them for a range of ``lmbda`` values:

    >>> rng = np.random.default_rng()
    >>> x = stats.loggamma.rvs(5, loc=10, size=1000, random_state=rng)
    >>> lmbdas = np.linspace(-2, 10)
    >>> llf = np.zeros(lmbdas.shape, dtype=float)
    >>> for ii, lmbda in enumerate(lmbdas):
    ...     llf[ii] = stats.boxcox_llf(lmbda, x)

    Also find the optimal lmbda value with `boxcox`:

    >>> x_most_normal, lmbda_optimal = stats.boxcox(x)

    Plot the log-likelihood as function of lmbda.  Add the optimal lmbda as a
    horizontal line to check that that's really the optimum:

    >>> fig = plt.figure()
    >>> ax = fig.add_subplot(111)
    >>> ax.plot(lmbdas, llf, 'b.-')
    >>> ax.axhline(stats.boxcox_llf(lmbda_optimal, x), color='r')
    >>> ax.set_xlabel('lmbda parameter')
    >>> ax.set_ylabel('Box-Cox log-likelihood')

    Now add some probability plots to show that where the log-likelihood is
    maximized the data transformed with `boxcox` looks closest to normal:

    >>> locs = [3, 10, 4]  # 'lower left', 'center', 'lower right'
    >>> for lmbda, loc in zip([-1, lmbda_optimal, 9], locs):
    ...     xt = stats.boxcox(x, lmbda=lmbda)
    ...     (osm, osr), (slope, intercept, r_sq) = stats.probplot(xt)
    ...     ax_inset = inset_axes(ax, width="20%", height="20%", loc=loc)
    ...     ax_inset.plot(osm, osr, 'c.', osm, slope*osm + intercept, 'k-')
    ...     ax_inset.set_xticklabels([])
    ...     ax_inset.set_yticklabels([])
    ...     ax_inset.set_title(r'$\lambda=%1.2f$' % lmbda)

    >>> plt.show()

    """
    data = np.asarray(data)
    N = data.shape[0]
    if N == 0:
        return np.nan

    logdata = np.log(data)

    # Compute the variance of the transformed data.
    if lmb == 0:
        logvar = np.log(np.var(logdata, axis=0))
    else:
        # Transform without the constant offset 1/lmb.  The offset does
        # not affect the variance, and the subtraction of the offset can
        # lead to loss of precision.
        # Division by lmb can be factored out to enhance numerical stability.
        logx = lmb * logdata
        logvar = _log_var(logx) - 2 * np.log(abs(lmb))

    return (lmb - 1) * np.sum(logdata, axis=0) - N/2 * logvar


def _boxcox_conf_interval(x, lmax, alpha):
    # Need to find the lambda for which
    #  f(x,lmbda) >= f(x,lmax) - 0.5*chi^2_alpha;1
    fac = 0.5 * distributions.chi2.ppf(1 - alpha, 1)
    target = boxcox_llf(lmax, x) - fac

    def rootfunc(lmbda, data, target):
        return boxcox_llf(lmbda, data) - target

    # Find positive endpoint of interval in which answer is to be found
    newlm = lmax + 0.5
    N = 0
    while (rootfunc(newlm, x, target) > 0.0) and (N < 500):
        newlm += 0.1
        N += 1

    if N == 500:
        raise RuntimeError("Could not find endpoint.")

    lmplus = optimize.brentq(rootfunc, lmax, newlm, args=(x, target))

    # Now find negative interval in the same way
    newlm = lmax - 0.5
    N = 0
    while (rootfunc(newlm, x, target) > 0.0) and (N < 500):
        newlm -= 0.1
        N += 1

    if N == 500:
        raise RuntimeError("Could not find endpoint.")

    lmminus = optimize.brentq(rootfunc, newlm, lmax, args=(x, target))
    return lmminus, lmplus


def boxcox(x, lmbda=None, alpha=None, optimizer=None):
    r"""Return a dataset transformed by a Box-Cox power transformation.

    Parameters
    ----------
    x : ndarray
        Input array to be transformed.

        If `lmbda` is not None, this is an alias of
        `scipy.special.boxcox`.
        Returns nan if ``x < 0``; returns -inf if ``x == 0 and lmbda < 0``.

        If `lmbda` is None, array must be positive, 1-dimensional, and
        non-constant.

    lmbda : scalar, optional
        If `lmbda` is None (default), find the value of `lmbda` that maximizes
        the log-likelihood function and return it as the second output
        argument.

        If `lmbda` is not None, do the transformation for that value.

    alpha : float, optional
        If `lmbda` is None and `alpha` is not None (default), return the
        ``100 * (1-alpha)%`` confidence  interval for `lmbda` as the third
        output argument. Must be between 0.0 and 1.0.

        If `lmbda` is not None, `alpha` is ignored.
    optimizer : callable, optional
        If `lmbda` is None, `optimizer` is the scalar optimizer used to find
        the value of `lmbda` that minimizes the negative log-likelihood
        function. `optimizer` is a callable that accepts one argument:

        fun : callable
            The objective function, which evaluates the negative
            log-likelihood function at a provided value of `lmbda`

        and returns an object, such as an instance of
        `scipy.optimize.OptimizeResult`, which holds the optimal value of
        `lmbda` in an attribute `x`.

        See the example in `boxcox_normmax` or the documentation of
        `scipy.optimize.minimize_scalar` for more information.

        If `lmbda` is not None, `optimizer` is ignored.

    Returns
    -------
    boxcox : ndarray
        Box-Cox power transformed array.
    maxlog : float, optional
        If the `lmbda` parameter is None, the second returned argument is
        the `lmbda` that maximizes the log-likelihood function.
    (min_ci, max_ci) : tuple of float, optional
        If `lmbda` parameter is None and `alpha` is not None, this returned
        tuple of floats represents the minimum and maximum confidence limits
        given `alpha`.

    See Also
    --------
    probplot, boxcox_normplot, boxcox_normmax, boxcox_llf

    Notes
    -----
    The Box-Cox transform is given by::

        y = (x**lmbda - 1) / lmbda,  for lmbda != 0
            log(x),                  for lmbda = 0

    `boxcox` requires the input data to be positive.  Sometimes a Box-Cox
    transformation provides a shift parameter to achieve this; `boxcox` does
    not.  Such a shift parameter is equivalent to adding a positive constant to
    `x` before calling `boxcox`.

    The confidence limits returned when `alpha` is provided give the interval
    where:

    .. math::

        llf(\hat{\lambda}) - llf(\lambda) < \frac{1}{2}\chi^2(1 - \alpha, 1),

    with ``llf`` the log-likelihood function and :math:`\chi^2` the chi-squared
    function.

    References
    ----------
    G.E.P. Box and D.R. Cox, "An Analysis of Transformations", Journal of the
    Royal Statistical Society B, 26, 211-252 (1964).

    Examples
    --------
    >>> from scipy import stats
    >>> import matplotlib.pyplot as plt

    We generate some random variates from a non-normal distribution and make a
    probability plot for it, to show it is non-normal in the tails:

    >>> fig = plt.figure()
    >>> ax1 = fig.add_subplot(211)
    >>> x = stats.loggamma.rvs(5, size=500) + 5
    >>> prob = stats.probplot(x, dist=stats.norm, plot=ax1)
    >>> ax1.set_xlabel('')
    >>> ax1.set_title('Probplot against normal distribution')

    We now use `boxcox` to transform the data so it's closest to normal:

    >>> ax2 = fig.add_subplot(212)
    >>> xt, _ = stats.boxcox(x)
    >>> prob = stats.probplot(xt, dist=stats.norm, plot=ax2)
    >>> ax2.set_title('Probplot after Box-Cox transformation')

    >>> plt.show()

    """
    x = np.asarray(x)

    if lmbda is not None:  # single transformation
        return special.boxcox(x, lmbda)

    if x.ndim != 1:
        raise ValueError("Data must be 1-dimensional.")

    if x.size == 0:
        return x

    if np.all(x == x[0]):
        raise ValueError("Data must not be constant.")

    if np.any(x <= 0):
        raise ValueError("Data must be positive.")

    # If lmbda=None, find the lmbda that maximizes the log-likelihood function.
    lmax = boxcox_normmax(x, method='mle', optimizer=optimizer)
    y = boxcox(x, lmax)

    if alpha is None:
        return y, lmax
    else:
        # Find confidence interval
        interval = _boxcox_conf_interval(x, lmax, alpha)
        return y, lmax, interval


def _boxcox_inv_lmbda(x, y):
    # compute lmbda given x and y for Box-Cox transformation
    num = special.lambertw(-(x ** (-1 / y)) * np.log(x) / y, k=-1)
    return np.real(-num / np.log(x) - 1 / y)


class _BigFloat:
    def __repr__(self):
        return "BIG_FLOAT"


def boxcox_normmax(
    x, brack=None, method='pearsonr', optimizer=None, *, ymax=_BigFloat()
):
    """Compute optimal Box-Cox transform parameter for input data.

    Parameters
    ----------
    x : array_like
        Input array. All entries must be positive, finite, real numbers.
    brack : 2-tuple, optional, default (-2.0, 2.0)
         The starting interval for a downhill bracket search for the default
         `optimize.brent` solver. Note that this is in most cases not
         critical; the final result is allowed to be outside this bracket.
         If `optimizer` is passed, `brack` must be None.
    method : str, optional
        The method to determine the optimal transform parameter (`boxcox`
        ``lmbda`` parameter). Options are:

        'pearsonr'  (default)
            Maximizes the Pearson correlation coefficient between
            ``y = boxcox(x)`` and the expected values for ``y`` if `x` would be
            normally-distributed.

        'mle'
            Maximizes the log-likelihood `boxcox_llf`.  This is the method used
            in `boxcox`.

        'all'
            Use all optimization methods available, and return all results.
            Useful to compare different methods.
    optimizer : callable, optional
        `optimizer` is a callable that accepts one argument:

        fun : callable
            The objective function to be minimized. `fun` accepts one argument,
            the Box-Cox transform parameter `lmbda`, and returns the value of
            the function (e.g., the negative log-likelihood) at the provided
            argument. The job of `optimizer` is to find the value of `lmbda`
            that *minimizes* `fun`.

        and returns an object, such as an instance of
        `scipy.optimize.OptimizeResult`, which holds the optimal value of
        `lmbda` in an attribute `x`.

        See the example below or the documentation of
        `scipy.optimize.minimize_scalar` for more information.
    ymax : float, optional
        The unconstrained optimal transform parameter may cause Box-Cox
        transformed data to have extreme magnitude or even overflow.
        This parameter constrains MLE optimization such that the magnitude
        of the transformed `x` does not exceed `ymax`. The default is
        the maximum value of the input dtype. If set to infinity,
        `boxcox_normmax` returns the unconstrained optimal lambda.
        Ignored when ``method='pearsonr'``.

    Returns
    -------
    maxlog : float or ndarray
        The optimal transform parameter found.  An array instead of a scalar
        for ``method='all'``.

    See Also
    --------
    boxcox, boxcox_llf, boxcox_normplot, scipy.optimize.minimize_scalar

    Examples
    --------
    >>> import numpy as np
    >>> from scipy import stats
    >>> import matplotlib.pyplot as plt

    We can generate some data and determine the optimal ``lmbda`` in various
    ways:

    >>> rng = np.random.default_rng()
    >>> x = stats.loggamma.rvs(5, size=30, random_state=rng) + 5
    >>> y, lmax_mle = stats.boxcox(x)
    >>> lmax_pearsonr = stats.boxcox_normmax(x)

    >>> lmax_mle
    2.217563431465757
    >>> lmax_pearsonr
    2.238318660200961
    >>> stats.boxcox_normmax(x, method='all')
    array([2.23831866, 2.21756343])

    >>> fig = plt.figure()
    >>> ax = fig.add_subplot(111)
    >>> prob = stats.boxcox_normplot(x, -10, 10, plot=ax)
    >>> ax.axvline(lmax_mle, color='r')
    >>> ax.axvline(lmax_pearsonr, color='g', ls='--')

    >>> plt.show()

    Alternatively, we can define our own `optimizer` function. Suppose we
    are only interested in values of `lmbda` on the interval [6, 7], we
    want to use `scipy.optimize.minimize_scalar` with ``method='bounded'``,
    and we want to use tighter tolerances when optimizing the log-likelihood
    function. To do this, we define a function that accepts positional argument
    `fun` and uses `scipy.optimize.minimize_scalar` to minimize `fun` subject
    to the provided bounds and tolerances:

    >>> from scipy import optimize
    >>> options = {'xatol': 1e-12}  # absolute tolerance on `x`
    >>> def optimizer(fun):
    ...     return optimize.minimize_scalar(fun, bounds=(6, 7),
    ...                                     method="bounded", options=options)
    >>> stats.boxcox_normmax(x, optimizer=optimizer)
    6.000...
    """
    x = np.asarray(x)
    end_msg = "exceed specified `ymax`."
    if isinstance(ymax, _BigFloat):
        dtype = x.dtype if np.issubdtype(x.dtype, np.floating) else np.float64
        # 10000 is a safety factor because `special.boxcox` overflows prematurely.
        ymax = np.finfo(dtype).max / 10000
        end_msg = f"overflow in {dtype}."
    elif ymax <= 0:
        raise ValueError("`ymax` must be strictly positive")

    # If optimizer is not given, define default 'brent' optimizer.
    if optimizer is None:

        # Set default value for `brack`.
        if brack is None:
            brack = (-2.0, 2.0)

        def _optimizer(func, args):
            return optimize.brent(func, args=args, brack=brack)

    # Otherwise check optimizer.
    else:
        if not callable(optimizer):
            raise ValueError("`optimizer` must be a callable")

        if brack is not None:
            raise ValueError("`brack` must be None if `optimizer` is given")

        # `optimizer` is expected to return a `OptimizeResult` object, we here
        # get the solution to the optimization problem.
        def _optimizer(func, args):
            def func_wrapped(x):
                return func(x, *args)
            return getattr(optimizer(func_wrapped), 'x', None)

    def _pearsonr(x):
        osm_uniform = _calc_uniform_order_statistic_medians(len(x))
        xvals = distributions.norm.ppf(osm_uniform)

        def _eval_pearsonr(lmbda, xvals, samps):
            # This function computes the x-axis values of the probability plot
            # and computes a linear regression (including the correlation) and
            # returns ``1 - r`` so that a minimization function maximizes the
            # correlation.
            y = boxcox(samps, lmbda)
            yvals = np.sort(y)
            r, prob = _stats_py.pearsonr(xvals, yvals)
            return 1 - r

        return _optimizer(_eval_pearsonr, args=(xvals, x))

    def _mle(x):
        def _eval_mle(lmb, data):
            # function to minimize
            return -boxcox_llf(lmb, data)

        return _optimizer(_eval_mle, args=(x,))

    def _all(x):
        maxlog = np.empty(2, dtype=float)
        maxlog[0] = _pearsonr(x)
        maxlog[1] = _mle(x)
        return maxlog

    methods = {'pearsonr': _pearsonr,
               'mle': _mle,
               'all': _all}
    if method not in methods.keys():
        raise ValueError("Method %s not recognized." % method)

    optimfunc = methods[method]

    try:
        res = optimfunc(x)
    except ValueError as e:
        if "infs or NaNs" in str(e):
            message = ("The `x` argument of `boxcox_normmax` must contain "
                       "only positive, finite, real numbers.")
            raise ValueError(message) from e
        else:
            raise e

    if res is None:
        message = ("The `optimizer` argument of `boxcox_normmax` must return "
                   "an object containing the optimal `lmbda` in attribute `x`.")
        raise ValueError(message)
    elif not np.isinf(ymax):  # adjust the final lambda
        # x > 1, boxcox(x) > 0; x < 1, boxcox(x) < 0
        xmax, xmin = np.max(x), np.min(x)
        if xmin >= 1:
            x_treme = xmax
        elif xmax <= 1:
            x_treme = xmin
        else:  # xmin < 1 < xmax
            indicator = special.boxcox(xmax, res) > abs(special.boxcox(xmin, res))
            if isinstance(res, np.ndarray):
                indicator = indicator[1]  # select corresponds with 'mle'
            x_treme = xmax if indicator else xmin

        mask = abs(special.boxcox(x_treme, res)) > ymax
        if np.any(mask):
            message = (
                f"The optimal lambda is {res}, but the returned lambda is the "
                f"constrained optimum to ensure that the maximum or the minimum "
                f"of the transformed data does not " + end_msg
            )
            warnings.warn(message, stacklevel=2)

            # Return the constrained lambda to ensure the transformation
            # does not cause overflow or exceed specified `ymax`
            constrained_res = _boxcox_inv_lmbda(x_treme, ymax * np.sign(x_treme - 1))

            if isinstance(res, np.ndarray):
                res[mask] = constrained_res
            else:
                res = constrained_res
    return res


def _normplot(method, x, la, lb, plot=None, N=80):
    """Compute parameters for a Box-Cox or Yeo-Johnson normality plot,
    optionally show it.

    See `boxcox_normplot` or `yeojohnson_normplot` for details.
    """

    if method == 'boxcox':
        title = 'Box-Cox Normality Plot'
        transform_func = boxcox
    else:
        title = 'Yeo-Johnson Normality Plot'
        transform_func = yeojohnson

    x = np.asarray(x)
    if x.size == 0:
        return x

    if lb <= la:
        raise ValueError("`lb` has to be larger than `la`.")

    if method == 'boxcox' and np.any(x <= 0):
        raise ValueError("Data must be positive.")

    lmbdas = np.linspace(la, lb, num=N)
    ppcc = lmbdas * 0.0
    for i, val in enumerate(lmbdas):
        # Determine for each lmbda the square root of correlation coefficient
        # of transformed x
        z = transform_func(x, lmbda=val)
        _, (_, _, r) = probplot(z, dist='norm', fit=True)
        ppcc[i] = r

    if plot is not None:
        plot.plot(lmbdas, ppcc, 'x')
        _add_axis_labels_title(plot, xlabel='$\\lambda$',
                               ylabel='Prob Plot Corr. Coef.',
                               title=title)

    return lmbdas, ppcc


def boxcox_normplot(x, la, lb, plot=None, N=80):
    """Compute parameters for a Box-Cox normality plot, optionally show it.

    A Box-Cox normality plot shows graphically what the best transformation
    parameter is to use in `boxcox` to obtain a distribution that is close
    to normal.

    Parameters
    ----------
    x : array_like
        Input array.
    la, lb : scalar
        The lower and upper bounds for the ``lmbda`` values to pass to `boxcox`
        for Box-Cox transformations.  These are also the limits of the
        horizontal axis of the plot if that is generated.
    plot : object, optional
        If given, plots the quantiles and least squares fit.
        `plot` is an object that has to have methods "plot" and "text".
        The `matplotlib.pyplot` module or a Matplotlib Axes object can be used,
        or a custom object with the same methods.
        Default is None, which means that no plot is created.
    N : int, optional
        Number of points on the horizontal axis (equally distributed from
        `la` to `lb`).

    Returns
    -------
    lmbdas : ndarray
        The ``lmbda`` values for which a Box-Cox transform was done.
    ppcc : ndarray
        Probability Plot Correlelation Coefficient, as obtained from `probplot`
        when fitting the Box-Cox transformed input `x` against a normal
        distribution.

    See Also
    --------
    probplot, boxcox, boxcox_normmax, boxcox_llf, ppcc_max

    Notes
    -----
    Even if `plot` is given, the figure is not shown or saved by
    `boxcox_normplot`; ``plt.show()`` or ``plt.savefig('figname.png')``
    should be used after calling `probplot`.

    Examples
    --------
    >>> from scipy import stats
    >>> import matplotlib.pyplot as plt

    Generate some non-normally distributed data, and create a Box-Cox plot:

    >>> x = stats.loggamma.rvs(5, size=500) + 5
    >>> fig = plt.figure()
    >>> ax = fig.add_subplot(111)
    >>> prob = stats.boxcox_normplot(x, -20, 20, plot=ax)

    Determine and plot the optimal ``lmbda`` to transform ``x`` and plot it in
    the same plot:

    >>> _, maxlog = stats.boxcox(x)
    >>> ax.axvline(maxlog, color='r')

    >>> plt.show()

    """
    return _normplot('boxcox', x, la, lb, plot, N)


def yeojohnson(x, lmbda=None):
    r"""Return a dataset transformed by a Yeo-Johnson power transformation.

    Parameters
    ----------
    x : ndarray
        Input array.  Should be 1-dimensional.
    lmbda : float, optional
        If ``lmbda`` is ``None``, find the lambda that maximizes the
        log-likelihood function and return it as the second output argument.
        Otherwise the transformation is done for the given value.

    Returns
    -------
    yeojohnson: ndarray
        Yeo-Johnson power transformed array.
    maxlog : float, optional
        If the `lmbda` parameter is None, the second returned argument is
        the lambda that maximizes the log-likelihood function.

    See Also
    --------
    probplot, yeojohnson_normplot, yeojohnson_normmax, yeojohnson_llf, boxcox

    Notes
    -----
    The Yeo-Johnson transform is given by::

        y = ((x + 1)**lmbda - 1) / lmbda,                for x >= 0, lmbda != 0
            log(x + 1),                                  for x >= 0, lmbda = 0
            -((-x + 1)**(2 - lmbda) - 1) / (2 - lmbda),  for x < 0, lmbda != 2
            -log(-x + 1),                                for x < 0, lmbda = 2

    Unlike `boxcox`, `yeojohnson` does not require the input data to be
    positive.

    .. versionadded:: 1.2.0


    References
    ----------
    I. Yeo and R.A. Johnson, "A New Family of Power Transformations to
    Improve Normality or Symmetry", Biometrika 87.4 (2000):


    Examples
    --------
    >>> from scipy import stats
    >>> import matplotlib.pyplot as plt

    We generate some random variates from a non-normal distribution and make a
    probability plot for it, to show it is non-normal in the tails:

    >>> fig = plt.figure()
    >>> ax1 = fig.add_subplot(211)
    >>> x = stats.loggamma.rvs(5, size=500) + 5
    >>> prob = stats.probplot(x, dist=stats.norm, plot=ax1)
    >>> ax1.set_xlabel('')
    >>> ax1.set_title('Probplot against normal distribution')

    We now use `yeojohnson` to transform the data so it's closest to normal:

    >>> ax2 = fig.add_subplot(212)
    >>> xt, lmbda = stats.yeojohnson(x)
    >>> prob = stats.probplot(xt, dist=stats.norm, plot=ax2)
    >>> ax2.set_title('Probplot after Yeo-Johnson transformation')

    >>> plt.show()

    """
    x = np.asarray(x)
    if x.size == 0:
        return x

    if np.issubdtype(x.dtype, np.complexfloating):
        raise ValueError('Yeo-Johnson transformation is not defined for '
                         'complex numbers.')

    if np.issubdtype(x.dtype, np.integer):
        x = x.astype(np.float64, copy=False)

    if lmbda is not None:
        return _yeojohnson_transform(x, lmbda)

    # if lmbda=None, find the lmbda that maximizes the log-likelihood function.
    lmax = yeojohnson_normmax(x)
    y = _yeojohnson_transform(x, lmax)

    return y, lmax


def _yeojohnson_transform(x, lmbda):
    """Returns `x` transformed by the Yeo-Johnson power transform with given
    parameter `lmbda`.
    """
    dtype = x.dtype if np.issubdtype(x.dtype, np.floating) else np.float64
    out = np.zeros_like(x, dtype=dtype)
    pos = x >= 0  # binary mask

    # when x >= 0
    if abs(lmbda) < np.spacing(1.):
        out[pos] = np.log1p(x[pos])
    else:  # lmbda != 0
        # more stable version of: ((x + 1) ** lmbda - 1) / lmbda
        out[pos] = np.expm1(lmbda * np.log1p(x[pos])) / lmbda

    # when x < 0
    if abs(lmbda - 2) > np.spacing(1.):
        out[~pos] = -np.expm1((2 - lmbda) * np.log1p(-x[~pos])) / (2 - lmbda)
    else:  # lmbda == 2
        out[~pos] = -np.log1p(-x[~pos])

    return out


def yeojohnson_llf(lmb, data):
    r"""The yeojohnson log-likelihood function.

    Parameters
    ----------
    lmb : scalar
        Parameter for Yeo-Johnson transformation. See `yeojohnson` for
        details.
    data : array_like
        Data to calculate Yeo-Johnson log-likelihood for. If `data` is
        multi-dimensional, the log-likelihood is calculated along the first
        axis.

    Returns
    -------
    llf : float
        Yeo-Johnson log-likelihood of `data` given `lmb`.

    See Also
    --------
    yeojohnson, probplot, yeojohnson_normplot, yeojohnson_normmax

    Notes
    -----
    The Yeo-Johnson log-likelihood function is defined here as

    .. math::

        llf = -N/2 \log(\hat{\sigma}^2) + (\lambda - 1)
              \sum_i \text{ sign }(x_i)\log(|x_i| + 1)

    where :math:`\hat{\sigma}^2` is estimated variance of the Yeo-Johnson
    transformed input data ``x``.

    .. versionadded:: 1.2.0

    Examples
    --------
    >>> import numpy as np
    >>> from scipy import stats
    >>> import matplotlib.pyplot as plt
    >>> from mpl_toolkits.axes_grid1.inset_locator import inset_axes

    Generate some random variates and calculate Yeo-Johnson log-likelihood
    values for them for a range of ``lmbda`` values:

    >>> x = stats.loggamma.rvs(5, loc=10, size=1000)
    >>> lmbdas = np.linspace(-2, 10)
    >>> llf = np.zeros(lmbdas.shape, dtype=float)
    >>> for ii, lmbda in enumerate(lmbdas):
    ...     llf[ii] = stats.yeojohnson_llf(lmbda, x)

    Also find the optimal lmbda value with `yeojohnson`:

    >>> x_most_normal, lmbda_optimal = stats.yeojohnson(x)

    Plot the log-likelihood as function of lmbda.  Add the optimal lmbda as a
    horizontal line to check that that's really the optimum:

    >>> fig = plt.figure()
    >>> ax = fig.add_subplot(111)
    >>> ax.plot(lmbdas, llf, 'b.-')
    >>> ax.axhline(stats.yeojohnson_llf(lmbda_optimal, x), color='r')
    >>> ax.set_xlabel('lmbda parameter')
    >>> ax.set_ylabel('Yeo-Johnson log-likelihood')

    Now add some probability plots to show that where the log-likelihood is
    maximized the data transformed with `yeojohnson` looks closest to normal:

    >>> locs = [3, 10, 4]  # 'lower left', 'center', 'lower right'
    >>> for lmbda, loc in zip([-1, lmbda_optimal, 9], locs):
    ...     xt = stats.yeojohnson(x, lmbda=lmbda)
    ...     (osm, osr), (slope, intercept, r_sq) = stats.probplot(xt)
    ...     ax_inset = inset_axes(ax, width="20%", height="20%", loc=loc)
    ...     ax_inset.plot(osm, osr, 'c.', osm, slope*osm + intercept, 'k-')
    ...     ax_inset.set_xticklabels([])
    ...     ax_inset.set_yticklabels([])
    ...     ax_inset.set_title(r'$\lambda=%1.2f$' % lmbda)

    >>> plt.show()

    """
    data = np.asarray(data)
    n_samples = data.shape[0]

    if n_samples == 0:
        return np.nan

    trans = _yeojohnson_transform(data, lmb)
    trans_var = trans.var(axis=0)
    loglike = np.empty_like(trans_var)

    # Avoid RuntimeWarning raised by np.log when the variance is too low
    tiny_variance = trans_var < np.finfo(trans_var.dtype).tiny
    loglike[tiny_variance] = np.inf

    loglike[~tiny_variance] = (
        -n_samples / 2 * np.log(trans_var[~tiny_variance]))
    loglike[~tiny_variance] += (
        (lmb - 1) * (np.sign(data) * np.log1p(np.abs(data))).sum(axis=0))
    return loglike


def yeojohnson_normmax(x, brack=None):
    """Compute optimal Yeo-Johnson transform parameter.

    Compute optimal Yeo-Johnson transform parameter for input data, using
    maximum likelihood estimation.

    Parameters
    ----------
    x : array_like
        Input array.
    brack : 2-tuple, optional
        The starting interval for a downhill bracket search with
        `optimize.brent`. Note that this is in most cases not critical; the
        final result is allowed to be outside this bracket. If None,
        `optimize.fminbound` is used with bounds that avoid overflow.

    Returns
    -------
    maxlog : float
        The optimal transform parameter found.

    See Also
    --------
    yeojohnson, yeojohnson_llf, yeojohnson_normplot

    Notes
    -----
    .. versionadded:: 1.2.0

    Examples
    --------
    >>> import numpy as np
    >>> from scipy import stats
    >>> import matplotlib.pyplot as plt

    Generate some data and determine optimal ``lmbda``

    >>> rng = np.random.default_rng()
    >>> x = stats.loggamma.rvs(5, size=30, random_state=rng) + 5
    >>> lmax = stats.yeojohnson_normmax(x)

    >>> fig = plt.figure()
    >>> ax = fig.add_subplot(111)
    >>> prob = stats.yeojohnson_normplot(x, -10, 10, plot=ax)
    >>> ax.axvline(lmax, color='r')

    >>> plt.show()

    """
    def _neg_llf(lmbda, data):
        llf = yeojohnson_llf(lmbda, data)
        # reject likelihoods that are inf which are likely due to small
        # variance in the transformed space
        llf[np.isinf(llf)] = -np.inf
        return -llf

    with np.errstate(invalid='ignore'):
        if not np.all(np.isfinite(x)):
            raise ValueError('Yeo-Johnson input must be finite.')
        if np.all(x == 0):
            return 1.0
        if brack is not None:
            return optimize.brent(_neg_llf, brack=brack, args=(x,))
        x = np.asarray(x)
        dtype = x.dtype if np.issubdtype(x.dtype, np.floating) else np.float64
        # Allow values up to 20 times the maximum observed value to be safely
        # transformed without over- or underflow.
        log1p_max_x = np.log1p(20 * np.max(np.abs(x)))
        # Use half of floating point's exponent range to allow safe computation
        # of the variance of the transformed data.
        log_eps = np.log(np.finfo(dtype).eps)
        log_tiny_float = (np.log(np.finfo(dtype).tiny) - log_eps) / 2
        log_max_float = (np.log(np.finfo(dtype).max) + log_eps) / 2
        # Compute the bounds by approximating the inverse of the Yeo-Johnson
        # transform on the smallest and largest floating point exponents, given
        # the largest data we expect to observe. See [1] for further details.
        # [1] https://github.com/scipy/scipy/pull/18852#issuecomment-1630286174
        lb = log_tiny_float / log1p_max_x
        ub = log_max_float / log1p_max_x
        # Convert the bounds if all or some of the data is negative.
        if np.all(x < 0):
            lb, ub = 2 - ub, 2 - lb
        elif np.any(x < 0):
            lb, ub = max(2 - ub, lb), min(2 - lb, ub)
        # Match `optimize.brent`'s tolerance.
        tol_brent = 1.48e-08
        return optimize.fminbound(_neg_llf, lb, ub, args=(x,), xtol=tol_brent)


def yeojohnson_normplot(x, la, lb, plot=None, N=80):
    """Compute parameters for a Yeo-Johnson normality plot, optionally show it.

    A Yeo-Johnson normality plot shows graphically what the best
    transformation parameter is to use in `yeojohnson` to obtain a
    distribution that is close to normal.

    Parameters
    ----------
    x : array_like
        Input array.
    la, lb : scalar
        The lower and upper bounds for the ``lmbda`` values to pass to
        `yeojohnson` for Yeo-Johnson transformations. These are also the
        limits of the horizontal axis of the plot if that is generated.
    plot : object, optional
        If given, plots the quantiles and least squares fit.
        `plot` is an object that has to have methods "plot" and "text".
        The `matplotlib.pyplot` module or a Matplotlib Axes object can be used,
        or a custom object with the same methods.
        Default is None, which means that no plot is created.
    N : int, optional
        Number of points on the horizontal axis (equally distributed from
        `la` to `lb`).

    Returns
    -------
    lmbdas : ndarray
        The ``lmbda`` values for which a Yeo-Johnson transform was done.
    ppcc : ndarray
        Probability Plot Correlelation Coefficient, as obtained from `probplot`
        when fitting the Box-Cox transformed input `x` against a normal
        distribution.

    See Also
    --------
    probplot, yeojohnson, yeojohnson_normmax, yeojohnson_llf, ppcc_max

    Notes
    -----
    Even if `plot` is given, the figure is not shown or saved by
    `boxcox_normplot`; ``plt.show()`` or ``plt.savefig('figname.png')``
    should be used after calling `probplot`.

    .. versionadded:: 1.2.0

    Examples
    --------
    >>> from scipy import stats
    >>> import matplotlib.pyplot as plt

    Generate some non-normally distributed data, and create a Yeo-Johnson plot:

    >>> x = stats.loggamma.rvs(5, size=500) + 5
    >>> fig = plt.figure()
    >>> ax = fig.add_subplot(111)
    >>> prob = stats.yeojohnson_normplot(x, -20, 20, plot=ax)

    Determine and plot the optimal ``lmbda`` to transform ``x`` and plot it in
    the same plot:

    >>> _, maxlog = stats.yeojohnson(x)
    >>> ax.axvline(maxlog, color='r')

    >>> plt.show()

    """
    return _normplot('yeojohnson', x, la, lb, plot, N)


ShapiroResult = namedtuple('ShapiroResult', ('statistic', 'pvalue'))


@_axis_nan_policy_factory(ShapiroResult, n_samples=1, too_small=2, default_axis=None)
def shapiro(x):
    r"""Perform the Shapiro-Wilk test for normality.

    The Shapiro-Wilk test tests the null hypothesis that the
    data was drawn from a normal distribution.

    Parameters
    ----------
    x : array_like
        Array of sample data.

    Returns
    -------
    statistic : float
        The test statistic.
    p-value : float
        The p-value for the hypothesis test.

    See Also
    --------
    anderson : The Anderson-Darling test for normality
    kstest : The Kolmogorov-Smirnov test for goodness of fit.

    Notes
    -----
    The algorithm used is described in [4]_ but censoring parameters as
    described are not implemented. For N > 5000 the W test statistic is
    accurate, but the p-value may not be.

    References
    ----------
    .. [1] https://www.itl.nist.gov/div898/handbook/prc/section2/prc213.htm
           :doi:`10.18434/M32189`
    .. [2] Shapiro, S. S. & Wilk, M.B, "An analysis of variance test for
           normality (complete samples)", Biometrika, 1965, Vol. 52,
           pp. 591-611, :doi:`10.2307/2333709`
    .. [3] Razali, N. M. & Wah, Y. B., "Power comparisons of Shapiro-Wilk,
           Kolmogorov-Smirnov, Lilliefors and Anderson-Darling tests", Journal
           of Statistical Modeling and Analytics, 2011, Vol. 2, pp. 21-33.
    .. [4] Royston P., "Remark AS R94: A Remark on Algorithm AS 181: The
           W-test for Normality", 1995, Applied Statistics, Vol. 44,
           :doi:`10.2307/2986146`
    .. [5] Phipson B., and Smyth, G. K., "Permutation P-values Should Never Be
           Zero: Calculating Exact P-values When Permutations Are Randomly
           Drawn", Statistical Applications in Genetics and Molecular Biology,
           2010, Vol.9, :doi:`10.2202/1544-6115.1585`
    .. [6] Panagiotakos, D. B., "The value of p-value in biomedical
           research", The Open Cardiovascular Medicine Journal, 2008, Vol.2,
           pp. 97-99, :doi:`10.2174/1874192400802010097`

    Examples
    --------
    Suppose we wish to infer from measurements whether the weights of adult
    human males in a medical study are not normally distributed [2]_.
    The weights (lbs) are recorded in the array ``x`` below.

    >>> import numpy as np
    >>> x = np.array([148, 154, 158, 160, 161, 162, 166, 170, 182, 195, 236])

    The normality test of [1]_ and [2]_ begins by computing a statistic based
    on the relationship between the observations and the expected order
    statistics of a normal distribution.

    >>> from scipy import stats
    >>> res = stats.shapiro(x)
    >>> res.statistic
    0.7888147830963135

    The value of this statistic tends to be high (close to 1) for samples drawn
    from a normal distribution.

    The test is performed by comparing the observed value of the statistic
    against the null distribution: the distribution of statistic values formed
    under the null hypothesis that the weights were drawn from a normal
    distribution. For this normality test, the null distribution is not easy to
    calculate exactly, so it is usually approximated by Monte Carlo methods,
    that is, drawing many samples of the same size as ``x`` from a normal
    distribution and computing the values of the statistic for each.

    >>> def statistic(x):
    ...     # Get only the `shapiro` statistic; ignore its p-value
    ...     return stats.shapiro(x).statistic
    >>> ref = stats.monte_carlo_test(x, stats.norm.rvs, statistic,
    ...                              alternative='less')
    >>> import matplotlib.pyplot as plt
    >>> fig, ax = plt.subplots(figsize=(8, 5))
    >>> bins = np.linspace(0.65, 1, 50)
    >>> def plot(ax):  # we'll reuse this
    ...     ax.hist(ref.null_distribution, density=True, bins=bins)
    ...     ax.set_title("Shapiro-Wilk Test Null Distribution \n"
    ...                  "(Monte Carlo Approximation, 11 Observations)")
    ...     ax.set_xlabel("statistic")
    ...     ax.set_ylabel("probability density")
    >>> plot(ax)
    >>> plt.show()

    The comparison is quantified by the p-value: the proportion of values in
    the null distribution less than or equal to the observed value of the
    statistic.

    >>> fig, ax = plt.subplots(figsize=(8, 5))
    >>> plot(ax)
    >>> annotation = (f'p-value={res.pvalue:.6f}\n(highlighted area)')
    >>> props = dict(facecolor='black', width=1, headwidth=5, headlength=8)
    >>> _ = ax.annotate(annotation, (0.75, 0.1), (0.68, 0.7), arrowprops=props)
    >>> i_extreme = np.where(bins <= res.statistic)[0]
    >>> for i in i_extreme:
    ...     ax.patches[i].set_color('C1')
    >>> plt.xlim(0.65, 0.9)
    >>> plt.ylim(0, 4)
    >>> plt.show
    >>> res.pvalue
    0.006703833118081093

    If the p-value is "small" - that is, if there is a low probability of
    sampling data from a normally distributed population that produces such an
    extreme value of the statistic - this may be taken as evidence against
    the null hypothesis in favor of the alternative: the weights were not
    drawn from a normal distribution. Note that:

    - The inverse is not true; that is, the test is not used to provide
      evidence *for* the null hypothesis.
    - The threshold for values that will be considered "small" is a choice that
      should be made before the data is analyzed [5]_ with consideration of the
      risks of both false positives (incorrectly rejecting the null hypothesis)
      and false negatives (failure to reject a false null hypothesis).

    """
    x = np.ravel(x).astype(np.float64)

    N = len(x)
    if N < 3:
        raise ValueError("Data must be at least length 3.")

    a = zeros(N//2, dtype=np.float64)
    init = 0

    y = sort(x)
    y -= x[N//2]  # subtract the median (or a nearby value); see gh-15777

    w, pw, ifault = swilk(y, a, init)
    if ifault not in [0, 2]:
        warnings.warn("scipy.stats.shapiro: Input data has range zero. The"
                      " results may not be accurate.", stacklevel=2)
    if N > 5000:
        warnings.warn("scipy.stats.shapiro: For N > 5000, computed p-value "
                      f"may not be accurate. Current N is {N}.",
                      stacklevel=2)

    # `w` and `pw` are always Python floats, which are double precision.
    # We want to ensure that they are NumPy floats, so until dtypes are
    # respected, we can explicitly convert each to float64 (faster than
    # `np.array([w, pw])`).
    return ShapiroResult(np.float64(w), np.float64(pw))


# Values from Stephens, M A, "EDF Statistics for Goodness of Fit and
#             Some Comparisons", Journal of the American Statistical
#             Association, Vol. 69, Issue 347, Sept. 1974, pp 730-737
_Avals_norm = array([0.576, 0.656, 0.787, 0.918, 1.092])
_Avals_expon = array([0.922, 1.078, 1.341, 1.606, 1.957])
# From Stephens, M A, "Goodness of Fit for the Extreme Value Distribution",
#             Biometrika, Vol. 64, Issue 3, Dec. 1977, pp 583-588.
_Avals_gumbel = array([0.474, 0.637, 0.757, 0.877, 1.038])
# From Stephens, M A, "Tests of Fit for the Logistic Distribution Based
#             on the Empirical Distribution Function.", Biometrika,
#             Vol. 66, Issue 3, Dec. 1979, pp 591-595.
_Avals_logistic = array([0.426, 0.563, 0.660, 0.769, 0.906, 1.010])
# From Richard A. Lockhart and Michael A. Stephens "Estimation and Tests of
#             Fit for the Three-Parameter Weibull Distribution"
#             Journal of the Royal Statistical Society.Series B(Methodological)
#             Vol. 56, No. 3 (1994), pp. 491-500, table 1. Keys are c*100
_Avals_weibull = [[0.292, 0.395, 0.467, 0.522, 0.617, 0.711, 0.836, 0.931],
                  [0.295, 0.399, 0.471, 0.527, 0.623, 0.719, 0.845, 0.941],
                  [0.298, 0.403, 0.476, 0.534, 0.631, 0.728, 0.856, 0.954],
                  [0.301, 0.408, 0.483, 0.541, 0.640, 0.738, 0.869, 0.969],
                  [0.305, 0.414, 0.490, 0.549, 0.650, 0.751, 0.885, 0.986],
                  [0.309, 0.421, 0.498, 0.559, 0.662, 0.765, 0.902, 1.007],
                  [0.314, 0.429, 0.508, 0.570, 0.676, 0.782, 0.923, 1.030],
                  [0.320, 0.438, 0.519, 0.583, 0.692, 0.802, 0.947, 1.057],
                  [0.327, 0.448, 0.532, 0.598, 0.711, 0.824, 0.974, 1.089],
                  [0.334, 0.469, 0.547, 0.615, 0.732, 0.850, 1.006, 1.125],
                  [0.342, 0.472, 0.563, 0.636, 0.757, 0.879, 1.043, 1.167]]
_Avals_weibull = np.array(_Avals_weibull)
_cvals_weibull = np.linspace(0, 0.5, 11)
_get_As_weibull = interpolate.interp1d(_cvals_weibull, _Avals_weibull.T,
                                       kind='linear', bounds_error=False,
                                       fill_value=_Avals_weibull[-1])


def _weibull_fit_check(params, x):
    # Refine the fit returned by `weibull_min.fit` to ensure that the first
    # order necessary conditions are satisfied. If not, raise an error.
    # Here, use `m` for the shape parameter to be consistent with [7]
    # and avoid confusion with `c` as defined in [7].
    n = len(x)
    m, u, s = params

    def dnllf_dm(m, u):
        # Partial w.r.t. shape w/ optimal scale. See [7] Equation 5.
        xu = x-u
        return (1/m - (xu**m*np.log(xu)).sum()/(xu**m).sum()
                + np.log(xu).sum()/n)

    def dnllf_du(m, u):
        # Partial w.r.t. loc w/ optimal scale. See [7] Equation 6.
        xu = x-u
        return (m-1)/m*(xu**-1).sum() - n*(xu**(m-1)).sum()/(xu**m).sum()

    def get_scale(m, u):
        # Partial w.r.t. scale solved in terms of shape and location.
        # See [7] Equation 7.
        return ((x-u)**m/n).sum()**(1/m)

    def dnllf(params):
        # Partial derivatives of the NLLF w.r.t. parameters, i.e.
        # first order necessary conditions for MLE fit.
        return [dnllf_dm(*params), dnllf_du(*params)]

    suggestion = ("Maximum likelihood estimation is known to be challenging "
                  "for the three-parameter Weibull distribution. Consider "
                  "performing a custom goodness-of-fit test using "
                  "`scipy.stats.monte_carlo_test`.")

    if np.allclose(u, np.min(x)) or m < 1:
        # The critical values provided by [7] don't seem to control the
        # Type I error rate in this case. Error out.
        message = ("Maximum likelihood estimation has converged to "
                   "a solution in which the location is equal to the minimum "
                   "of the data, the shape parameter is less than 2, or both. "
                   "The table of critical values in [7] does not "
                   "include this case. " + suggestion)
        raise ValueError(message)

    try:
        # Refine the MLE / verify that first-order necessary conditions are
        # satisfied. If so, the critical values provided in [7] seem reliable.
        with np.errstate(over='raise', invalid='raise'):
            res = optimize.root(dnllf, params[:-1])

        message = ("Solution of MLE first-order conditions failed: "
                   f"{res.message}. `anderson` cannot continue. " + suggestion)
        if not res.success:
            raise ValueError(message)

    except (FloatingPointError, ValueError) as e:
        message = ("An error occurred while fitting the Weibull distribution "
                   "to the data, so `anderson` cannot continue. " + suggestion)
        raise ValueError(message) from e

    m, u = res.x
    s = get_scale(m, u)
    return m, u, s


AndersonResult = _make_tuple_bunch('AndersonResult',
                                   ['statistic', 'critical_values',
                                    'significance_level'], ['fit_result'])


def anderson(x, dist='norm'):
    """Anderson-Darling test for data coming from a particular distribution.

    The Anderson-Darling test tests the null hypothesis that a sample is
    drawn from a population that follows a particular distribution.
    For the Anderson-Darling test, the critical values depend on
    which distribution is being tested against.  This function works
    for normal, exponential, logistic, weibull_min, or Gumbel (Extreme Value
    Type I) distributions.

    Parameters
    ----------
    x : array_like
        Array of sample data.
    dist : {'norm', 'expon', 'logistic', 'gumbel', 'gumbel_l', 'gumbel_r', 'extreme1', 'weibull_min'}, optional
        The type of distribution to test against.  The default is 'norm'.
        The names 'extreme1', 'gumbel_l' and 'gumbel' are synonyms for the
        same distribution.

    Returns
    -------
    result : AndersonResult
        An object with the following attributes:

        statistic : float
            The Anderson-Darling test statistic.
        critical_values : list
            The critical values for this distribution.
        significance_level : list
            The significance levels for the corresponding critical values
            in percents.  The function returns critical values for a
            differing set of significance levels depending on the
            distribution that is being tested against.
        fit_result : `~scipy.stats._result_classes.FitResult`
            An object containing the results of fitting the distribution to
            the data.

    See Also
    --------
    kstest : The Kolmogorov-Smirnov test for goodness-of-fit.

    Notes
    -----
    Critical values provided are for the following significance levels:

    normal/exponential
        15%, 10%, 5%, 2.5%, 1%
    logistic
        25%, 10%, 5%, 2.5%, 1%, 0.5%
    gumbel_l / gumbel_r
        25%, 10%, 5%, 2.5%, 1%
    weibull_min
        50%, 25%, 15%, 10%, 5%, 2.5%, 1%, 0.5%

    If the returned statistic is larger than these critical values then
    for the corresponding significance level, the null hypothesis that
    the data come from the chosen distribution can be rejected.
    The returned statistic is referred to as 'A2' in the references.

    For `weibull_min`, maximum likelihood estimation is known to be
    challenging. If the test returns successfully, then the first order
    conditions for a maximum likehood estimate have been verified and
    the critical values correspond relatively well to the significance levels,
    provided that the sample is sufficiently large (>10 observations [7]).
    However, for some data - especially data with no left tail - `anderson`
    is likely to result in an error message. In this case, consider
    performing a custom goodness of fit test using
    `scipy.stats.monte_carlo_test`.

    References
    ----------
    .. [1] https://www.itl.nist.gov/div898/handbook/prc/section2/prc213.htm
    .. [2] Stephens, M. A. (1974). EDF Statistics for Goodness of Fit and
           Some Comparisons, Journal of the American Statistical Association,
           Vol. 69, pp. 730-737.
    .. [3] Stephens, M. A. (1976). Asymptotic Results for Goodness-of-Fit
           Statistics with Unknown Parameters, Annals of Statistics, Vol. 4,
           pp. 357-369.
    .. [4] Stephens, M. A. (1977). Goodness of Fit for the Extreme Value
           Distribution, Biometrika, Vol. 64, pp. 583-588.
    .. [5] Stephens, M. A. (1977). Goodness of Fit with Special Reference
           to Tests for Exponentiality , Technical Report No. 262,
           Department of Statistics, Stanford University, Stanford, CA.
    .. [6] Stephens, M. A. (1979). Tests of Fit for the Logistic Distribution
           Based on the Empirical Distribution Function, Biometrika, Vol. 66,
           pp. 591-595.
    .. [7] Richard A. Lockhart and Michael A. Stephens "Estimation and Tests of
           Fit for the Three-Parameter Weibull Distribution"
           Journal of the Royal Statistical Society.Series B(Methodological)
           Vol. 56, No. 3 (1994), pp. 491-500, Table 0.

    Examples
    --------
    Test the null hypothesis that a random sample was drawn from a normal
    distribution (with unspecified mean and standard deviation).

    >>> import numpy as np
    >>> from scipy.stats import anderson
    >>> rng = np.random.default_rng()
    >>> data = rng.random(size=35)
    >>> res = anderson(data)
    >>> res.statistic
    0.8398018749744764
    >>> res.critical_values
    array([0.527, 0.6  , 0.719, 0.839, 0.998])
    >>> res.significance_level
    array([15. , 10. ,  5. ,  2.5,  1. ])

    The value of the statistic (barely) exceeds the critical value associated
    with a significance level of 2.5%, so the null hypothesis may be rejected
    at a significance level of 2.5%, but not at a significance level of 1%.

    """ # numpy/numpydoc#87  # noqa: E501
    dist = dist.lower()
    if dist in {'extreme1', 'gumbel'}:
        dist = 'gumbel_l'
    dists = {'norm', 'expon', 'gumbel_l',
             'gumbel_r', 'logistic', 'weibull_min'}

    if dist not in dists:
        raise ValueError(f"Invalid distribution; dist must be in {dists}.")
    y = sort(x)
    xbar = np.mean(x, axis=0)
    N = len(y)
    if dist == 'norm':
        s = np.std(x, ddof=1, axis=0)
        w = (y - xbar) / s
        fit_params = xbar, s
        logcdf = distributions.norm.logcdf(w)
        logsf = distributions.norm.logsf(w)
        sig = array([15, 10, 5, 2.5, 1])
        critical = around(_Avals_norm / (1.0 + 4.0/N - 25.0/N/N), 3)
    elif dist == 'expon':
        w = y / xbar
        fit_params = 0, xbar
        logcdf = distributions.expon.logcdf(w)
        logsf = distributions.expon.logsf(w)
        sig = array([15, 10, 5, 2.5, 1])
        critical = around(_Avals_expon / (1.0 + 0.6/N), 3)
    elif dist == 'logistic':
        def rootfunc(ab, xj, N):
            a, b = ab
            tmp = (xj - a) / b
            tmp2 = exp(tmp)
            val = [np.sum(1.0/(1+tmp2), axis=0) - 0.5*N,
                   np.sum(tmp*(1.0-tmp2)/(1+tmp2), axis=0) + N]
            return array(val)

        sol0 = array([xbar, np.std(x, ddof=1, axis=0)])
        sol = optimize.fsolve(rootfunc, sol0, args=(x, N), xtol=1e-5)
        w = (y - sol[0]) / sol[1]
        fit_params = sol
        logcdf = distributions.logistic.logcdf(w)
        logsf = distributions.logistic.logsf(w)
        sig = array([25, 10, 5, 2.5, 1, 0.5])
        critical = around(_Avals_logistic / (1.0 + 0.25/N), 3)
    elif dist == 'gumbel_r':
        xbar, s = distributions.gumbel_r.fit(x)
        w = (y - xbar) / s
        fit_params = xbar, s
        logcdf = distributions.gumbel_r.logcdf(w)
        logsf = distributions.gumbel_r.logsf(w)
        sig = array([25, 10, 5, 2.5, 1])
        critical = around(_Avals_gumbel / (1.0 + 0.2/sqrt(N)), 3)
    elif dist == 'gumbel_l':
        xbar, s = distributions.gumbel_l.fit(x)
        w = (y - xbar) / s
        fit_params = xbar, s
        logcdf = distributions.gumbel_l.logcdf(w)
        logsf = distributions.gumbel_l.logsf(w)
        sig = array([25, 10, 5, 2.5, 1])
        critical = around(_Avals_gumbel / (1.0 + 0.2/sqrt(N)), 3)
    elif dist == 'weibull_min':
        message = ("Critical values of the test statistic are given for the "
                   "asymptotic distribution. These may not be accurate for "
                   "samples with fewer than 10 observations. Consider using "
                   "`scipy.stats.monte_carlo_test`.")
        if N < 10:
            warnings.warn(message, stacklevel=2)
        # [7] writes our 'c' as 'm', and they write `c = 1/m`. Use their names.
        m, loc, scale = distributions.weibull_min.fit(y)
        m, loc, scale = _weibull_fit_check((m, loc, scale), y)
        fit_params = m, loc, scale
        logcdf = stats.weibull_min(*fit_params).logcdf(y)
        logsf = stats.weibull_min(*fit_params).logsf(y)
        c = 1 / m  # m and c are as used in [7]
        sig = array([0.5, 0.75, 0.85, 0.9, 0.95, 0.975, 0.99, 0.995])
        critical = _get_As_weibull(c)
        # Goodness-of-fit tests should only be used to provide evidence
        # _against_ the null hypothesis. Be conservative and round up.
        critical = np.round(critical + 0.0005, decimals=3)

    i = arange(1, N + 1)
    A2 = -N - np.sum((2*i - 1.0) / N * (logcdf + logsf[::-1]), axis=0)

    # FitResult initializer expects an optimize result, so let's work with it
    message = '`anderson` successfully fit the distribution to the data.'
    res = optimize.OptimizeResult(success=True, message=message)
    res.x = np.array(fit_params)
    fit_result = FitResult(getattr(distributions, dist), y,
                           discrete=False, res=res)

    return AndersonResult(A2, critical, sig, fit_result=fit_result)


def _anderson_ksamp_midrank(samples, Z, Zstar, k, n, N):
    """Compute A2akN equation 7 of Scholz and Stephens.

    Parameters
    ----------
    samples : sequence of 1-D array_like
        Array of sample arrays.
    Z : array_like
        Sorted array of all observations.
    Zstar : array_like
        Sorted array of unique observations.
    k : int
        Number of samples.
    n : array_like
        Number of observations in each sample.
    N : int
        Total number of observations.

    Returns
    -------
    A2aKN : float
        The A2aKN statistics of Scholz and Stephens 1987.

    """
    A2akN = 0.
    Z_ssorted_left = Z.searchsorted(Zstar, 'left')
    if N == Zstar.size:
        lj = 1.
    else:
        lj = Z.searchsorted(Zstar, 'right') - Z_ssorted_left
    Bj = Z_ssorted_left + lj / 2.
    for i in arange(0, k):
        s = np.sort(samples[i])
        s_ssorted_right = s.searchsorted(Zstar, side='right')
        Mij = s_ssorted_right.astype(float)
        fij = s_ssorted_right - s.searchsorted(Zstar, 'left')
        Mij -= fij / 2.
        inner = lj / float(N) * (N*Mij - Bj*n[i])**2 / (Bj*(N - Bj) - N*lj/4.)
        A2akN += inner.sum() / n[i]
    A2akN *= (N - 1.) / N
    return A2akN


def _anderson_ksamp_right(samples, Z, Zstar, k, n, N):
    """Compute A2akN equation 6 of Scholz & Stephens.

    Parameters
    ----------
    samples : sequence of 1-D array_like
        Array of sample arrays.
    Z : array_like
        Sorted array of all observations.
    Zstar : array_like
        Sorted array of unique observations.
    k : int
        Number of samples.
    n : array_like
        Number of observations in each sample.
    N : int
        Total number of observations.

    Returns
    -------
    A2KN : float
        The A2KN statistics of Scholz and Stephens 1987.

    """
    A2kN = 0.
    lj = Z.searchsorted(Zstar[:-1], 'right') - Z.searchsorted(Zstar[:-1],
                                                              'left')
    Bj = lj.cumsum()
    for i in arange(0, k):
        s = np.sort(samples[i])
        Mij = s.searchsorted(Zstar[:-1], side='right')
        inner = lj / float(N) * (N * Mij - Bj * n[i])**2 / (Bj * (N - Bj))
        A2kN += inner.sum() / n[i]
    return A2kN


Anderson_ksampResult = _make_tuple_bunch(
    'Anderson_ksampResult',
    ['statistic', 'critical_values', 'pvalue'], []
)


def anderson_ksamp(samples, midrank=True, *, method=None):
    """The Anderson-Darling test for k-samples.

    The k-sample Anderson-Darling test is a modification of the
    one-sample Anderson-Darling test. It tests the null hypothesis
    that k-samples are drawn from the same population without having
    to specify the distribution function of that population. The
    critical values depend on the number of samples.

    Parameters
    ----------
    samples : sequence of 1-D array_like
        Array of sample data in arrays.
    midrank : bool, optional
        Type of Anderson-Darling test which is computed. Default
        (True) is the midrank test applicable to continuous and
        discrete populations. If False, the right side empirical
        distribution is used.
    method : PermutationMethod, optional
        Defines the method used to compute the p-value. If `method` is an
        instance of `PermutationMethod`, the p-value is computed using
        `scipy.stats.permutation_test` with the provided configuration options
        and other appropriate settings. Otherwise, the p-value is interpolated
        from tabulated values.

    Returns
    -------
    res : Anderson_ksampResult
        An object containing attributes:

        statistic : float
            Normalized k-sample Anderson-Darling test statistic.
        critical_values : array
            The critical values for significance levels 25%, 10%, 5%, 2.5%, 1%,
            0.5%, 0.1%.
        pvalue : float
            The approximate p-value of the test. If `method` is not
            provided, the value is floored / capped at 0.1% / 25%.

    Raises
    ------
    ValueError
        If fewer than 2 samples are provided, a sample is empty, or no
        distinct observations are in the samples.

    See Also
    --------
    ks_2samp : 2 sample Kolmogorov-Smirnov test
    anderson : 1 sample Anderson-Darling test

    Notes
    -----
    [1]_ defines three versions of the k-sample Anderson-Darling test:
    one for continuous distributions and two for discrete
    distributions, in which ties between samples may occur. The
    default of this routine is to compute the version based on the
    midrank empirical distribution function. This test is applicable
    to continuous and discrete data. If midrank is set to False, the
    right side empirical distribution is used for a test for discrete
    data. According to [1]_, the two discrete test statistics differ
    only slightly if a few collisions due to round-off errors occur in
    the test not adjusted for ties between samples.

    The critical values corresponding to the significance levels from 0.01
    to 0.25 are taken from [1]_. p-values are floored / capped
    at 0.1% / 25%. Since the range of critical values might be extended in
    future releases, it is recommended not to test ``p == 0.25``, but rather
    ``p >= 0.25`` (analogously for the lower bound).

    .. versionadded:: 0.14.0

    References
    ----------
    .. [1] Scholz, F. W and Stephens, M. A. (1987), K-Sample
           Anderson-Darling Tests, Journal of the American Statistical
           Association, Vol. 82, pp. 918-924.

    Examples
    --------
    >>> import numpy as np
    >>> from scipy import stats
    >>> rng = np.random.default_rng()
    >>> res = stats.anderson_ksamp([rng.normal(size=50),
    ... rng.normal(loc=0.5, size=30)])
    >>> res.statistic, res.pvalue
    (1.974403288713695, 0.04991293614572478)
    >>> res.critical_values
    array([0.325, 1.226, 1.961, 2.718, 3.752, 4.592, 6.546])

    The null hypothesis that the two random samples come from the same
    distribution can be rejected at the 5% level because the returned
    test value is greater than the critical value for 5% (1.961) but
    not at the 2.5% level. The interpolation gives an approximate
    p-value of 4.99%.

    >>> samples = [rng.normal(size=50), rng.normal(size=30),
    ...            rng.normal(size=20)]
    >>> res = stats.anderson_ksamp(samples)
    >>> res.statistic, res.pvalue
    (-0.29103725200789504, 0.25)
    >>> res.critical_values
    array([ 0.44925884,  1.3052767 ,  1.9434184 ,  2.57696569,  3.41634856,
      4.07210043, 5.56419101])

    The null hypothesis cannot be rejected for three samples from an
    identical distribution. The reported p-value (25%) has been capped and
    may not be very accurate (since it corresponds to the value 0.449
    whereas the statistic is -0.291).

    In such cases where the p-value is capped or when sample sizes are
    small, a permutation test may be more accurate.

    >>> method = stats.PermutationMethod(n_resamples=9999, random_state=rng)
    >>> res = stats.anderson_ksamp(samples, method=method)
    >>> res.pvalue
    0.5254

    """
    k = len(samples)
    if (k < 2):
        raise ValueError("anderson_ksamp needs at least two samples")

    samples = list(map(np.asarray, samples))
    Z = np.sort(np.hstack(samples))
    N = Z.size
    Zstar = np.unique(Z)
    if Zstar.size < 2:
        raise ValueError("anderson_ksamp needs more than one distinct "
                         "observation")

    n = np.array([sample.size for sample in samples])
    if np.any(n == 0):
        raise ValueError("anderson_ksamp encountered sample without "
                         "observations")

    if midrank:
        A2kN_fun = _anderson_ksamp_midrank
    else:
        A2kN_fun = _anderson_ksamp_right
    A2kN = A2kN_fun(samples, Z, Zstar, k, n, N)

    def statistic(*samples):
        return A2kN_fun(samples, Z, Zstar, k, n, N)

    if method is not None:
        res = stats.permutation_test(samples, statistic, **method._asdict(),
                                     alternative='greater')

    H = (1. / n).sum()
    hs_cs = (1. / arange(N - 1, 1, -1)).cumsum()
    h = hs_cs[-1] + 1
    g = (hs_cs / arange(2, N)).sum()

    a = (4*g - 6) * (k - 1) + (10 - 6*g)*H
    b = (2*g - 4)*k**2 + 8*h*k + (2*g - 14*h - 4)*H - 8*h + 4*g - 6
    c = (6*h + 2*g - 2)*k**2 + (4*h - 4*g + 6)*k + (2*h - 6)*H + 4*h
    d = (2*h + 6)*k**2 - 4*h*k
    sigmasq = (a*N**3 + b*N**2 + c*N + d) / ((N - 1.) * (N - 2.) * (N - 3.))
    m = k - 1
    A2 = (A2kN - m) / math.sqrt(sigmasq)

    # The b_i values are the interpolation coefficients from Table 2
    # of Scholz and Stephens 1987
    b0 = np.array([0.675, 1.281, 1.645, 1.96, 2.326, 2.573, 3.085])
    b1 = np.array([-0.245, 0.25, 0.678, 1.149, 1.822, 2.364, 3.615])
    b2 = np.array([-0.105, -0.305, -0.362, -0.391, -0.396, -0.345, -0.154])
    critical = b0 + b1 / math.sqrt(m) + b2 / m

    sig = np.array([0.25, 0.1, 0.05, 0.025, 0.01, 0.005, 0.001])

    if A2 < critical.min() and method is None:
        p = sig.max()
        msg = (f"p-value capped: true value larger than {p}. Consider "
               "specifying `method` "
               "(e.g. `method=stats.PermutationMethod()`.)")
        warnings.warn(msg, stacklevel=2)
    elif A2 > critical.max() and method is None:
        p = sig.min()
        msg = (f"p-value floored: true value smaller than {p}. Consider "
               "specifying `method` "
               "(e.g. `method=stats.PermutationMethod()`.)")
        warnings.warn(msg, stacklevel=2)
    elif method is None:
        # interpolation of probit of significance level
        pf = np.polyfit(critical, log(sig), 2)
        p = math.exp(np.polyval(pf, A2))
    else:
        p = res.pvalue if method is not None else p

    # create result object with alias for backward compatibility
    res = Anderson_ksampResult(A2, critical, p)
    res.significance_level = p
    return res


AnsariResult = namedtuple('AnsariResult', ('statistic', 'pvalue'))


class _ABW:
    """Distribution of Ansari-Bradley W-statistic under the null hypothesis."""
    # TODO: calculate exact distribution considering ties
    # We could avoid summing over more than half the frequencies,
    # but initially it doesn't seem worth the extra complexity

    def __init__(self):
        """Minimal initializer."""
        self.m = None
        self.n = None
        self.astart = None
        self.total = None
        self.freqs = None

    def _recalc(self, n, m):
        """When necessary, recalculate exact distribution."""
        if n != self.n or m != self.m:
            self.n, self.m = n, m
            # distribution is NOT symmetric when m + n is odd
            # n is len(x), m is len(y), and ratio of scales is defined x/y
            astart, a1, _ = gscale(n, m)
            self.astart = astart  # minimum value of statistic
            # Exact distribution of test statistic under null hypothesis
            # expressed as frequencies/counts/integers to maintain precision.
            # Stored as floats to avoid overflow of sums.
            self.freqs = a1.astype(np.float64)
            self.total = self.freqs.sum()  # could calculate from m and n
            # probability mass is self.freqs / self.total;

    def pmf(self, k, n, m):
        """Probability mass function."""
        self._recalc(n, m)
        # The convention here is that PMF at k = 12.5 is the same as at k = 12,
        # -> use `floor` in case of ties.
        ind = np.floor(k - self.astart).astype(int)
        return self.freqs[ind] / self.total

    def cdf(self, k, n, m):
        """Cumulative distribution function."""
        self._recalc(n, m)
        # Null distribution derived without considering ties is
        # approximate. Round down to avoid Type I error.
        ind = np.ceil(k - self.astart).astype(int)
        return self.freqs[:ind+1].sum() / self.total

    def sf(self, k, n, m):
        """Survival function."""
        self._recalc(n, m)
        # Null distribution derived without considering ties is
        # approximate. Round down to avoid Type I error.
        ind = np.floor(k - self.astart).astype(int)
        return self.freqs[ind:].sum() / self.total


# Maintain state for faster repeat calls to ansari w/ method='exact'
_abw_state = _ABW()


@_axis_nan_policy_factory(AnsariResult, n_samples=2)
def ansari(x, y, alternative='two-sided'):
    """Perform the Ansari-Bradley test for equal scale parameters.

    The Ansari-Bradley test ([1]_, [2]_) is a non-parametric test
    for the equality of the scale parameter of the distributions
    from which two samples were drawn. The null hypothesis states that
    the ratio of the scale of the distribution underlying `x` to the scale
    of the distribution underlying `y` is 1.

    Parameters
    ----------
    x, y : array_like
        Arrays of sample data.
    alternative : {'two-sided', 'less', 'greater'}, optional
        Defines the alternative hypothesis. Default is 'two-sided'.
        The following options are available:

        * 'two-sided': the ratio of scales is not equal to 1.
        * 'less': the ratio of scales is less than 1.
        * 'greater': the ratio of scales is greater than 1.

        .. versionadded:: 1.7.0

    Returns
    -------
    statistic : float
        The Ansari-Bradley test statistic.
    pvalue : float
        The p-value of the hypothesis test.

    See Also
    --------
    fligner : A non-parametric test for the equality of k variances
    mood : A non-parametric test for the equality of two scale parameters

    Notes
    -----
    The p-value given is exact when the sample sizes are both less than
    55 and there are no ties, otherwise a normal approximation for the
    p-value is used.

    References
    ----------
    .. [1] Ansari, A. R. and Bradley, R. A. (1960) Rank-sum tests for
           dispersions, Annals of Mathematical Statistics, 31, 1174-1189.
    .. [2] Sprent, Peter and N.C. Smeeton.  Applied nonparametric
           statistical methods.  3rd ed. Chapman and Hall/CRC. 2001.
           Section 5.8.2.
    .. [3] Nathaniel E. Helwig "Nonparametric Dispersion and Equality
           Tests" at http://users.stat.umn.edu/~helwig/notes/npde-Notes.pdf

    Examples
    --------
    >>> import numpy as np
    >>> from scipy.stats import ansari
    >>> rng = np.random.default_rng()

    For these examples, we'll create three random data sets.  The first
    two, with sizes 35 and 25, are drawn from a normal distribution with
    mean 0 and standard deviation 2.  The third data set has size 25 and
    is drawn from a normal distribution with standard deviation 1.25.

    >>> x1 = rng.normal(loc=0, scale=2, size=35)
    >>> x2 = rng.normal(loc=0, scale=2, size=25)
    >>> x3 = rng.normal(loc=0, scale=1.25, size=25)

    First we apply `ansari` to `x1` and `x2`.  These samples are drawn
    from the same distribution, so we expect the Ansari-Bradley test
    should not lead us to conclude that the scales of the distributions
    are different.

    >>> ansari(x1, x2)
    AnsariResult(statistic=541.0, pvalue=0.9762532927399098)

    With a p-value close to 1, we cannot conclude that there is a
    significant difference in the scales (as expected).

    Now apply the test to `x1` and `x3`:

    >>> ansari(x1, x3)
    AnsariResult(statistic=425.0, pvalue=0.0003087020407974518)

    The probability of observing such an extreme value of the statistic
    under the null hypothesis of equal scales is only 0.03087%. We take this
    as evidence against the null hypothesis in favor of the alternative:
    the scales of the distributions from which the samples were drawn
    are not equal.

    We can use the `alternative` parameter to perform a one-tailed test.
    In the above example, the scale of `x1` is greater than `x3` and so
    the ratio of scales of `x1` and `x3` is greater than 1. This means
    that the p-value when ``alternative='greater'`` should be near 0 and
    hence we should be able to reject the null hypothesis:

    >>> ansari(x1, x3, alternative='greater')
    AnsariResult(statistic=425.0, pvalue=0.0001543510203987259)

    As we can see, the p-value is indeed quite low. Use of
    ``alternative='less'`` should thus yield a large p-value:

    >>> ansari(x1, x3, alternative='less')
    AnsariResult(statistic=425.0, pvalue=0.9998643258449039)

    """
    if alternative not in {'two-sided', 'greater', 'less'}:
        raise ValueError("'alternative' must be 'two-sided',"
                         " 'greater', or 'less'.")
    x, y = asarray(x), asarray(y)
    n = len(x)
    m = len(y)
    if m < 1:
        raise ValueError("Not enough other observations.")
    if n < 1:
        raise ValueError("Not enough test observations.")

    N = m + n
    xy = r_[x, y]  # combine
    rank = _stats_py.rankdata(xy)
    symrank = amin(array((rank, N - rank + 1)), 0)
    AB = np.sum(symrank[:n], axis=0)
    uxy = unique(xy)
    repeats = (len(uxy) != len(xy))
    exact = ((m < 55) and (n < 55) and not repeats)
    if repeats and (m < 55 or n < 55):
        warnings.warn("Ties preclude use of exact statistic.", stacklevel=2)
    if exact:
        if alternative == 'two-sided':
            pval = 2.0 * np.minimum(_abw_state.cdf(AB, n, m),
                                    _abw_state.sf(AB, n, m))
        elif alternative == 'greater':
            # AB statistic is _smaller_ when ratio of scales is larger,
            # so this is the opposite of the usual calculation
            pval = _abw_state.cdf(AB, n, m)
        else:
            pval = _abw_state.sf(AB, n, m)
        return AnsariResult(AB, min(1.0, pval))

    # otherwise compute normal approximation
    if N % 2:  # N odd
        mnAB = n * (N+1.0)**2 / 4.0 / N
        varAB = n * m * (N+1.0) * (3+N**2) / (48.0 * N**2)
    else:
        mnAB = n * (N+2.0) / 4.0
        varAB = m * n * (N+2) * (N-2.0) / 48 / (N-1.0)
    if repeats:   # adjust variance estimates
        # compute np.sum(tj * rj**2,axis=0)
        fac = np.sum(symrank**2, axis=0)
        if N % 2:  # N odd
            varAB = m * n * (16*N*fac - (N+1)**4) / (16.0 * N**2 * (N-1))
        else:  # N even
            varAB = m * n * (16*fac - N*(N+2)**2) / (16.0 * N * (N-1))

    # Small values of AB indicate larger dispersion for the x sample.
    # Large values of AB indicate larger dispersion for the y sample.
    # This is opposite to the way we define the ratio of scales. see [1]_.
    z = (mnAB - AB) / sqrt(varAB)
    pvalue = _get_pvalue(z, distributions.norm, alternative)
    return AnsariResult(AB[()], pvalue[()])


BartlettResult = namedtuple('BartlettResult', ('statistic', 'pvalue'))


@_axis_nan_policy_factory(BartlettResult, n_samples=None)
def bartlett(*samples):
    r"""Perform Bartlett's test for equal variances.

    Bartlett's test tests the null hypothesis that all input samples
    are from populations with equal variances.  For samples
    from significantly non-normal populations, Levene's test
    `levene` is more robust.

    Parameters
    ----------
    sample1, sample2, ... : array_like
        arrays of sample data.  Only 1d arrays are accepted, they may have
        different lengths.

    Returns
    -------
    statistic : float
        The test statistic.
    pvalue : float
        The p-value of the test.

    See Also
    --------
    fligner : A non-parametric test for the equality of k variances
    levene : A robust parametric test for equality of k variances

    Notes
    -----
    Conover et al. (1981) examine many of the existing parametric and
    nonparametric tests by extensive simulations and they conclude that the
    tests proposed by Fligner and Killeen (1976) and Levene (1960) appear to be
    superior in terms of robustness of departures from normality and power
    ([3]_).

    References
    ----------
    .. [1]  https://www.itl.nist.gov/div898/handbook/eda/section3/eda357.htm
    .. [2]  Snedecor, George W. and Cochran, William G. (1989), Statistical
              Methods, Eighth Edition, Iowa State University Press.
    .. [3] Park, C. and Lindsay, B. G. (1999). Robust Scale Estimation and
           Hypothesis Testing based on Quadratic Inference Function. Technical
           Report #99-03, Center for Likelihood Studies, Pennsylvania State
           University.
    .. [4] Bartlett, M. S. (1937). Properties of Sufficiency and Statistical
           Tests. Proceedings of the Royal Society of London. Series A,
           Mathematical and Physical Sciences, Vol. 160, No.901, pp. 268-282.
    .. [5] C.I. BLISS (1952), The Statistics of Bioassay: With Special
           Reference to the Vitamins, pp 499-503,
           :doi:`10.1016/C2013-0-12584-6`.
    .. [6] B. Phipson and G. K. Smyth. "Permutation P-values Should Never Be
           Zero: Calculating Exact P-values When Permutations Are Randomly
           Drawn." Statistical Applications in Genetics and Molecular Biology
           9.1 (2010).
    .. [7] Ludbrook, J., & Dudley, H. (1998). Why permutation tests are
           superior to t and F tests in biomedical research. The American
           Statistician, 52(2), 127-132.

    Examples
    --------
    In [5]_, the influence of vitamin C on the tooth growth of guinea pigs
    was investigated. In a control study, 60 subjects were divided into
    small dose, medium dose, and large dose groups that received
    daily doses of 0.5, 1.0 and 2.0 mg of vitamin C, respectively.
    After 42 days, the tooth growth was measured.

    The ``small_dose``, ``medium_dose``, and ``large_dose`` arrays below record
    tooth growth measurements of the three groups in microns.

    >>> import numpy as np
    >>> small_dose = np.array([
    ...     4.2, 11.5, 7.3, 5.8, 6.4, 10, 11.2, 11.2, 5.2, 7,
    ...     15.2, 21.5, 17.6, 9.7, 14.5, 10, 8.2, 9.4, 16.5, 9.7
    ... ])
    >>> medium_dose = np.array([
    ...     16.5, 16.5, 15.2, 17.3, 22.5, 17.3, 13.6, 14.5, 18.8, 15.5,
    ...     19.7, 23.3, 23.6, 26.4, 20, 25.2, 25.8, 21.2, 14.5, 27.3
    ... ])
    >>> large_dose = np.array([
    ...     23.6, 18.5, 33.9, 25.5, 26.4, 32.5, 26.7, 21.5, 23.3, 29.5,
    ...     25.5, 26.4, 22.4, 24.5, 24.8, 30.9, 26.4, 27.3, 29.4, 23
    ... ])

    The `bartlett` statistic is sensitive to differences in variances
    between the samples.

    >>> from scipy import stats
    >>> res = stats.bartlett(small_dose, medium_dose, large_dose)
    >>> res.statistic
    0.6654670663030519

    The value of the statistic tends to be high when there is a large
    difference in variances.

    We can test for inequality of variance among the groups by comparing the
    observed value of the statistic against the null distribution: the
    distribution of statistic values derived under the null hypothesis that
    the population variances of the three groups are equal.

    For this test, the null distribution follows the chi-square distribution
    as shown below.

    >>> import matplotlib.pyplot as plt
    >>> k = 3  # number of samples
    >>> dist = stats.chi2(df=k-1)
    >>> val = np.linspace(0, 5, 100)
    >>> pdf = dist.pdf(val)
    >>> fig, ax = plt.subplots(figsize=(8, 5))
    >>> def plot(ax):  # we'll reuse this
    ...     ax.plot(val, pdf, color='C0')
    ...     ax.set_title("Bartlett Test Null Distribution")
    ...     ax.set_xlabel("statistic")
    ...     ax.set_ylabel("probability density")
    ...     ax.set_xlim(0, 5)
    ...     ax.set_ylim(0, 1)
    >>> plot(ax)
    >>> plt.show()

    The comparison is quantified by the p-value: the proportion of values in
    the null distribution greater than or equal to the observed value of the
    statistic.

    >>> fig, ax = plt.subplots(figsize=(8, 5))
    >>> plot(ax)
    >>> pvalue = dist.sf(res.statistic)
    >>> annotation = (f'p-value={pvalue:.3f}\n(shaded area)')
    >>> props = dict(facecolor='black', width=1, headwidth=5, headlength=8)
    >>> _ = ax.annotate(annotation, (1.5, 0.22), (2.25, 0.3), arrowprops=props)
    >>> i = val >= res.statistic
    >>> ax.fill_between(val[i], y1=0, y2=pdf[i], color='C0')
    >>> plt.show()

    >>> res.pvalue
    0.71696121509966

    If the p-value is "small" - that is, if there is a low probability of
    sampling data from distributions with identical variances that produces
    such an extreme value of the statistic - this may be taken as evidence
    against the null hypothesis in favor of the alternative: the variances of
    the groups are not equal. Note that:

    - The inverse is not true; that is, the test is not used to provide
      evidence for the null hypothesis.
    - The threshold for values that will be considered "small" is a choice that
      should be made before the data is analyzed [6]_ with consideration of the
      risks of both false positives (incorrectly rejecting the null hypothesis)
      and false negatives (failure to reject a false null hypothesis).
    - Small p-values are not evidence for a *large* effect; rather, they can
      only provide evidence for a "significant" effect, meaning that they are
      unlikely to have occurred under the null hypothesis.

    Note that the chi-square distribution provides the null distribution
    when the observations are normally distributed. For small samples
    drawn from non-normal populations, it may be more appropriate to
    perform a
    permutation test: Under the null hypothesis that all three samples were
    drawn from the same population, each of the measurements is equally likely
    to have been observed in any of the three samples. Therefore, we can form
    a randomized null distribution by calculating the statistic under many
    randomly-generated partitionings of the observations into the three
    samples.

    >>> def statistic(*samples):
    ...     return stats.bartlett(*samples).statistic
    >>> ref = stats.permutation_test(
    ...     (small_dose, medium_dose, large_dose), statistic,
    ...     permutation_type='independent', alternative='greater'
    ... )
    >>> fig, ax = plt.subplots(figsize=(8, 5))
    >>> plot(ax)
    >>> bins = np.linspace(0, 5, 25)
    >>> ax.hist(
    ...     ref.null_distribution, bins=bins, density=True, facecolor="C1"
    ... )
    >>> ax.legend(['aymptotic approximation\n(many observations)',
    ...            'randomized null distribution'])
    >>> plot(ax)
    >>> plt.show()

    >>> ref.pvalue  # randomized test p-value
    0.5387  # may vary

    Note that there is significant disagreement between the p-value calculated
    here and the asymptotic approximation returned by `bartlett` above.
    The statistical inferences that can be drawn rigorously from a permutation
    test are limited; nonetheless, they may be the preferred approach in many
    circumstances [7]_.

    Following is another generic example where the null hypothesis would be
    rejected.

    Test whether the lists `a`, `b` and `c` come from populations
    with equal variances.

    >>> a = [8.88, 9.12, 9.04, 8.98, 9.00, 9.08, 9.01, 8.85, 9.06, 8.99]
    >>> b = [8.88, 8.95, 9.29, 9.44, 9.15, 9.58, 8.36, 9.18, 8.67, 9.05]
    >>> c = [8.95, 9.12, 8.95, 8.85, 9.03, 8.84, 9.07, 8.98, 8.86, 8.98]
    >>> stat, p = stats.bartlett(a, b, c)
    >>> p
    1.1254782518834628e-05

    The very small p-value suggests that the populations do not have equal
    variances.

    This is not surprising, given that the sample variance of `b` is much
    larger than that of `a` and `c`:

    >>> [np.var(x, ddof=1) for x in [a, b, c]]
    [0.007054444444444413, 0.13073888888888888, 0.008890000000000002]

    """
    k = len(samples)
    if k < 2:
        raise ValueError("Must enter at least two input sample vectors.")

    # Handle empty input and input that is not 1d
    for sample in samples:
        if np.asanyarray(sample).size == 0:
            NaN = _get_nan(*samples)  # get NaN of result_dtype of all samples
            return BartlettResult(NaN, NaN)

    Ni = np.empty(k)
    ssq = np.empty(k, 'd')
    for j in range(k):
        Ni[j] = len(samples[j])
        ssq[j] = np.var(samples[j], ddof=1)
    Ntot = np.sum(Ni, axis=0)
    spsq = np.sum((Ni - 1)*ssq, axis=0) / (1.0*(Ntot - k))
    numer = (Ntot*1.0 - k) * log(spsq) - np.sum((Ni - 1.0)*log(ssq), axis=0)
    denom = 1.0 + 1.0/(3*(k - 1)) * ((np.sum(1.0/(Ni - 1.0), axis=0)) -
                                     1.0/(Ntot - k))
    T = numer / denom
    pval = distributions.chi2.sf(T, k - 1)  # 1 - cdf

    return BartlettResult(T, pval)


LeveneResult = namedtuple('LeveneResult', ('statistic', 'pvalue'))


@_axis_nan_policy_factory(LeveneResult, n_samples=None)
def levene(*samples, center='median', proportiontocut=0.05):
    r"""Perform Levene test for equal variances.

    The Levene test tests the null hypothesis that all input samples
    are from populations with equal variances.  Levene's test is an
    alternative to Bartlett's test `bartlett` in the case where
    there are significant deviations from normality.

    Parameters
    ----------
    sample1, sample2, ... : array_like
        The sample data, possibly with different lengths. Only one-dimensional
        samples are accepted.
    center : {'mean', 'median', 'trimmed'}, optional
        Which function of the data to use in the test.  The default
        is 'median'.
    proportiontocut : float, optional
        When `center` is 'trimmed', this gives the proportion of data points
        to cut from each end. (See `scipy.stats.trim_mean`.)
        Default is 0.05.

    Returns
    -------
    statistic : float
        The test statistic.
    pvalue : float
        The p-value for the test.

    See Also
    --------
    fligner : A non-parametric test for the equality of k variances
    bartlett : A parametric test for equality of k variances in normal samples

    Notes
    -----
    Three variations of Levene's test are possible.  The possibilities
    and their recommended usages are:

      * 'median' : Recommended for skewed (non-normal) distributions>
      * 'mean' : Recommended for symmetric, moderate-tailed distributions.
      * 'trimmed' : Recommended for heavy-tailed distributions.

    The test version using the mean was proposed in the original article
    of Levene ([2]_) while the median and trimmed mean have been studied by
    Brown and Forsythe ([3]_), sometimes also referred to as Brown-Forsythe
    test.

    References
    ----------
    .. [1] https://www.itl.nist.gov/div898/handbook/eda/section3/eda35a.htm
    .. [2] Levene, H. (1960). In Contributions to Probability and Statistics:
           Essays in Honor of Harold Hotelling, I. Olkin et al. eds.,
           Stanford University Press, pp. 278-292.
    .. [3] Brown, M. B. and Forsythe, A. B. (1974), Journal of the American
           Statistical Association, 69, 364-367
    .. [4] C.I. BLISS (1952), The Statistics of Bioassay: With Special
           Reference to the Vitamins, pp 499-503,
           :doi:`10.1016/C2013-0-12584-6`.
    .. [5] B. Phipson and G. K. Smyth. "Permutation P-values Should Never Be
           Zero: Calculating Exact P-values When Permutations Are Randomly
           Drawn." Statistical Applications in Genetics and Molecular Biology
           9.1 (2010).
    .. [6] Ludbrook, J., & Dudley, H. (1998). Why permutation tests are
           superior to t and F tests in biomedical research. The American
           Statistician, 52(2), 127-132.

    Examples
    --------
    In [4]_, the influence of vitamin C on the tooth growth of guinea pigs
    was investigated. In a control study, 60 subjects were divided into
    small dose, medium dose, and large dose groups that received
    daily doses of 0.5, 1.0 and 2.0 mg of vitamin C, respectively.
    After 42 days, the tooth growth was measured.

    The ``small_dose``, ``medium_dose``, and ``large_dose`` arrays below record
    tooth growth measurements of the three groups in microns.

    >>> import numpy as np
    >>> small_dose = np.array([
    ...     4.2, 11.5, 7.3, 5.8, 6.4, 10, 11.2, 11.2, 5.2, 7,
    ...     15.2, 21.5, 17.6, 9.7, 14.5, 10, 8.2, 9.4, 16.5, 9.7
    ... ])
    >>> medium_dose = np.array([
    ...     16.5, 16.5, 15.2, 17.3, 22.5, 17.3, 13.6, 14.5, 18.8, 15.5,
    ...     19.7, 23.3, 23.6, 26.4, 20, 25.2, 25.8, 21.2, 14.5, 27.3
    ... ])
    >>> large_dose = np.array([
    ...     23.6, 18.5, 33.9, 25.5, 26.4, 32.5, 26.7, 21.5, 23.3, 29.5,
    ...     25.5, 26.4, 22.4, 24.5, 24.8, 30.9, 26.4, 27.3, 29.4, 23
    ... ])

    The `levene` statistic is sensitive to differences in variances
    between the samples.

    >>> from scipy import stats
    >>> res = stats.levene(small_dose, medium_dose, large_dose)
    >>> res.statistic
    0.6457341109631506

    The value of the statistic tends to be high when there is a large
    difference in variances.

    We can test for inequality of variance among the groups by comparing the
    observed value of the statistic against the null distribution: the
    distribution of statistic values derived under the null hypothesis that
    the population variances of the three groups are equal.

    For this test, the null distribution follows the F distribution as shown
    below.

    >>> import matplotlib.pyplot as plt
    >>> k, n = 3, 60   # number of samples, total number of observations
    >>> dist = stats.f(dfn=k-1, dfd=n-k)
    >>> val = np.linspace(0, 5, 100)
    >>> pdf = dist.pdf(val)
    >>> fig, ax = plt.subplots(figsize=(8, 5))
    >>> def plot(ax):  # we'll reuse this
    ...     ax.plot(val, pdf, color='C0')
    ...     ax.set_title("Levene Test Null Distribution")
    ...     ax.set_xlabel("statistic")
    ...     ax.set_ylabel("probability density")
    ...     ax.set_xlim(0, 5)
    ...     ax.set_ylim(0, 1)
    >>> plot(ax)
    >>> plt.show()

    The comparison is quantified by the p-value: the proportion of values in
    the null distribution greater than or equal to the observed value of the
    statistic.

    >>> fig, ax = plt.subplots(figsize=(8, 5))
    >>> plot(ax)
    >>> pvalue = dist.sf(res.statistic)
    >>> annotation = (f'p-value={pvalue:.3f}\n(shaded area)')
    >>> props = dict(facecolor='black', width=1, headwidth=5, headlength=8)
    >>> _ = ax.annotate(annotation, (1.5, 0.22), (2.25, 0.3), arrowprops=props)
    >>> i = val >= res.statistic
    >>> ax.fill_between(val[i], y1=0, y2=pdf[i], color='C0')
    >>> plt.show()

    >>> res.pvalue
    0.5280694573759905

    If the p-value is "small" - that is, if there is a low probability of
    sampling data from distributions with identical variances that produces
    such an extreme value of the statistic - this may be taken as evidence
    against the null hypothesis in favor of the alternative: the variances of
    the groups are not equal. Note that:

    - The inverse is not true; that is, the test is not used to provide
      evidence for the null hypothesis.
    - The threshold for values that will be considered "small" is a choice that
      should be made before the data is analyzed [5]_ with consideration of the
      risks of both false positives (incorrectly rejecting the null hypothesis)
      and false negatives (failure to reject a false null hypothesis).
    - Small p-values are not evidence for a *large* effect; rather, they can
      only provide evidence for a "significant" effect, meaning that they are
      unlikely to have occurred under the null hypothesis.

    Note that the F distribution provides an asymptotic approximation of the
    null distribution.
    For small samples, it may be more appropriate to perform a permutation
    test: Under the null hypothesis that all three samples were drawn from
    the same population, each of the measurements is equally likely to have
    been observed in any of the three samples. Therefore, we can form a
    randomized null distribution by calculating the statistic under many
    randomly-generated partitionings of the observations into the three
    samples.

    >>> def statistic(*samples):
    ...     return stats.levene(*samples).statistic
    >>> ref = stats.permutation_test(
    ...     (small_dose, medium_dose, large_dose), statistic,
    ...     permutation_type='independent', alternative='greater'
    ... )
    >>> fig, ax = plt.subplots(figsize=(8, 5))
    >>> plot(ax)
    >>> bins = np.linspace(0, 5, 25)
    >>> ax.hist(
    ...     ref.null_distribution, bins=bins, density=True, facecolor="C1"
    ... )
    >>> ax.legend(['aymptotic approximation\n(many observations)',
    ...            'randomized null distribution'])
    >>> plot(ax)
    >>> plt.show()

    >>> ref.pvalue  # randomized test p-value
    0.4559  # may vary

    Note that there is significant disagreement between the p-value calculated
    here and the asymptotic approximation returned by `levene` above.
    The statistical inferences that can be drawn rigorously from a permutation
    test are limited; nonetheless, they may be the preferred approach in many
    circumstances [6]_.

    Following is another generic example where the null hypothesis would be
    rejected.

    Test whether the lists `a`, `b` and `c` come from populations
    with equal variances.

    >>> a = [8.88, 9.12, 9.04, 8.98, 9.00, 9.08, 9.01, 8.85, 9.06, 8.99]
    >>> b = [8.88, 8.95, 9.29, 9.44, 9.15, 9.58, 8.36, 9.18, 8.67, 9.05]
    >>> c = [8.95, 9.12, 8.95, 8.85, 9.03, 8.84, 9.07, 8.98, 8.86, 8.98]
    >>> stat, p = stats.levene(a, b, c)
    >>> p
    0.002431505967249681

    The small p-value suggests that the populations do not have equal
    variances.

    This is not surprising, given that the sample variance of `b` is much
    larger than that of `a` and `c`:

    >>> [np.var(x, ddof=1) for x in [a, b, c]]
    [0.007054444444444413, 0.13073888888888888, 0.008890000000000002]

    """
    if center not in ['mean', 'median', 'trimmed']:
        raise ValueError("center must be 'mean', 'median' or 'trimmed'.")

    k = len(samples)
    if k < 2:
        raise ValueError("Must enter at least two input sample vectors.")

    Ni = np.empty(k)
    Yci = np.empty(k, 'd')

    if center == 'median':

        def func(x):
            return np.median(x, axis=0)

    elif center == 'mean':

        def func(x):
            return np.mean(x, axis=0)

    else:  # center == 'trimmed'
        samples = tuple(_stats_py.trimboth(np.sort(sample), proportiontocut)
                        for sample in samples)

        def func(x):
            return np.mean(x, axis=0)

    for j in range(k):
        Ni[j] = len(samples[j])
        Yci[j] = func(samples[j])
    Ntot = np.sum(Ni, axis=0)

    # compute Zij's
    Zij = [None] * k
    for i in range(k):
        Zij[i] = abs(asarray(samples[i]) - Yci[i])

    # compute Zbari
    Zbari = np.empty(k, 'd')
    Zbar = 0.0
    for i in range(k):
        Zbari[i] = np.mean(Zij[i], axis=0)
        Zbar += Zbari[i] * Ni[i]

    Zbar /= Ntot
    numer = (Ntot - k) * np.sum(Ni * (Zbari - Zbar)**2, axis=0)

    # compute denom_variance
    dvar = 0.0
    for i in range(k):
        dvar += np.sum((Zij[i] - Zbari[i])**2, axis=0)

    denom = (k - 1.0) * dvar

    W = numer / denom
    pval = distributions.f.sf(W, k-1, Ntot-k)  # 1 - cdf
    return LeveneResult(W, pval)


def _apply_func(x, g, func):
    # g is list of indices into x
    #  separating x into different groups
    #  func should be applied over the groups
    g = unique(r_[0, g, len(x)])
    output = [func(x[g[k]:g[k+1]]) for k in range(len(g) - 1)]

    return asarray(output)


FlignerResult = namedtuple('FlignerResult', ('statistic', 'pvalue'))


@_axis_nan_policy_factory(FlignerResult, n_samples=None)
def fligner(*samples, center='median', proportiontocut=0.05):
    r"""Perform Fligner-Killeen test for equality of variance.

    Fligner's test tests the null hypothesis that all input samples
    are from populations with equal variances.  Fligner-Killeen's test is
    distribution free when populations are identical [2]_.

    Parameters
    ----------
    sample1, sample2, ... : array_like
        Arrays of sample data.  Need not be the same length.
    center : {'mean', 'median', 'trimmed'}, optional
        Keyword argument controlling which function of the data is used in
        computing the test statistic.  The default is 'median'.
    proportiontocut : float, optional
        When `center` is 'trimmed', this gives the proportion of data points
        to cut from each end. (See `scipy.stats.trim_mean`.)
        Default is 0.05.

    Returns
    -------
    statistic : float
        The test statistic.
    pvalue : float
        The p-value for the hypothesis test.

    See Also
    --------
    bartlett : A parametric test for equality of k variances in normal samples
    levene : A robust parametric test for equality of k variances

    Notes
    -----
    As with Levene's test there are three variants of Fligner's test that
    differ by the measure of central tendency used in the test.  See `levene`
    for more information.

    Conover et al. (1981) examine many of the existing parametric and
    nonparametric tests by extensive simulations and they conclude that the
    tests proposed by Fligner and Killeen (1976) and Levene (1960) appear to be
    superior in terms of robustness of departures from normality and power
    [3]_.

    References
    ----------
    .. [1] Park, C. and Lindsay, B. G. (1999). Robust Scale Estimation and
           Hypothesis Testing based on Quadratic Inference Function. Technical
           Report #99-03, Center for Likelihood Studies, Pennsylvania State
           University.
           https://cecas.clemson.edu/~cspark/cv/paper/qif/draftqif2.pdf
    .. [2] Fligner, M.A. and Killeen, T.J. (1976). Distribution-free two-sample
           tests for scale. 'Journal of the American Statistical Association.'
           71(353), 210-213.
    .. [3] Park, C. and Lindsay, B. G. (1999). Robust Scale Estimation and
           Hypothesis Testing based on Quadratic Inference Function. Technical
           Report #99-03, Center for Likelihood Studies, Pennsylvania State
           University.
    .. [4] Conover, W. J., Johnson, M. E. and Johnson M. M. (1981). A
           comparative study of tests for homogeneity of variances, with
           applications to the outer continental shelf bidding data.
           Technometrics, 23(4), 351-361.
    .. [5] C.I. BLISS (1952), The Statistics of Bioassay: With Special
           Reference to the Vitamins, pp 499-503,
           :doi:`10.1016/C2013-0-12584-6`.
    .. [6] B. Phipson and G. K. Smyth. "Permutation P-values Should Never Be
           Zero: Calculating Exact P-values When Permutations Are Randomly
           Drawn." Statistical Applications in Genetics and Molecular Biology
           9.1 (2010).
    .. [7] Ludbrook, J., & Dudley, H. (1998). Why permutation tests are
           superior to t and F tests in biomedical research. The American
           Statistician, 52(2), 127-132.

    Examples
    --------
    In [5]_, the influence of vitamin C on the tooth growth of guinea pigs
    was investigated. In a control study, 60 subjects were divided into
    small dose, medium dose, and large dose groups that received
    daily doses of 0.5, 1.0 and 2.0 mg of vitamin C, respectively.
    After 42 days, the tooth growth was measured.

    The ``small_dose``, ``medium_dose``, and ``large_dose`` arrays below record
    tooth growth measurements of the three groups in microns.

    >>> import numpy as np
    >>> small_dose = np.array([
    ...     4.2, 11.5, 7.3, 5.8, 6.4, 10, 11.2, 11.2, 5.2, 7,
    ...     15.2, 21.5, 17.6, 9.7, 14.5, 10, 8.2, 9.4, 16.5, 9.7
    ... ])
    >>> medium_dose = np.array([
    ...     16.5, 16.5, 15.2, 17.3, 22.5, 17.3, 13.6, 14.5, 18.8, 15.5,
    ...     19.7, 23.3, 23.6, 26.4, 20, 25.2, 25.8, 21.2, 14.5, 27.3
    ... ])
    >>> large_dose = np.array([
    ...     23.6, 18.5, 33.9, 25.5, 26.4, 32.5, 26.7, 21.5, 23.3, 29.5,
    ...     25.5, 26.4, 22.4, 24.5, 24.8, 30.9, 26.4, 27.3, 29.4, 23
    ... ])

    The `fligner` statistic is sensitive to differences in variances
    between the samples.

    >>> from scipy import stats
    >>> res = stats.fligner(small_dose, medium_dose, large_dose)
    >>> res.statistic
    1.3878943408857916

    The value of the statistic tends to be high when there is a large
    difference in variances.

    We can test for inequality of variance among the groups by comparing the
    observed value of the statistic against the null distribution: the
    distribution of statistic values derived under the null hypothesis that
    the population variances of the three groups are equal.

    For this test, the null distribution follows the chi-square distribution
    as shown below.

    >>> import matplotlib.pyplot as plt
    >>> k = 3  # number of samples
    >>> dist = stats.chi2(df=k-1)
    >>> val = np.linspace(0, 8, 100)
    >>> pdf = dist.pdf(val)
    >>> fig, ax = plt.subplots(figsize=(8, 5))
    >>> def plot(ax):  # we'll reuse this
    ...     ax.plot(val, pdf, color='C0')
    ...     ax.set_title("Fligner Test Null Distribution")
    ...     ax.set_xlabel("statistic")
    ...     ax.set_ylabel("probability density")
    ...     ax.set_xlim(0, 8)
    ...     ax.set_ylim(0, 0.5)
    >>> plot(ax)
    >>> plt.show()

    The comparison is quantified by the p-value: the proportion of values in
    the null distribution greater than or equal to the observed value of the
    statistic.

    >>> fig, ax = plt.subplots(figsize=(8, 5))
    >>> plot(ax)
    >>> pvalue = dist.sf(res.statistic)
    >>> annotation = (f'p-value={pvalue:.4f}\n(shaded area)')
    >>> props = dict(facecolor='black', width=1, headwidth=5, headlength=8)
    >>> _ = ax.annotate(annotation, (1.5, 0.22), (2.25, 0.3), arrowprops=props)
    >>> i = val >= res.statistic
    >>> ax.fill_between(val[i], y1=0, y2=pdf[i], color='C0')
    >>> plt.show()

    >>> res.pvalue
    0.49960016501182125

    If the p-value is "small" - that is, if there is a low probability of
    sampling data from distributions with identical variances that produces
    such an extreme value of the statistic - this may be taken as evidence
    against the null hypothesis in favor of the alternative: the variances of
    the groups are not equal. Note that:

    - The inverse is not true; that is, the test is not used to provide
      evidence for the null hypothesis.
    - The threshold for values that will be considered "small" is a choice that
      should be made before the data is analyzed [6]_ with consideration of the
      risks of both false positives (incorrectly rejecting the null hypothesis)
      and false negatives (failure to reject a false null hypothesis).
    - Small p-values are not evidence for a *large* effect; rather, they can
      only provide evidence for a "significant" effect, meaning that they are
      unlikely to have occurred under the null hypothesis.

    Note that the chi-square distribution provides an asymptotic approximation
    of the null distribution.
    For small samples, it may be more appropriate to perform a
    permutation test: Under the null hypothesis that all three samples were
    drawn from the same population, each of the measurements is equally likely
    to have been observed in any of the three samples. Therefore, we can form
    a randomized null distribution by calculating the statistic under many
    randomly-generated partitionings of the observations into the three
    samples.

    >>> def statistic(*samples):
    ...     return stats.fligner(*samples).statistic
    >>> ref = stats.permutation_test(
    ...     (small_dose, medium_dose, large_dose), statistic,
    ...     permutation_type='independent', alternative='greater'
    ... )
    >>> fig, ax = plt.subplots(figsize=(8, 5))
    >>> plot(ax)
    >>> bins = np.linspace(0, 8, 25)
    >>> ax.hist(
    ...     ref.null_distribution, bins=bins, density=True, facecolor="C1"
    ... )
    >>> ax.legend(['aymptotic approximation\n(many observations)',
    ...            'randomized null distribution'])
    >>> plot(ax)
    >>> plt.show()

    >>> ref.pvalue  # randomized test p-value
    0.4332  # may vary

    Note that there is significant disagreement between the p-value calculated
    here and the asymptotic approximation returned by `fligner` above.
    The statistical inferences that can be drawn rigorously from a permutation
    test are limited; nonetheless, they may be the preferred approach in many
    circumstances [7]_.

    Following is another generic example where the null hypothesis would be
    rejected.

    Test whether the lists `a`, `b` and `c` come from populations
    with equal variances.

    >>> a = [8.88, 9.12, 9.04, 8.98, 9.00, 9.08, 9.01, 8.85, 9.06, 8.99]
    >>> b = [8.88, 8.95, 9.29, 9.44, 9.15, 9.58, 8.36, 9.18, 8.67, 9.05]
    >>> c = [8.95, 9.12, 8.95, 8.85, 9.03, 8.84, 9.07, 8.98, 8.86, 8.98]
    >>> stat, p = stats.fligner(a, b, c)
    >>> p
    0.00450826080004775

    The small p-value suggests that the populations do not have equal
    variances.

    This is not surprising, given that the sample variance of `b` is much
    larger than that of `a` and `c`:

    >>> [np.var(x, ddof=1) for x in [a, b, c]]
    [0.007054444444444413, 0.13073888888888888, 0.008890000000000002]

    """
    if center not in ['mean', 'median', 'trimmed']:
        raise ValueError("center must be 'mean', 'median' or 'trimmed'.")

    k = len(samples)
    if k < 2:
        raise ValueError("Must enter at least two input sample vectors.")

    # Handle empty input
    for sample in samples:
        if sample.size == 0:
            NaN = _get_nan(*samples)
            return FlignerResult(NaN, NaN)

    if center == 'median':

        def func(x):
            return np.median(x, axis=0)

    elif center == 'mean':

        def func(x):
            return np.mean(x, axis=0)

    else:  # center == 'trimmed'
        samples = tuple(_stats_py.trimboth(sample, proportiontocut)
                        for sample in samples)

        def func(x):
            return np.mean(x, axis=0)

    Ni = asarray([len(samples[j]) for j in range(k)])
    Yci = asarray([func(samples[j]) for j in range(k)])
    Ntot = np.sum(Ni, axis=0)
    # compute Zij's
    Zij = [abs(asarray(samples[i]) - Yci[i]) for i in range(k)]
    allZij = []
    g = [0]
    for i in range(k):
        allZij.extend(list(Zij[i]))
        g.append(len(allZij))

    ranks = _stats_py.rankdata(allZij)
    sample = distributions.norm.ppf(ranks / (2*(Ntot + 1.0)) + 0.5)

    # compute Aibar
    Aibar = _apply_func(sample, g, np.sum) / Ni
    anbar = np.mean(sample, axis=0)
    varsq = np.var(sample, axis=0, ddof=1)
    Xsq = np.sum(Ni * (asarray(Aibar) - anbar)**2.0, axis=0) / varsq
    pval = distributions.chi2.sf(Xsq, k - 1)  # 1 - cdf
    return FlignerResult(Xsq, pval)


@_axis_nan_policy_factory(lambda x1: (x1,), n_samples=4, n_outputs=1)
def _mood_inner_lc(xy, x, diffs, sorted_xy, n, m, N) -> float:
    # Obtain the unique values and their frequencies from the pooled samples.
    # "a_j, + b_j, = t_j, for j = 1, ... k" where `k` is the number of unique
    # classes, and "[t]he number of values associated with the x's and y's in
    # the jth class will be denoted by a_j, and b_j respectively."
    # (Mielke, 312)
    # Reuse previously computed sorted array and `diff` arrays to obtain the
    # unique values and counts. Prepend `diffs` with a non-zero to indicate
    # that the first element should be marked as not matching what preceded it.
    diffs_prep = np.concatenate(([1], diffs))
    # Unique elements are where the was a difference between elements in the
    # sorted array
    uniques = sorted_xy[diffs_prep != 0]
    # The count of each element is the bin size for each set of consecutive
    # differences where the difference is zero. Replace nonzero differences
    # with 1 and then use the cumulative sum to count the indices.
    t = np.bincount(np.cumsum(np.asarray(diffs_prep != 0, dtype=int)))[1:]
    k = len(uniques)
    js = np.arange(1, k + 1, dtype=int)
    # the `b` array mentioned in the paper is not used, outside of the
    # calculation of `t`, so we do not need to calculate it separately. Here
    # we calculate `a`. In plain language, `a[j]` is the number of values in
    # `x` that equal `uniques[j]`.
    sorted_xyx = np.sort(np.concatenate((xy, x)))
    diffs = np.diff(sorted_xyx)
    diffs_prep = np.concatenate(([1], diffs))
    diff_is_zero = np.asarray(diffs_prep != 0, dtype=int)
    xyx_counts = np.bincount(np.cumsum(diff_is_zero))[1:]
    a = xyx_counts - t
    # "Define .. a_0 = b_0 = t_0 = S_0 = 0" (Mielke 312) so we shift  `a`
    # and `t` arrays over 1 to allow a first element of 0 to accommodate this
    # indexing.
    t = np.concatenate(([0], t))
    a = np.concatenate(([0], a))
    # S is built from `t`, so it does not need a preceding zero added on.
    S = np.cumsum(t)
    # define a copy of `S` with a prepending zero for later use to avoid
    # the need for indexing.
    S_i_m1 = np.concatenate(([0], S[:-1]))

    # Psi, as defined by the 6th unnumbered equation on page 313 (Mielke).
    # Note that in the paper there is an error where the denominator `2` is
    # squared when it should be the entire equation.
    def psi(indicator):
        return (indicator - (N + 1)/2)**2

    # define summation range for use in calculation of phi, as seen in sum
    # in the unnumbered equation on the bottom of page 312 (Mielke).
    s_lower = S[js - 1] + 1
    s_upper = S[js] + 1
    phi_J = [np.arange(s_lower[idx], s_upper[idx]) for idx in range(k)]

    # for every range in the above array, determine the sum of psi(I) for
    # every element in the range. Divide all the sums by `t`. Following the
    # last unnumbered equation on page 312.
    phis = [np.sum(psi(I_j)) for I_j in phi_J] / t[js]

    # `T` is equal to a[j] * phi[j], per the first unnumbered equation on
    # page 312. `phis` is already in the order based on `js`, so we index
    # into `a` with `js` as well.
    T = sum(phis * a[js])

    # The approximate statistic
    E_0_T = n * (N * N - 1) / 12

    varM = (m * n * (N + 1.0) * (N ** 2 - 4) / 180 -
            m * n / (180 * N * (N - 1)) * np.sum(
                t * (t**2 - 1) * (t**2 - 4 + (15 * (N - S - S_i_m1) ** 2))
            ))

    return ((T - E_0_T) / np.sqrt(varM),)


def _mood_too_small(samples, kwargs, axis=-1):
    x, y = samples
    n = x.shape[axis]
    m = y.shape[axis]
    N = m + n
    return N < 3


@_axis_nan_policy_factory(SignificanceResult, n_samples=2, too_small=_mood_too_small)
def mood(x, y, axis=0, alternative="two-sided"):
    """Perform Mood's test for equal scale parameters.

    Mood's two-sample test for scale parameters is a non-parametric
    test for the null hypothesis that two samples are drawn from the
    same distribution with the same scale parameter.

    Parameters
    ----------
    x, y : array_like
        Arrays of sample data.
    axis : int, optional
        The axis along which the samples are tested.  `x` and `y` can be of
        different length along `axis`.
        If `axis` is None, `x` and `y` are flattened and the test is done on
        all values in the flattened arrays.
    alternative : {'two-sided', 'less', 'greater'}, optional
        Defines the alternative hypothesis. Default is 'two-sided'.
        The following options are available:

        * 'two-sided': the scales of the distributions underlying `x` and `y`
          are different.
        * 'less': the scale of the distribution underlying `x` is less than
          the scale of the distribution underlying `y`.
        * 'greater': the scale of the distribution underlying `x` is greater
          than the scale of the distribution underlying `y`.

        .. versionadded:: 1.7.0

    Returns
    -------
    res : SignificanceResult
        An object containing attributes:

        statistic : scalar or ndarray
            The z-score for the hypothesis test.  For 1-D inputs a scalar is
            returned.
        pvalue : scalar ndarray
            The p-value for the hypothesis test.

    See Also
    --------
    fligner : A non-parametric test for the equality of k variances
    ansari : A non-parametric test for the equality of 2 variances
    bartlett : A parametric test for equality of k variances in normal samples
    levene : A parametric test for equality of k variances

    Notes
    -----
    The data are assumed to be drawn from probability distributions ``f(x)``
    and ``f(x/s) / s`` respectively, for some probability density function f.
    The null hypothesis is that ``s == 1``.

    For multi-dimensional arrays, if the inputs are of shapes
    ``(n0, n1, n2, n3)``  and ``(n0, m1, n2, n3)``, then if ``axis=1``, the
    resulting z and p values will have shape ``(n0, n2, n3)``.  Note that
    ``n1`` and ``m1`` don't have to be equal, but the other dimensions do.

    References
    ----------
    [1] Mielke, Paul W. "Note on Some Squared Rank Tests with Existing Ties."
        Technometrics, vol. 9, no. 2, 1967, pp. 312-14. JSTOR,
        https://doi.org/10.2307/1266427. Accessed 18 May 2022.

    Examples
    --------
    >>> import numpy as np
    >>> from scipy import stats
    >>> rng = np.random.default_rng()
    >>> x2 = rng.standard_normal((2, 45, 6, 7))
    >>> x1 = rng.standard_normal((2, 30, 6, 7))
    >>> res = stats.mood(x1, x2, axis=1)
    >>> res.pvalue.shape
    (2, 6, 7)

    Find the number of points where the difference in scale is not significant:

    >>> (res.pvalue > 0.1).sum()
    78

    Perform the test with different scales:

    >>> x1 = rng.standard_normal((2, 30))
    >>> x2 = rng.standard_normal((2, 35)) * 10.0
    >>> stats.mood(x1, x2, axis=1)
    SignificanceResult(statistic=array([-5.76174136, -6.12650783]),
                       pvalue=array([8.32505043e-09, 8.98287869e-10]))

    """
    x = np.asarray(x, dtype=float)
    y = np.asarray(y, dtype=float)

    if axis < 0:
        axis = x.ndim + axis

    # Determine shape of the result arrays
    res_shape = tuple([x.shape[ax] for ax in range(len(x.shape)) if ax != axis])
    if not (res_shape == tuple([y.shape[ax] for ax in range(len(y.shape)) if
                                ax != axis])):
        raise ValueError("Dimensions of x and y on all axes except `axis` "
                         "should match")

    n = x.shape[axis]
    m = y.shape[axis]
    N = m + n
    if N < 3:
        raise ValueError("Not enough observations.")

    xy = np.concatenate((x, y), axis=axis)
    # determine if any of the samples contain ties
    sorted_xy = np.sort(xy, axis=axis)
    diffs = np.diff(sorted_xy, axis=axis)
    if 0 in diffs:
        z = np.asarray(_mood_inner_lc(xy, x, diffs, sorted_xy, n, m, N,
                                      axis=axis))
    else:
        if axis != 0:
            xy = np.moveaxis(xy, axis, 0)

        xy = xy.reshape(xy.shape[0], -1)
        # Generalized to the n-dimensional case by adding the axis argument,
        # and using for loops, since rankdata is not vectorized.  For improving
        # performance consider vectorizing rankdata function.
        all_ranks = np.empty_like(xy)
        for j in range(xy.shape[1]):
            all_ranks[:, j] = _stats_py.rankdata(xy[:, j])

        Ri = all_ranks[:n]
        M = np.sum((Ri - (N + 1.0) / 2) ** 2, axis=0)
        # Approx stat.
        mnM = n * (N * N - 1.0) / 12
        varM = m * n * (N + 1.0) * (N + 2) * (N - 2) / 180
        z = (M - mnM) / sqrt(varM)
    pval = _get_pvalue(z, distributions.norm, alternative)

    if res_shape == ():
        # Return scalars, not 0-D arrays
        z = z[0]
        pval = pval[0]
    else:
        z.shape = res_shape
        pval.shape = res_shape
    return SignificanceResult(z[()], pval[()])


WilcoxonResult = _make_tuple_bunch('WilcoxonResult', ['statistic', 'pvalue'])


def wilcoxon_result_unpacker(res):
    if hasattr(res, 'zstatistic'):
        return res.statistic, res.pvalue, res.zstatistic
    else:
        return res.statistic, res.pvalue


def wilcoxon_result_object(statistic, pvalue, zstatistic=None):
    res = WilcoxonResult(statistic, pvalue)
    if zstatistic is not None:
        res.zstatistic = zstatistic
    return res


def wilcoxon_outputs(kwds):
    method = kwds.get('method', 'auto')
    if method == 'approx':
        return 3
    return 2


@_rename_parameter("mode", "method")
@_axis_nan_policy_factory(
    wilcoxon_result_object, paired=True,
    n_samples=lambda kwds: 2 if kwds.get('y', None) is not None else 1,
    result_to_tuple=wilcoxon_result_unpacker, n_outputs=wilcoxon_outputs,
)
def wilcoxon(x, y=None, zero_method="wilcox", correction=False,
             alternative="two-sided", method='auto', *, axis=0):
    """Calculate the Wilcoxon signed-rank test.

    The Wilcoxon signed-rank test tests the null hypothesis that two
    related paired samples come from the same distribution. In particular,
    it tests whether the distribution of the differences ``x - y`` is symmetric
    about zero. It is a non-parametric version of the paired T-test.

    Parameters
    ----------
    x : array_like
        Either the first set of measurements (in which case ``y`` is the second
        set of measurements), or the differences between two sets of
        measurements (in which case ``y`` is not to be specified.)  Must be
        one-dimensional.
    y : array_like, optional
        Either the second set of measurements (if ``x`` is the first set of
        measurements), or not specified (if ``x`` is the differences between
        two sets of measurements.)  Must be one-dimensional.

        .. warning::
            When `y` is provided, `wilcoxon` calculates the test statistic
            based on the ranks of the absolute values of ``d = x - y``.
            Roundoff error in the subtraction can result in elements of ``d``
            being assigned different ranks even when they would be tied with
            exact arithmetic. Rather than passing `x` and `y` separately,
            consider computing the difference ``x - y``, rounding as needed to
            ensure that only truly unique elements are numerically distinct,
            and passing the result as `x`, leaving `y` at the default (None).

    zero_method : {"wilcox", "pratt", "zsplit"}, optional
        There are different conventions for handling pairs of observations
        with equal values ("zero-differences", or "zeros").

        * "wilcox": Discards all zero-differences (default); see [4]_.
        * "pratt": Includes zero-differences in the ranking process,
          but drops the ranks of the zeros (more conservative); see [3]_.
          In this case, the normal approximation is adjusted as in [5]_.
        * "zsplit": Includes zero-differences in the ranking process and
          splits the zero rank between positive and negative ones.

    correction : bool, optional
        If True, apply continuity correction by adjusting the Wilcoxon rank
        statistic by 0.5 towards the mean value when computing the
        z-statistic if a normal approximation is used.  Default is False.
    alternative : {"two-sided", "greater", "less"}, optional
        Defines the alternative hypothesis. Default is 'two-sided'.
        In the following, let ``d`` represent the difference between the paired
        samples: ``d = x - y`` if both ``x`` and ``y`` are provided, or
        ``d = x`` otherwise.

        * 'two-sided': the distribution underlying ``d`` is not symmetric
          about zero.
        * 'less': the distribution underlying ``d`` is stochastically less
          than a distribution symmetric about zero.
        * 'greater': the distribution underlying ``d`` is stochastically
          greater than a distribution symmetric about zero.

    method : {"auto", "exact", "approx"} or `PermutationMethod` instance, optional
        Method to calculate the p-value, see Notes. Default is "auto".

    axis : int or None, default: 0
        If an int, the axis of the input along which to compute the statistic.
        The statistic of each axis-slice (e.g. row) of the input will appear
        in a corresponding element of the output. If ``None``, the input will
        be raveled before computing the statistic.

    Returns
    -------
    An object with the following attributes.

    statistic : array_like
        If `alternative` is "two-sided", the sum of the ranks of the
        differences above or below zero, whichever is smaller.
        Otherwise the sum of the ranks of the differences above zero.
    pvalue : array_like
        The p-value for the test depending on `alternative` and `method`.
    zstatistic : array_like
        When ``method = 'approx'``, this is the normalized z-statistic::

            z = (T - mn - d) / se

        where ``T`` is `statistic` as defined above, ``mn`` is the mean of the
        distribution under the null hypothesis, ``d`` is a continuity
        correction, and ``se`` is the standard error.
        When ``method != 'approx'``, this attribute is not available.

    See Also
    --------
    kruskal, mannwhitneyu

    Notes
    -----
    In the following, let ``d`` represent the difference between the paired
    samples: ``d = x - y`` if both ``x`` and ``y`` are provided, or ``d = x``
    otherwise. Assume that all elements of ``d`` are independent and
    identically distributed observations, and all are distinct and nonzero.

    - When ``len(d)`` is sufficiently large, the null distribution of the
      normalized test statistic (`zstatistic` above) is approximately normal,
      and ``method = 'approx'`` can be used to compute the p-value.

    - When ``len(d)`` is small, the normal approximation may not be accurate,
      and ``method='exact'`` is preferred (at the cost of additional
      execution time).

    - The default, ``method='auto'``, selects between the two: when
      ``len(d) <= 50`` and there are no zeros, the exact method is used;
      otherwise, the approximate method is used.

    The presence of "ties" (i.e. not all elements of ``d`` are unique) or
    "zeros" (i.e. elements of ``d`` are zero) changes the null distribution
    of the test statistic, and ``method='exact'`` no longer calculates
    the exact p-value. If ``method='approx'``, the z-statistic is adjusted
    for more accurate comparison against the standard normal, but still,
    for finite sample sizes, the standard normal is only an approximation of
    the true null distribution of the z-statistic. For such situations, the
    `method` parameter also accepts instances `PermutationMethod`. In this
    case, the p-value is computed using `permutation_test` with the provided
    configuration options and other appropriate settings.

    References
    ----------
    .. [1] https://en.wikipedia.org/wiki/Wilcoxon_signed-rank_test
    .. [2] Conover, W.J., Practical Nonparametric Statistics, 1971.
    .. [3] Pratt, J.W., Remarks on Zeros and Ties in the Wilcoxon Signed
       Rank Procedures, Journal of the American Statistical Association,
       Vol. 54, 1959, pp. 655-667. :doi:`10.1080/01621459.1959.10501526`
    .. [4] Wilcoxon, F., Individual Comparisons by Ranking Methods,
       Biometrics Bulletin, Vol. 1, 1945, pp. 80-83. :doi:`10.2307/3001968`
    .. [5] Cureton, E.E., The Normal Approximation to the Signed-Rank
       Sampling Distribution When Zero Differences are Present,
       Journal of the American Statistical Association, Vol. 62, 1967,
       pp. 1068-1069. :doi:`10.1080/01621459.1967.10500917`

    Examples
    --------
    In [4]_, the differences in height between cross- and self-fertilized
    corn plants is given as follows:

    >>> d = [6, 8, 14, 16, 23, 24, 28, 29, 41, -48, 49, 56, 60, -67, 75]

    Cross-fertilized plants appear to be higher. To test the null
    hypothesis that there is no height difference, we can apply the
    two-sided test:

    >>> from scipy.stats import wilcoxon
    >>> res = wilcoxon(d)
    >>> res.statistic, res.pvalue
    (24.0, 0.041259765625)

    Hence, we would reject the null hypothesis at a confidence level of 5%,
    concluding that there is a difference in height between the groups.
    To confirm that the median of the differences can be assumed to be
    positive, we use:

    >>> res = wilcoxon(d, alternative='greater')
    >>> res.statistic, res.pvalue
    (96.0, 0.0206298828125)

    This shows that the null hypothesis that the median is negative can be
    rejected at a confidence level of 5% in favor of the alternative that
    the median is greater than zero. The p-values above are exact. Using the
    normal approximation gives very similar values:

    >>> res = wilcoxon(d, method='approx')
    >>> res.statistic, res.pvalue
    (24.0, 0.04088813291185591)

    Note that the statistic changed to 96 in the one-sided case (the sum
    of ranks of positive differences) whereas it is 24 in the two-sided
    case (the minimum of sum of ranks above and below zero).

    In the example above, the differences in height between paired plants are
    provided to `wilcoxon` directly. Alternatively, `wilcoxon` accepts two
    samples of equal length, calculates the differences between paired
    elements, then performs the test. Consider the samples ``x`` and ``y``:

    >>> import numpy as np
    >>> x = np.array([0.5, 0.825, 0.375, 0.5])
    >>> y = np.array([0.525, 0.775, 0.325, 0.55])
    >>> res = wilcoxon(x, y, alternative='greater')
    >>> res
    WilcoxonResult(statistic=5.0, pvalue=0.5625)

    Note that had we calculated the differences by hand, the test would have
    produced different results:

    >>> d = [-0.025, 0.05, 0.05, -0.05]
    >>> ref = wilcoxon(d, alternative='greater')
    >>> ref
    WilcoxonResult(statistic=6.0, pvalue=0.4375)

    The substantial difference is due to roundoff error in the results of
    ``x-y``:

    >>> d - (x-y)
    array([2.08166817e-17, 6.93889390e-17, 1.38777878e-17, 4.16333634e-17])

    Even though we expected all the elements of ``(x-y)[1:]`` to have the same
    magnitude ``0.05``, they have slightly different magnitudes in practice,
    and therefore are assigned different ranks in the test. Before performing
    the test, consider calculating ``d`` and adjusting it as necessary to
    ensure that theoretically identically values are not numerically distinct.
    For example:

    >>> d2 = np.around(x - y, decimals=3)
    >>> wilcoxon(d2, alternative='greater')
    WilcoxonResult(statistic=6.0, pvalue=0.4375)

    """
    return _wilcoxon._wilcoxon_nd(x, y, zero_method, correction, alternative,
                                  method, axis)


MedianTestResult = _make_tuple_bunch(
    'MedianTestResult',
    ['statistic', 'pvalue', 'median', 'table'], []
)


def median_test(*samples, ties='below', correction=True, lambda_=1,
                nan_policy='propagate'):
    """Perform a Mood's median test.

    Test that two or more samples come from populations with the same median.

    Let ``n = len(samples)`` be the number of samples.  The "grand median" of
    all the data is computed, and a contingency table is formed by
    classifying the values in each sample as being above or below the grand
    median.  The contingency table, along with `correction` and `lambda_`,
    are passed to `scipy.stats.chi2_contingency` to compute the test statistic
    and p-value.

    Parameters
    ----------
    sample1, sample2, ... : array_like
        The set of samples.  There must be at least two samples.
        Each sample must be a one-dimensional sequence containing at least
        one value.  The samples are not required to have the same length.
    ties : str, optional
        Determines how values equal to the grand median are classified in
        the contingency table.  The string must be one of::

            "below":
                Values equal to the grand median are counted as "below".
            "above":
                Values equal to the grand median are counted as "above".
            "ignore":
                Values equal to the grand median are not counted.

        The default is "below".
    correction : bool, optional
        If True, *and* there are just two samples, apply Yates' correction
        for continuity when computing the test statistic associated with
        the contingency table.  Default is True.
    lambda_ : float or str, optional
        By default, the statistic computed in this test is Pearson's
        chi-squared statistic.  `lambda_` allows a statistic from the
        Cressie-Read power divergence family to be used instead.  See
        `power_divergence` for details.
        Default is 1 (Pearson's chi-squared statistic).
    nan_policy : {'propagate', 'raise', 'omit'}, optional
        Defines how to handle when input contains nan. 'propagate' returns nan,
        'raise' throws an error, 'omit' performs the calculations ignoring nan
        values. Default is 'propagate'.

    Returns
    -------
    res : MedianTestResult
        An object containing attributes:

        statistic : float
            The test statistic.  The statistic that is returned is determined
            by `lambda_`.  The default is Pearson's chi-squared statistic.
        pvalue : float
            The p-value of the test.
        median : float
            The grand median.
        table : ndarray
            The contingency table.  The shape of the table is (2, n), where
            n is the number of samples.  The first row holds the counts of the
            values above the grand median, and the second row holds the counts
            of the values below the grand median.  The table allows further
            analysis with, for example, `scipy.stats.chi2_contingency`, or with
            `scipy.stats.fisher_exact` if there are two samples, without having
            to recompute the table.  If ``nan_policy`` is "propagate" and there
            are nans in the input, the return value for ``table`` is ``None``.

    See Also
    --------
    kruskal : Compute the Kruskal-Wallis H-test for independent samples.
    mannwhitneyu : Computes the Mann-Whitney rank test on samples x and y.

    Notes
    -----
    .. versionadded:: 0.15.0

    References
    ----------
    .. [1] Mood, A. M., Introduction to the Theory of Statistics. McGraw-Hill
        (1950), pp. 394-399.
    .. [2] Zar, J. H., Biostatistical Analysis, 5th ed. Prentice Hall (2010).
        See Sections 8.12 and 10.15.

    Examples
    --------
    A biologist runs an experiment in which there are three groups of plants.
    Group 1 has 16 plants, group 2 has 15 plants, and group 3 has 17 plants.
    Each plant produces a number of seeds.  The seed counts for each group
    are::

        Group 1: 10 14 14 18 20 22 24 25 31 31 32 39 43 43 48 49
        Group 2: 28 30 31 33 34 35 36 40 44 55 57 61 91 92 99
        Group 3:  0  3  9 22 23 25 25 33 34 34 40 45 46 48 62 67 84

    The following code applies Mood's median test to these samples.

    >>> g1 = [10, 14, 14, 18, 20, 22, 24, 25, 31, 31, 32, 39, 43, 43, 48, 49]
    >>> g2 = [28, 30, 31, 33, 34, 35, 36, 40, 44, 55, 57, 61, 91, 92, 99]
    >>> g3 = [0, 3, 9, 22, 23, 25, 25, 33, 34, 34, 40, 45, 46, 48, 62, 67, 84]
    >>> from scipy.stats import median_test
    >>> res = median_test(g1, g2, g3)

    The median is

    >>> res.median
    34.0

    and the contingency table is

    >>> res.table
    array([[ 5, 10,  7],
           [11,  5, 10]])

    `p` is too large to conclude that the medians are not the same:

    >>> res.pvalue
    0.12609082774093244

    The "G-test" can be performed by passing ``lambda_="log-likelihood"`` to
    `median_test`.

    >>> res = median_test(g1, g2, g3, lambda_="log-likelihood")
    >>> res.pvalue
    0.12224779737117837

    The median occurs several times in the data, so we'll get a different
    result if, for example, ``ties="above"`` is used:

    >>> res = median_test(g1, g2, g3, ties="above")
    >>> res.pvalue
    0.063873276069553273

    >>> res.table
    array([[ 5, 11,  9],
           [11,  4,  8]])

    This example demonstrates that if the data set is not large and there
    are values equal to the median, the p-value can be sensitive to the
    choice of `ties`.

    """
    if len(samples) < 2:
        raise ValueError('median_test requires two or more samples.')

    ties_options = ['below', 'above', 'ignore']
    if ties not in ties_options:
        raise ValueError(f"invalid 'ties' option '{ties}'; 'ties' must be one "
                         f"of: {str(ties_options)[1:-1]}")

    data = [np.asarray(sample) for sample in samples]

    # Validate the sizes and shapes of the arguments.
    for k, d in enumerate(data):
        if d.size == 0:
            raise ValueError("Sample %d is empty. All samples must "
                             "contain at least one value." % (k + 1))
        if d.ndim != 1:
            raise ValueError("Sample %d has %d dimensions.  All "
                             "samples must be one-dimensional sequences." %
                             (k + 1, d.ndim))

    cdata = np.concatenate(data)
    contains_nan, nan_policy = _contains_nan(cdata, nan_policy)
    if contains_nan and nan_policy == 'propagate':
        return MedianTestResult(np.nan, np.nan, np.nan, None)

    if contains_nan:
        grand_median = np.median(cdata[~np.isnan(cdata)])
    else:
        grand_median = np.median(cdata)
    # When the minimum version of numpy supported by scipy is 1.9.0,
    # the above if/else statement can be replaced by the single line:
    #     grand_median = np.nanmedian(cdata)

    # Create the contingency table.
    table = np.zeros((2, len(data)), dtype=np.int64)
    for k, sample in enumerate(data):
        sample = sample[~np.isnan(sample)]

        nabove = count_nonzero(sample > grand_median)
        nbelow = count_nonzero(sample < grand_median)
        nequal = sample.size - (nabove + nbelow)
        table[0, k] += nabove
        table[1, k] += nbelow
        if ties == "below":
            table[1, k] += nequal
        elif ties == "above":
            table[0, k] += nequal

    # Check that no row or column of the table is all zero.
    # Such a table can not be given to chi2_contingency, because it would have
    # a zero in the table of expected frequencies.
    rowsums = table.sum(axis=1)
    if rowsums[0] == 0:
        raise ValueError("All values are below the grand median (%r)." %
                         grand_median)
    if rowsums[1] == 0:
        raise ValueError("All values are above the grand median (%r)." %
                         grand_median)
    if ties == "ignore":
        # We already checked that each sample has at least one value, but it
        # is possible that all those values equal the grand median.  If `ties`
        # is "ignore", that would result in a column of zeros in `table`.  We
        # check for that case here.
        zero_cols = np.nonzero((table == 0).all(axis=0))[0]
        if len(zero_cols) > 0:
            msg = ("All values in sample %d are equal to the grand "
                   "median (%r), so they are ignored, resulting in an "
                   "empty sample." % (zero_cols[0] + 1, grand_median))
            raise ValueError(msg)

    stat, p, dof, expected = chi2_contingency(table, lambda_=lambda_,
                                              correction=correction)
    return MedianTestResult(stat, p, grand_median, table)


def _circfuncs_common(samples, high, low):
    # Ensure samples are array-like and size is not zero
    if samples.size == 0:
        NaN = _get_nan(samples)
        return NaN, NaN, NaN

    # Recast samples as radians that range between 0 and 2 pi and calculate
    # the sine and cosine
    sin_samp = sin((samples - low)*2.*pi / (high - low))
    cos_samp = cos((samples - low)*2.*pi / (high - low))

    return samples, sin_samp, cos_samp


@_axis_nan_policy_factory(
    lambda x: x, n_outputs=1, default_axis=None,
    result_to_tuple=lambda x: (x,)
)
def circmean(samples, high=2*pi, low=0, axis=None, nan_policy='propagate'):
    """Compute the circular mean for samples in a range.

    Parameters
    ----------
    samples : array_like
        Input array.
    high : float or int, optional
        High boundary for the sample range. Default is ``2*pi``.
    low : float or int, optional
        Low boundary for the sample range. Default is 0.

    Returns
    -------
    circmean : float
        Circular mean.

    See Also
    --------
    circstd : Circular standard deviation.
    circvar : Circular variance.

    Examples
    --------
    For simplicity, all angles are printed out in degrees.

    >>> import numpy as np
    >>> from scipy.stats import circmean
    >>> import matplotlib.pyplot as plt
    >>> angles = np.deg2rad(np.array([20, 30, 330]))
    >>> circmean = circmean(angles)
    >>> np.rad2deg(circmean)
    7.294976657784009

    >>> mean = angles.mean()
    >>> np.rad2deg(mean)
    126.66666666666666

    Plot and compare the circular mean against the arithmetic mean.

    >>> plt.plot(np.cos(np.linspace(0, 2*np.pi, 500)),
    ...          np.sin(np.linspace(0, 2*np.pi, 500)),
    ...          c='k')
    >>> plt.scatter(np.cos(angles), np.sin(angles), c='k')
    >>> plt.scatter(np.cos(circmean), np.sin(circmean), c='b',
    ...             label='circmean')
    >>> plt.scatter(np.cos(mean), np.sin(mean), c='r', label='mean')
    >>> plt.legend()
    >>> plt.axis('equal')
    >>> plt.show()

    """
    samples, sin_samp, cos_samp = _circfuncs_common(samples, high, low)
    sin_sum = sin_samp.sum(axis)
    cos_sum = cos_samp.sum(axis)
    res = arctan2(sin_sum, cos_sum)

    res = np.asarray(res)
    res[res < 0] += 2*pi
    res = res[()]

    return res*(high - low)/2.0/pi + low


@_axis_nan_policy_factory(
    lambda x: x, n_outputs=1, default_axis=None,
    result_to_tuple=lambda x: (x,)
)
def circvar(samples, high=2*pi, low=0, axis=None, nan_policy='propagate'):
    """Compute the circular variance for samples assumed to be in a range.

    Parameters
    ----------
    samples : array_like
        Input array.
    high : float or int, optional
        High boundary for the sample range. Default is ``2*pi``.
    low : float or int, optional
        Low boundary for the sample range. Default is 0.

    Returns
    -------
    circvar : float
        Circular variance.

    See Also
    --------
    circmean : Circular mean.
    circstd : Circular standard deviation.

    Notes
    -----
    This uses the following definition of circular variance: ``1-R``, where
    ``R`` is the mean resultant vector. The
    returned value is in the range [0, 1], 0 standing for no variance, and 1
    for a large variance. In the limit of small angles, this value is similar
    to half the 'linear' variance.

    References
    ----------
    .. [1] Fisher, N.I. *Statistical analysis of circular data*. Cambridge
          University Press, 1993.

    Examples
    --------
    >>> import numpy as np
    >>> from scipy.stats import circvar
    >>> import matplotlib.pyplot as plt
    >>> samples_1 = np.array([0.072, -0.158, 0.077, 0.108, 0.286,
    ...                       0.133, -0.473, -0.001, -0.348, 0.131])
    >>> samples_2 = np.array([0.111, -0.879, 0.078, 0.733, 0.421,
    ...                       0.104, -0.136, -0.867,  0.012,  0.105])
    >>> circvar_1 = circvar(samples_1)
    >>> circvar_2 = circvar(samples_2)

    Plot the samples.

    >>> fig, (left, right) = plt.subplots(ncols=2)
    >>> for image in (left, right):
    ...     image.plot(np.cos(np.linspace(0, 2*np.pi, 500)),
    ...                np.sin(np.linspace(0, 2*np.pi, 500)),
    ...                c='k')
    ...     image.axis('equal')
    ...     image.axis('off')
    >>> left.scatter(np.cos(samples_1), np.sin(samples_1), c='k', s=15)
    >>> left.set_title(f"circular variance: {np.round(circvar_1, 2)!r}")
    >>> right.scatter(np.cos(samples_2), np.sin(samples_2), c='k', s=15)
    >>> right.set_title(f"circular variance: {np.round(circvar_2, 2)!r}")
    >>> plt.show()

    """
    samples, sin_samp, cos_samp = _circfuncs_common(samples, high, low)
    sin_mean = sin_samp.mean(axis)
    cos_mean = cos_samp.mean(axis)
    # hypot can go slightly above 1 due to rounding errors
    with np.errstate(invalid='ignore'):
        R = np.minimum(1, hypot(sin_mean, cos_mean))

    res = 1. - R
    return res


@_axis_nan_policy_factory(
    lambda x: x, n_outputs=1, default_axis=None,
    result_to_tuple=lambda x: (x,)
)
def circstd(samples, high=2*pi, low=0, axis=None, nan_policy='propagate', *,
            normalize=False):
    """
    Compute the circular standard deviation for samples assumed to be in the
    range [low to high].

    Parameters
    ----------
    samples : array_like
        Input array.
    high : float or int, optional
        High boundary for the sample range. Default is ``2*pi``.
    low : float or int, optional
        Low boundary for the sample range. Default is 0.
    normalize : boolean, optional
        If True, the returned value is equal to ``sqrt(-2*log(R))`` and does
        not depend on the variable units. If False (default), the returned
        value is scaled by ``((high-low)/(2*pi))``.

    Returns
    -------
    circstd : float
        Circular standard deviation.

    See Also
    --------
    circmean : Circular mean.
    circvar : Circular variance.

    Notes
    -----
    This uses a definition of circular standard deviation from [1]_.
    Essentially, the calculation is as follows.

    .. code-block:: python

        import numpy as np
        C = np.cos(samples).mean()
        S = np.sin(samples).mean()
        R = np.sqrt(C**2 + S**2)
        l = 2*np.pi / (high-low)
        circstd = np.sqrt(-2*np.log(R)) / l

    In the limit of small angles, it returns a number close to the 'linear'
    standard deviation.

    References
    ----------
    .. [1] Mardia, K. V. (1972). 2. In *Statistics of Directional Data*
       (pp. 18-24). Academic Press. :doi:`10.1016/C2013-0-07425-7`.

    Examples
    --------
    >>> import numpy as np
    >>> from scipy.stats import circstd
    >>> import matplotlib.pyplot as plt
    >>> samples_1 = np.array([0.072, -0.158, 0.077, 0.108, 0.286,
    ...                       0.133, -0.473, -0.001, -0.348, 0.131])
    >>> samples_2 = np.array([0.111, -0.879, 0.078, 0.733, 0.421,
    ...                       0.104, -0.136, -0.867,  0.012,  0.105])
    >>> circstd_1 = circstd(samples_1)
    >>> circstd_2 = circstd(samples_2)

    Plot the samples.

    >>> fig, (left, right) = plt.subplots(ncols=2)
    >>> for image in (left, right):
    ...     image.plot(np.cos(np.linspace(0, 2*np.pi, 500)),
    ...                np.sin(np.linspace(0, 2*np.pi, 500)),
    ...                c='k')
    ...     image.axis('equal')
    ...     image.axis('off')
    >>> left.scatter(np.cos(samples_1), np.sin(samples_1), c='k', s=15)
    >>> left.set_title(f"circular std: {np.round(circstd_1, 2)!r}")
    >>> right.plot(np.cos(np.linspace(0, 2*np.pi, 500)),
    ...            np.sin(np.linspace(0, 2*np.pi, 500)),
    ...            c='k')
    >>> right.scatter(np.cos(samples_2), np.sin(samples_2), c='k', s=15)
    >>> right.set_title(f"circular std: {np.round(circstd_2, 2)!r}")
    >>> plt.show()

    """
    samples, sin_samp, cos_samp = _circfuncs_common(samples, high, low)
    sin_mean = sin_samp.mean(axis)  # [1] (2.2.3)
    cos_mean = cos_samp.mean(axis)  # [1] (2.2.3)
    # hypot can go slightly above 1 due to rounding errors
    with np.errstate(invalid='ignore'):
        R = np.minimum(1, hypot(sin_mean, cos_mean))  # [1] (2.2.4)

    res = sqrt(-2*log(R))
    if not normalize:
        res *= (high-low)/(2.*pi)  # [1] (2.3.14) w/ (2.3.7)
    return res


class DirectionalStats:
    def __init__(self, mean_direction, mean_resultant_length):
        self.mean_direction = mean_direction
        self.mean_resultant_length = mean_resultant_length

    def __repr__(self):
        return (f"DirectionalStats(mean_direction={self.mean_direction},"
                f" mean_resultant_length={self.mean_resultant_length})")


def directional_stats(samples, *, axis=0, normalize=True):
    """
    Computes sample statistics for directional data.

    Computes the directional mean (also called the mean direction vector) and
    mean resultant length of a sample of vectors.

    The directional mean is a measure of "preferred direction" of vector data.
    It is analogous to the sample mean, but it is for use when the length of
    the data is irrelevant (e.g. unit vectors).

    The mean resultant length is a value between 0 and 1 used to quantify the
    dispersion of directional data: the smaller the mean resultant length, the
    greater the dispersion. Several definitions of directional variance
    involving the mean resultant length are given in [1]_ and [2]_.

    Parameters
    ----------
    samples : array_like
        Input array. Must be at least two-dimensional, and the last axis of the
        input must correspond with the dimensionality of the vector space.
        When the input is exactly two dimensional, this means that each row
        of the data is a vector observation.
    axis : int, default: 0
        Axis along which the directional mean is computed.
    normalize: boolean, default: True
        If True, normalize the input to ensure that each observation is a
        unit vector. It the observations are already unit vectors, consider
        setting this to False to avoid unnecessary computation.

    Returns
    -------
    res : DirectionalStats
        An object containing attributes:

        mean_direction : ndarray
            Directional mean.
        mean_resultant_length : ndarray
            The mean resultant length [1]_.

    See Also
    --------
    circmean: circular mean; i.e. directional mean for 2D *angles*
    circvar: circular variance; i.e. directional variance for 2D *angles*

    Notes
    -----
    This uses a definition of directional mean from [1]_.
    Assuming the observations are unit vectors, the calculation is as follows.

    .. code-block:: python

        mean = samples.mean(axis=0)
        mean_resultant_length = np.linalg.norm(mean)
        mean_direction = mean / mean_resultant_length

    This definition is appropriate for *directional* data (i.e. vector data
    for which the magnitude of each observation is irrelevant) but not
    for *axial* data (i.e. vector data for which the magnitude and *sign* of
    each observation is irrelevant).

    Several definitions of directional variance involving the mean resultant
    length ``R`` have been proposed, including ``1 - R`` [1]_, ``1 - R**2``
    [2]_, and ``2 * (1 - R)`` [2]_. Rather than choosing one, this function
    returns ``R`` as attribute `mean_resultant_length` so the user can compute
    their preferred measure of dispersion.

    References
    ----------
    .. [1] Mardia, Jupp. (2000). *Directional Statistics*
       (p. 163). Wiley.

    .. [2] https://en.wikipedia.org/wiki/Directional_statistics

    Examples
    --------
    >>> import numpy as np
    >>> from scipy.stats import directional_stats
    >>> data = np.array([[3, 4],    # first observation, 2D vector space
    ...                  [6, -8]])  # second observation
    >>> dirstats = directional_stats(data)
    >>> dirstats.mean_direction
    array([1., 0.])

    In contrast, the regular sample mean of the vectors would be influenced
    by the magnitude of each observation. Furthermore, the result would not be
    a unit vector.

    >>> data.mean(axis=0)
    array([4.5, -2.])

    An exemplary use case for `directional_stats` is to find a *meaningful*
    center for a set of observations on a sphere, e.g. geographical locations.

    >>> data = np.array([[0.8660254, 0.5, 0.],
    ...                  [0.8660254, -0.5, 0.]])
    >>> dirstats = directional_stats(data)
    >>> dirstats.mean_direction
    array([1., 0., 0.])

    The regular sample mean on the other hand yields a result which does not
    lie on the surface of the sphere.

    >>> data.mean(axis=0)
    array([0.8660254, 0., 0.])

    The function also returns the mean resultant length, which
    can be used to calculate a directional variance. For example, using the
    definition ``Var(z) = 1 - R`` from [2]_ where ``R`` is the
    mean resultant length, we can calculate the directional variance of the
    vectors in the above example as:

    >>> 1 - dirstats.mean_resultant_length
    0.13397459716167093
    """
    samples = np.asarray(samples)
    if samples.ndim < 2:
        raise ValueError("samples must at least be two-dimensional. "
                         f"Instead samples has shape: {samples.shape!r}")
    samples = np.moveaxis(samples, axis, 0)
    if normalize:
        vectornorms = np.linalg.norm(samples, axis=-1, keepdims=True)
        samples = samples/vectornorms
    mean = np.mean(samples, axis=0)
    mean_resultant_length = np.linalg.norm(mean, axis=-1, keepdims=True)
    mean_direction = mean / mean_resultant_length
    return DirectionalStats(mean_direction,
                            mean_resultant_length.squeeze(-1)[()])


def false_discovery_control(ps, *, axis=0, method='bh'):
    """Adjust p-values to control the false discovery rate.

    The false discovery rate (FDR) is the expected proportion of rejected null
    hypotheses that are actually true.
    If the null hypothesis is rejected when the *adjusted* p-value falls below
    a specified level, the false discovery rate is controlled at that level.

    Parameters
    ----------
    ps : 1D array_like
        The p-values to adjust. Elements must be real numbers between 0 and 1.
    axis : int
        The axis along which to perform the adjustment. The adjustment is
        performed independently along each axis-slice. If `axis` is None, `ps`
        is raveled before performing the adjustment.
    method : {'bh', 'by'}
        The false discovery rate control procedure to apply: ``'bh'`` is for
        Benjamini-Hochberg [1]_ (Eq. 1), ``'by'`` is for Benjaminini-Yekutieli
        [2]_ (Theorem 1.3). The latter is more conservative, but it is
        guaranteed to control the FDR even when the p-values are not from
        independent tests.

    Returns
    -------
    ps_adusted : array_like
        The adjusted p-values. If the null hypothesis is rejected where these
        fall below a specified level, the false discovery rate is controlled
        at that level.

    See Also
    --------
    combine_pvalues
    statsmodels.stats.multitest.multipletests

    Notes
    -----
    In multiple hypothesis testing, false discovery control procedures tend to
    offer higher power than familywise error rate control procedures (e.g.
    Bonferroni correction [1]_).

    If the p-values correspond with independent tests (or tests with
    "positive regression dependencies" [2]_), rejecting null hypotheses
    corresponding with Benjamini-Hochberg-adjusted p-values below :math:`q`
    controls the false discovery rate at a level less than or equal to
    :math:`q m_0 / m`, where :math:`m_0` is the number of true null hypotheses
    and :math:`m` is the total number of null hypotheses tested. The same is
    true even for dependent tests when the p-values are adjusted accorded to
    the more conservative Benjaminini-Yekutieli procedure.

    The adjusted p-values produced by this function are comparable to those
    produced by the R function ``p.adjust`` and the statsmodels function
    `statsmodels.stats.multitest.multipletests`. Please consider the latter
    for more advanced methods of multiple comparison correction.

    References
    ----------
    .. [1] Benjamini, Yoav, and Yosef Hochberg. "Controlling the false
           discovery rate: a practical and powerful approach to multiple
           testing." Journal of the Royal statistical society: series B
           (Methodological) 57.1 (1995): 289-300.

    .. [2] Benjamini, Yoav, and Daniel Yekutieli. "The control of the false
           discovery rate in multiple testing under dependency." Annals of
           statistics (2001): 1165-1188.

    .. [3] TileStats. FDR - Benjamini-Hochberg explained - Youtube.
           https://www.youtube.com/watch?v=rZKa4tW2NKs.

    .. [4] Neuhaus, Karl-Ludwig, et al. "Improved thrombolysis in acute
           myocardial infarction with front-loaded administration of alteplase:
           results of the rt-PA-APSAC patency study (TAPS)." Journal of the
           American College of Cardiology 19.5 (1992): 885-891.

    Examples
    --------
    We follow the example from [1]_.

        Thrombolysis with recombinant tissue-type plasminogen activator (rt-PA)
        and anisoylated plasminogen streptokinase activator (APSAC) in
        myocardial infarction has been proved to reduce mortality. [4]_
        investigated the effects of a new front-loaded administration of rt-PA
        versus those obtained with a standard regimen of APSAC, in a randomized
        multicentre trial in 421 patients with acute myocardial infarction.

    There were four families of hypotheses tested in the study, the last of
    which was "cardiac and other events after the start of thrombolitic
    treatment". FDR control may be desired in this family of hypotheses
    because it would not be appropriate to conclude that the front-loaded
    treatment is better if it is merely equivalent to the previous treatment.

    The p-values corresponding with the 15 hypotheses in this family were

    >>> ps = [0.0001, 0.0004, 0.0019, 0.0095, 0.0201, 0.0278, 0.0298, 0.0344,
    ...       0.0459, 0.3240, 0.4262, 0.5719, 0.6528, 0.7590, 1.000]

    If the chosen significance level is 0.05, we may be tempted to reject the
    null hypotheses for the tests corresponding with the first nine p-values,
    as the first nine p-values fall below the chosen significance level.
    However, this would ignore the problem of "multiplicity": if we fail to
    correct for the fact that multiple comparisons are being performed, we
    are more likely to incorrectly reject true null hypotheses.

    One approach to the multiplicity problem is to control the family-wise
    error rate (FWER), that is, the rate at which the null hypothesis is
    rejected when it is actually true. A common procedure of this kind is the
    Bonferroni correction [1]_.  We begin by multiplying the p-values by the
    number of hypotheses tested.

    >>> import numpy as np
    >>> np.array(ps) * len(ps)
    array([1.5000e-03, 6.0000e-03, 2.8500e-02, 1.4250e-01, 3.0150e-01,
           4.1700e-01, 4.4700e-01, 5.1600e-01, 6.8850e-01, 4.8600e+00,
           6.3930e+00, 8.5785e+00, 9.7920e+00, 1.1385e+01, 1.5000e+01])

    To control the FWER at 5%, we reject only the hypotheses corresponding
    with adjusted p-values less than 0.05. In this case, only the hypotheses
    corresponding with the first three p-values can be rejected. According to
    [1]_, these three hypotheses concerned "allergic reaction" and "two
    different aspects of bleeding."

    An alternative approach is to control the false discovery rate: the
    expected fraction of rejected null hypotheses that are actually true. The
    advantage of this approach is that it typically affords greater power: an
    increased rate of rejecting the null hypothesis when it is indeed false. To
    control the false discovery rate at 5%, we apply the Benjamini-Hochberg
    p-value adjustment.

    >>> from scipy import stats
    >>> stats.false_discovery_control(ps)
    array([0.0015    , 0.003     , 0.0095    , 0.035625  , 0.0603    ,
           0.06385714, 0.06385714, 0.0645    , 0.0765    , 0.486     ,
           0.58118182, 0.714875  , 0.75323077, 0.81321429, 1.        ])

    Now, the first *four* adjusted p-values fall below 0.05, so we would reject
    the null hypotheses corresponding with these *four* p-values. Rejection
    of the fourth null hypothesis was particularly important to the original
    study as it led to the conclusion that the new treatment had a
    "substantially lower in-hospital mortality rate."

    """
    # Input Validation and Special Cases
    ps = np.asarray(ps)

    ps_in_range = (np.issubdtype(ps.dtype, np.number)
                   and np.all(ps == np.clip(ps, 0, 1)))
    if not ps_in_range:
        raise ValueError("`ps` must include only numbers between 0 and 1.")

    methods = {'bh', 'by'}
    if method.lower() not in methods:
        raise ValueError(f"Unrecognized `method` '{method}'."
                         f"Method must be one of {methods}.")
    method = method.lower()

    if axis is None:
        axis = 0
        ps = ps.ravel()

    axis = np.asarray(axis)[()]
    if not np.issubdtype(axis.dtype, np.integer) or axis.size != 1:
        raise ValueError("`axis` must be an integer or `None`")

    if ps.size <= 1 or ps.shape[axis] <= 1:
        return ps[()]

    ps = np.moveaxis(ps, axis, -1)
    m = ps.shape[-1]

    # Main Algorithm
    # Equivalent to the ideas of [1] and [2], except that this adjusts the
    # p-values as described in [3]. The results are similar to those produced
    # by R's p.adjust.

    # "Let [ps] be the ordered observed p-values..."
    order = np.argsort(ps, axis=-1)
    ps = np.take_along_axis(ps, order, axis=-1)  # this copies ps

    # Equation 1 of [1] rearranged to reject when p is less than specified q
    i = np.arange(1, m+1)
    ps *= m / i

    # Theorem 1.3 of [2]
    if method == 'by':
        ps *= np.sum(1 / i)

    # accounts for rejecting all null hypotheses i for i < k, where k is
    # defined in Eq. 1 of either [1] or [2]. See [3]. Starting with the index j
    # of the second to last element, we replace element j with element j+1 if
    # the latter is smaller.
    np.minimum.accumulate(ps[..., ::-1], out=ps[..., ::-1], axis=-1)

    # Restore original order of axes and data
    np.put_along_axis(ps, order, values=ps.copy(), axis=-1)
    ps = np.moveaxis(ps, -1, axis)

    return np.clip(ps, 0, 1)