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

from collections import namedtuple
from dataclasses import dataclass
from math import comb
import numpy as np
import warnings
from itertools import combinations
import scipy.stats
from scipy.optimize import shgo
from . import distributions
from ._common import ConfidenceInterval
from ._continuous_distns import chi2, norm
from scipy.special import gamma, kv, gammaln
from scipy.fft import ifft
from ._stats_pythran import _a_ij_Aij_Dij2
from ._stats_pythran import (
    _concordant_pairs as _P, _discordant_pairs as _Q
)
from ._axis_nan_policy import _axis_nan_policy_factory
from scipy.stats import _stats_py

__all__ = ['epps_singleton_2samp', 'cramervonmises', 'somersd',
           'barnard_exact', 'boschloo_exact', 'cramervonmises_2samp',
           'tukey_hsd', 'poisson_means_test']

Epps_Singleton_2sampResult = namedtuple('Epps_Singleton_2sampResult',
                                        ('statistic', 'pvalue'))


@_axis_nan_policy_factory(Epps_Singleton_2sampResult, n_samples=2, too_small=4)
def epps_singleton_2samp(x, y, t=(0.4, 0.8)):
    """Compute the Epps-Singleton (ES) test statistic.

    Test the null hypothesis that two samples have the same underlying
    probability distribution.

    Parameters
    ----------
    x, y : array-like
        The two samples of observations to be tested. Input must not have more
        than one dimension. Samples can have different lengths.
    t : array-like, optional
        The points (t1, ..., tn) where the empirical characteristic function is
        to be evaluated. It should be positive distinct numbers. The default
        value (0.4, 0.8) is proposed in [1]_. Input must not have more than
        one dimension.

    Returns
    -------
    statistic : float
        The test statistic.
    pvalue : float
        The associated p-value based on the asymptotic chi2-distribution.

    See Also
    --------
    ks_2samp, anderson_ksamp

    Notes
    -----
    Testing whether two samples are generated by the same underlying
    distribution is a classical question in statistics. A widely used test is
    the Kolmogorov-Smirnov (KS) test which relies on the empirical
    distribution function. Epps and Singleton introduce a test based on the
    empirical characteristic function in [1]_.

    One advantage of the ES test compared to the KS test is that is does
    not assume a continuous distribution. In [1]_, the authors conclude
    that the test also has a higher power than the KS test in many
    examples. They recommend the use of the ES test for discrete samples as
    well as continuous samples with at least 25 observations each, whereas
    `anderson_ksamp` is recommended for smaller sample sizes in the
    continuous case.

    The p-value is computed from the asymptotic distribution of the test
    statistic which follows a `chi2` distribution. If the sample size of both
    `x` and `y` is below 25, the small sample correction proposed in [1]_ is
    applied to the test statistic.

    The default values of `t` are determined in [1]_ by considering
    various distributions and finding good values that lead to a high power
    of the test in general. Table III in [1]_ gives the optimal values for
    the distributions tested in that study. The values of `t` are scaled by
    the semi-interquartile range in the implementation, see [1]_.

    References
    ----------
    .. [1] T. W. Epps and K. J. Singleton, "An omnibus test for the two-sample
       problem using the empirical characteristic function", Journal of
       Statistical Computation and Simulation 26, p. 177--203, 1986.

    .. [2] S. J. Goerg and J. Kaiser, "Nonparametric testing of distributions
       - the Epps-Singleton two-sample test using the empirical characteristic
       function", The Stata Journal 9(3), p. 454--465, 2009.

    """
    # x and y are converted to arrays by the decorator
    t = np.asarray(t)
    # check if x and y are valid inputs
    nx, ny = len(x), len(y)
    if (nx < 5) or (ny < 5):
        raise ValueError('x and y should have at least 5 elements, but len(x) '
                         f'= {nx} and len(y) = {ny}.')
    if not np.isfinite(x).all():
        raise ValueError('x must not contain nonfinite values.')
    if not np.isfinite(y).all():
        raise ValueError('y must not contain nonfinite values.')
    n = nx + ny

    # check if t is valid
    if t.ndim > 1:
        raise ValueError(f't must be 1d, but t.ndim equals {t.ndim}.')
    if np.less_equal(t, 0).any():
        raise ValueError('t must contain positive elements only.')

    # rescale t with semi-iqr as proposed in [1]; import iqr here to avoid
    # circular import
    from scipy.stats import iqr
    sigma = iqr(np.hstack((x, y))) / 2
    ts = np.reshape(t, (-1, 1)) / sigma

    # covariance estimation of ES test
    gx = np.vstack((np.cos(ts*x), np.sin(ts*x))).T  # shape = (nx, 2*len(t))
    gy = np.vstack((np.cos(ts*y), np.sin(ts*y))).T
    cov_x = np.cov(gx.T, bias=True)  # the test uses biased cov-estimate
    cov_y = np.cov(gy.T, bias=True)
    est_cov = (n/nx)*cov_x + (n/ny)*cov_y
    est_cov_inv = np.linalg.pinv(est_cov)
    r = np.linalg.matrix_rank(est_cov_inv)
    if r < 2*len(t):
        warnings.warn('Estimated covariance matrix does not have full rank. '
                      'This indicates a bad choice of the input t and the '
                      'test might not be consistent.', # see p. 183 in [1]_
                      stacklevel=2)

    # compute test statistic w distributed asympt. as chisquare with df=r
    g_diff = np.mean(gx, axis=0) - np.mean(gy, axis=0)
    w = n*np.dot(g_diff.T, np.dot(est_cov_inv, g_diff))

    # apply small-sample correction
    if (max(nx, ny) < 25):
        corr = 1.0/(1.0 + n**(-0.45) + 10.1*(nx**(-1.7) + ny**(-1.7)))
        w = corr * w

    p = chi2.sf(w, r)

    return Epps_Singleton_2sampResult(w, p)


def poisson_means_test(k1, n1, k2, n2, *, diff=0, alternative='two-sided'):
    r"""
    Performs the Poisson means test, AKA the "E-test".

    This is a test of the null hypothesis that the difference between means of
    two Poisson distributions is `diff`. The samples are provided as the
    number of events `k1` and `k2` observed within measurement intervals
    (e.g. of time, space, number of observations) of sizes `n1` and `n2`.

    Parameters
    ----------
    k1 : int
        Number of events observed from distribution 1.
    n1: float
        Size of sample from distribution 1.
    k2 : int
        Number of events observed from distribution 2.
    n2 : float
        Size of sample from distribution 2.
    diff : float, default=0
        The hypothesized difference in means between the distributions
        underlying the samples.
    alternative : {'two-sided', 'less', 'greater'}, optional
        Defines the alternative hypothesis.
        The following options are available (default is 'two-sided'):

          * 'two-sided': the difference between distribution means is not
            equal to `diff`
          * 'less': the difference between distribution means is less than
            `diff`
          * 'greater': the difference between distribution means is greater
            than `diff`

    Returns
    -------
    statistic : float
        The test statistic (see [1]_ equation 3.3).
    pvalue : float
        The probability of achieving such an extreme value of the test
        statistic under the null hypothesis.

    Notes
    -----

    Let:

    .. math:: X_1 \sim \mbox{Poisson}(\mathtt{n1}\lambda_1)

    be a random variable independent of

    .. math:: X_2  \sim \mbox{Poisson}(\mathtt{n2}\lambda_2)

    and let ``k1`` and ``k2`` be the observed values of :math:`X_1`
    and :math:`X_2`, respectively. Then `poisson_means_test` uses the number
    of observed events ``k1`` and ``k2`` from samples of size ``n1`` and
    ``n2``, respectively, to test the null hypothesis that

    .. math::
       H_0: \lambda_1 - \lambda_2 = \mathtt{diff}

    A benefit of the E-test is that it has good power for small sample sizes,
    which can reduce sampling costs [1]_. It has been evaluated and determined
    to be more powerful than the comparable C-test, sometimes referred to as
    the Poisson exact test.

    References
    ----------
    .. [1]  Krishnamoorthy, K., & Thomson, J. (2004). A more powerful test for
       comparing two Poisson means. Journal of Statistical Planning and
       Inference, 119(1), 23-35.

    .. [2]  Przyborowski, J., & Wilenski, H. (1940). Homogeneity of results in
       testing samples from Poisson series: With an application to testing
       clover seed for dodder. Biometrika, 31(3/4), 313-323.

    Examples
    --------

    Suppose that a gardener wishes to test the number of dodder (weed) seeds
    in a sack of clover seeds that they buy from a seed company. It has
    previously been established that the number of dodder seeds in clover
    follows the Poisson distribution.

    A 100 gram sample is drawn from the sack before being shipped to the
    gardener. The sample is analyzed, and it is found to contain no dodder
    seeds; that is, `k1` is 0. However, upon arrival, the gardener draws
    another 100 gram sample from the sack. This time, three dodder seeds are
    found in the sample; that is, `k2` is 3. The gardener would like to
    know if the difference is significant and not due to chance. The
    null hypothesis is that the difference between the two samples is merely
    due to chance, or that :math:`\lambda_1 - \lambda_2 = \mathtt{diff}`
    where :math:`\mathtt{diff} = 0`. The alternative hypothesis is that the
    difference is not due to chance, or :math:`\lambda_1 - \lambda_2 \ne 0`.
    The gardener selects a significance level of 5% to reject the null
    hypothesis in favor of the alternative [2]_.

    >>> import scipy.stats as stats
    >>> res = stats.poisson_means_test(0, 100, 3, 100)
    >>> res.statistic, res.pvalue
    (-1.7320508075688772, 0.08837900929018157)

    The p-value is .088, indicating a near 9% chance of observing a value of
    the test statistic under the null hypothesis. This exceeds 5%, so the
    gardener does not reject the null hypothesis as the difference cannot be
    regarded as significant at this level.
    """

    _poisson_means_test_iv(k1, n1, k2, n2, diff, alternative)

    # "for a given k_1 and k_2, an estimate of \lambda_2 is given by" [1] (3.4)
    lmbd_hat2 = ((k1 + k2) / (n1 + n2) - diff * n1 / (n1 + n2))

    # "\hat{\lambda_{2k}} may be less than or equal to zero ... and in this
    # case the null hypothesis cannot be rejected ... [and] it is not necessary
    # to compute the p-value". [1] page 26 below eq. (3.6).
    if lmbd_hat2 <= 0:
        return _stats_py.SignificanceResult(0, 1)

    # The unbiased variance estimate [1] (3.2)
    var = k1 / (n1 ** 2) + k2 / (n2 ** 2)

    # The _observed_ pivot statistic from the input. It follows the
    # unnumbered equation following equation (3.3) This is used later in
    # comparison with the computed pivot statistics in an indicator function.
    t_k1k2 = (k1 / n1 - k2 / n2 - diff) / np.sqrt(var)

    # Equation (3.5) of [1] is lengthy, so it is broken into several parts,
    # beginning here. Note that the probability mass function of poisson is
    # exp^(-\mu)*\mu^k/k!, so and this is called with shape \mu, here noted
    # here as nlmbd_hat*. The strategy for evaluating the double summation in
    # (3.5) is to create two arrays of the values of the two products inside
    # the summation and then broadcast them together into a matrix, and then
    # sum across the entire matrix.

    # Compute constants (as seen in the first and second separated products in
    # (3.5).). (This is the shape (\mu) parameter of the poisson distribution.)
    nlmbd_hat1 = n1 * (lmbd_hat2 + diff)
    nlmbd_hat2 = n2 * lmbd_hat2

    # Determine summation bounds for tail ends of distribution rather than
    # summing to infinity. `x1*` is for the outer sum and `x2*` is the inner
    # sum.
    x1_lb, x1_ub = distributions.poisson.ppf([1e-10, 1 - 1e-16], nlmbd_hat1)
    x2_lb, x2_ub = distributions.poisson.ppf([1e-10, 1 - 1e-16], nlmbd_hat2)

    # Construct arrays to function as the x_1 and x_2 counters on the summation
    # in (3.5). `x1` is in columns and `x2` is in rows to allow for
    # broadcasting.
    x1 = np.arange(x1_lb, x1_ub + 1)
    x2 = np.arange(x2_lb, x2_ub + 1)[:, None]

    # These are the two products in equation (3.5) with `prob_x1` being the
    # first (left side) and `prob_x2` being the second (right side). (To
    # make as clear as possible: the 1st contains a "+ d" term, the 2nd does
    # not.)
    prob_x1 = distributions.poisson.pmf(x1, nlmbd_hat1)
    prob_x2 = distributions.poisson.pmf(x2, nlmbd_hat2)

    # compute constants for use in the "pivot statistic" per the
    # unnumbered equation following (3.3).
    lmbd_x1 = x1 / n1
    lmbd_x2 = x2 / n2
    lmbds_diff = lmbd_x1 - lmbd_x2 - diff
    var_x1x2 = lmbd_x1 / n1 + lmbd_x2 / n2

    # This is the 'pivot statistic' for use in the indicator of the summation
    # (left side of "I[.]").
    with np.errstate(invalid='ignore', divide='ignore'):
        t_x1x2 = lmbds_diff / np.sqrt(var_x1x2)

    # `[indicator]` implements the "I[.] ... the indicator function" per
    # the paragraph following equation (3.5).
    if alternative == 'two-sided':
        indicator = np.abs(t_x1x2) >= np.abs(t_k1k2)
    elif alternative == 'less':
        indicator = t_x1x2 <= t_k1k2
    else:
        indicator = t_x1x2 >= t_k1k2

    # Multiply all combinations of the products together, exclude terms
    # based on the `indicator` and then sum. (3.5)
    pvalue = np.sum((prob_x1 * prob_x2)[indicator])
    return _stats_py.SignificanceResult(t_k1k2, pvalue)


def _poisson_means_test_iv(k1, n1, k2, n2, diff, alternative):
    # """check for valid types and values of input to `poisson_mean_test`."""
    if k1 != int(k1) or k2 != int(k2):
        raise TypeError('`k1` and `k2` must be integers.')

    count_err = '`k1` and `k2` must be greater than or equal to 0.'
    if k1 < 0 or k2 < 0:
        raise ValueError(count_err)

    if n1 <= 0 or n2 <= 0:
        raise ValueError('`n1` and `n2` must be greater than 0.')

    if diff < 0:
        raise ValueError('diff must be greater than or equal to 0.')

    alternatives = {'two-sided', 'less', 'greater'}
    if alternative.lower() not in alternatives:
        raise ValueError(f"Alternative must be one of '{alternatives}'.")


class CramerVonMisesResult:
    def __init__(self, statistic, pvalue):
        self.statistic = statistic
        self.pvalue = pvalue

    def __repr__(self):
        return (f"{self.__class__.__name__}(statistic={self.statistic}, "
                f"pvalue={self.pvalue})")


def _psi1_mod(x):
    """
    psi1 is defined in equation 1.10 in Csörgő, S. and Faraway, J. (1996).
    This implements a modified version by excluding the term V(x) / 12
    (here: _cdf_cvm_inf(x) / 12) to avoid evaluating _cdf_cvm_inf(x)
    twice in _cdf_cvm.

    Implementation based on MAPLE code of Julian Faraway and R code of the
    function pCvM in the package goftest (v1.1.1), permission granted
    by Adrian Baddeley. Main difference in the implementation: the code
    here keeps adding terms of the series until the terms are small enough.
    """

    def _ed2(y):
        z = y**2 / 4
        b = kv(1/4, z) + kv(3/4, z)
        return np.exp(-z) * (y/2)**(3/2) * b / np.sqrt(np.pi)

    def _ed3(y):
        z = y**2 / 4
        c = np.exp(-z) / np.sqrt(np.pi)
        return c * (y/2)**(5/2) * (2*kv(1/4, z) + 3*kv(3/4, z) - kv(5/4, z))

    def _Ak(k, x):
        m = 2*k + 1
        sx = 2 * np.sqrt(x)
        y1 = x**(3/4)
        y2 = x**(5/4)

        e1 = m * gamma(k + 1/2) * _ed2((4 * k + 3)/sx) / (9 * y1)
        e2 = gamma(k + 1/2) * _ed3((4 * k + 1) / sx) / (72 * y2)
        e3 = 2 * (m + 2) * gamma(k + 3/2) * _ed3((4 * k + 5) / sx) / (12 * y2)
        e4 = 7 * m * gamma(k + 1/2) * _ed2((4 * k + 1) / sx) / (144 * y1)
        e5 = 7 * m * gamma(k + 1/2) * _ed2((4 * k + 5) / sx) / (144 * y1)

        return e1 + e2 + e3 + e4 + e5

    x = np.asarray(x)
    tot = np.zeros_like(x, dtype='float')
    cond = np.ones_like(x, dtype='bool')
    k = 0
    while np.any(cond):
        z = -_Ak(k, x[cond]) / (np.pi * gamma(k + 1))
        tot[cond] = tot[cond] + z
        cond[cond] = np.abs(z) >= 1e-7
        k += 1

    return tot


def _cdf_cvm_inf(x):
    """
    Calculate the cdf of the Cramér-von Mises statistic (infinite sample size).

    See equation 1.2 in Csörgő, S. and Faraway, J. (1996).

    Implementation based on MAPLE code of Julian Faraway and R code of the
    function pCvM in the package goftest (v1.1.1), permission granted
    by Adrian Baddeley. Main difference in the implementation: the code
    here keeps adding terms of the series until the terms are small enough.

    The function is not expected to be accurate for large values of x, say
    x > 4, when the cdf is very close to 1.
    """
    x = np.asarray(x)

    def term(x, k):
        # this expression can be found in [2], second line of (1.3)
        u = np.exp(gammaln(k + 0.5) - gammaln(k+1)) / (np.pi**1.5 * np.sqrt(x))
        y = 4*k + 1
        q = y**2 / (16*x)
        b = kv(0.25, q)
        return u * np.sqrt(y) * np.exp(-q) * b

    tot = np.zeros_like(x, dtype='float')
    cond = np.ones_like(x, dtype='bool')
    k = 0
    while np.any(cond):
        z = term(x[cond], k)
        tot[cond] = tot[cond] + z
        cond[cond] = np.abs(z) >= 1e-7
        k += 1

    return tot


def _cdf_cvm(x, n=None):
    """
    Calculate the cdf of the Cramér-von Mises statistic for a finite sample
    size n. If N is None, use the asymptotic cdf (n=inf).

    See equation 1.8 in Csörgő, S. and Faraway, J. (1996) for finite samples,
    1.2 for the asymptotic cdf.

    The function is not expected to be accurate for large values of x, say
    x > 2, when the cdf is very close to 1 and it might return values > 1
    in that case, e.g. _cdf_cvm(2.0, 12) = 1.0000027556716846. Moreover, it
    is not accurate for small values of n, especially close to the bounds of
    the distribution's domain, [1/(12*n), n/3], where the value jumps to 0
    and 1, respectively. These are limitations of the approximation by Csörgő
    and Faraway (1996) implemented in this function.
    """
    x = np.asarray(x)
    if n is None:
        y = _cdf_cvm_inf(x)
    else:
        # support of the test statistic is [12/n, n/3], see 1.1 in [2]
        y = np.zeros_like(x, dtype='float')
        sup = (1./(12*n) < x) & (x < n/3.)
        # note: _psi1_mod does not include the term _cdf_cvm_inf(x) / 12
        # therefore, we need to add it here
        y[sup] = _cdf_cvm_inf(x[sup]) * (1 + 1./(12*n)) + _psi1_mod(x[sup]) / n
        y[x >= n/3] = 1

    if y.ndim == 0:
        return y[()]
    return y


def _cvm_result_to_tuple(res):
    return res.statistic, res.pvalue


@_axis_nan_policy_factory(CramerVonMisesResult, n_samples=1, too_small=1,
                          result_to_tuple=_cvm_result_to_tuple)
def cramervonmises(rvs, cdf, args=()):
    """Perform the one-sample Cramér-von Mises test for goodness of fit.

    This performs a test of the goodness of fit of a cumulative distribution
    function (cdf) :math:`F` compared to the empirical distribution function
    :math:`F_n` of observed random variates :math:`X_1, ..., X_n` that are
    assumed to be independent and identically distributed ([1]_).
    The null hypothesis is that the :math:`X_i` have cumulative distribution
    :math:`F`.

    Parameters
    ----------
    rvs : array_like
        A 1-D array of observed values of the random variables :math:`X_i`.
    cdf : str or callable
        The cumulative distribution function :math:`F` to test the
        observations against. If a string, it should be the name of a
        distribution in `scipy.stats`. If a callable, that callable is used
        to calculate the cdf: ``cdf(x, *args) -> float``.
    args : tuple, optional
        Distribution parameters. These are assumed to be known; see Notes.

    Returns
    -------
    res : object with attributes
        statistic : float
            Cramér-von Mises statistic.
        pvalue : float
            The p-value.

    See Also
    --------
    kstest, cramervonmises_2samp

    Notes
    -----
    .. versionadded:: 1.6.0

    The p-value relies on the approximation given by equation 1.8 in [2]_.
    It is important to keep in mind that the p-value is only accurate if
    one tests a simple hypothesis, i.e. the parameters of the reference
    distribution are known. If the parameters are estimated from the data
    (composite hypothesis), the computed p-value is not reliable.

    References
    ----------
    .. [1] Cramér-von Mises criterion, Wikipedia,
           https://en.wikipedia.org/wiki/Cram%C3%A9r%E2%80%93von_Mises_criterion
    .. [2] Csörgő, S. and Faraway, J. (1996). The Exact and Asymptotic
           Distribution of Cramér-von Mises Statistics. Journal of the
           Royal Statistical Society, pp. 221-234.

    Examples
    --------

    Suppose we wish to test whether data generated by ``scipy.stats.norm.rvs``
    were, in fact, drawn from the standard normal distribution. We choose a
    significance level of ``alpha=0.05``.

    >>> import numpy as np
    >>> from scipy import stats
    >>> rng = np.random.default_rng(165417232101553420507139617764912913465)
    >>> x = stats.norm.rvs(size=500, random_state=rng)
    >>> res = stats.cramervonmises(x, 'norm')
    >>> res.statistic, res.pvalue
    (0.1072085112565724, 0.5508482238203407)

    The p-value exceeds our chosen significance level, so we do not
    reject the null hypothesis that the observed sample is drawn from the
    standard normal distribution.

    Now suppose we wish to check whether the same samples shifted by 2.1 is
    consistent with being drawn from a normal distribution with a mean of 2.

    >>> y = x + 2.1
    >>> res = stats.cramervonmises(y, 'norm', args=(2,))
    >>> res.statistic, res.pvalue
    (0.8364446265294695, 0.00596286797008283)

    Here we have used the `args` keyword to specify the mean (``loc``)
    of the normal distribution to test the data against. This is equivalent
    to the following, in which we create a frozen normal distribution with
    mean 2.1, then pass its ``cdf`` method as an argument.

    >>> frozen_dist = stats.norm(loc=2)
    >>> res = stats.cramervonmises(y, frozen_dist.cdf)
    >>> res.statistic, res.pvalue
    (0.8364446265294695, 0.00596286797008283)

    In either case, we would reject the null hypothesis that the observed
    sample is drawn from a normal distribution with a mean of 2 (and default
    variance of 1) because the p-value is less than our chosen
    significance level.

    """
    if isinstance(cdf, str):
        cdf = getattr(distributions, cdf).cdf

    vals = np.sort(np.asarray(rvs))

    if vals.size <= 1:
        raise ValueError('The sample must contain at least two observations.')

    n = len(vals)
    cdfvals = cdf(vals, *args)

    u = (2*np.arange(1, n+1) - 1)/(2*n)
    w = 1/(12*n) + np.sum((u - cdfvals)**2)

    # avoid small negative values that can occur due to the approximation
    p = max(0, 1. - _cdf_cvm(w, n))

    return CramerVonMisesResult(statistic=w, pvalue=p)


def _get_wilcoxon_distr(n):
    """
    Distribution of probability of the Wilcoxon ranksum statistic r_plus (sum
    of ranks of positive differences).
    Returns an array with the probabilities of all the possible ranks
    r = 0, ..., n*(n+1)/2
    """
    c = np.ones(1, dtype=np.float64)
    for k in range(1, n + 1):
        prev_c = c
        c = np.zeros(k * (k + 1) // 2 + 1, dtype=np.float64)
        m = len(prev_c)
        c[:m] = prev_c * 0.5
        c[-m:] += prev_c * 0.5
    return c


def _get_wilcoxon_distr2(n):
    """
    Distribution of probability of the Wilcoxon ranksum statistic r_plus (sum
    of ranks of positive differences).
    Returns an array with the probabilities of all the possible ranks
    r = 0, ..., n*(n+1)/2
    This is a slower reference function
    References
    ----------
    .. [1] 1. Harris T, Hardin JW. Exact Wilcoxon Signed-Rank and Wilcoxon
        Mann-Whitney Ranksum Tests. The Stata Journal. 2013;13(2):337-343.
    """
    ai = np.arange(1, n+1)[:, None]
    t = n*(n+1)/2
    q = 2*t
    j = np.arange(q)
    theta = 2*np.pi/q*j
    phi_sp = np.prod(np.cos(theta*ai), axis=0)
    phi_s = np.exp(1j*theta*t) * phi_sp
    p = np.real(ifft(phi_s))
    res = np.zeros(int(t)+1)
    res[:-1:] = p[::2]
    res[0] /= 2
    res[-1] = res[0]
    return res


def _tau_b(A):
    """Calculate Kendall's tau-b and p-value from contingency table."""
    # See [2] 2.2 and 4.2

    # contingency table must be truly 2D
    if A.shape[0] == 1 or A.shape[1] == 1:
        return np.nan, np.nan

    NA = A.sum()
    PA = _P(A)
    QA = _Q(A)
    Sri2 = (A.sum(axis=1)**2).sum()
    Scj2 = (A.sum(axis=0)**2).sum()
    denominator = (NA**2 - Sri2)*(NA**2 - Scj2)

    tau = (PA-QA)/(denominator)**0.5

    numerator = 4*(_a_ij_Aij_Dij2(A) - (PA - QA)**2 / NA)
    s02_tau_b = numerator/denominator
    if s02_tau_b == 0:  # Avoid divide by zero
        return tau, 0
    Z = tau/s02_tau_b**0.5
    p = 2*norm.sf(abs(Z))  # 2-sided p-value

    return tau, p


def _somers_d(A, alternative='two-sided'):
    """Calculate Somers' D and p-value from contingency table."""
    # See [3] page 1740

    # contingency table must be truly 2D
    if A.shape[0] <= 1 or A.shape[1] <= 1:
        return np.nan, np.nan

    NA = A.sum()
    NA2 = NA**2
    PA = _P(A)
    QA = _Q(A)
    Sri2 = (A.sum(axis=1)**2).sum()

    d = (PA - QA)/(NA2 - Sri2)

    S = _a_ij_Aij_Dij2(A) - (PA-QA)**2/NA

    with np.errstate(divide='ignore'):
        Z = (PA - QA)/(4*(S))**0.5

    p = scipy.stats._stats_py._get_pvalue(Z, distributions.norm, alternative)

    return d, p


@dataclass
class SomersDResult:
    statistic: float
    pvalue: float
    table: np.ndarray


def somersd(x, y=None, alternative='two-sided'):
    r"""Calculates Somers' D, an asymmetric measure of ordinal association.

    Like Kendall's :math:`\tau`, Somers' :math:`D` is a measure of the
    correspondence between two rankings. Both statistics consider the
    difference between the number of concordant and discordant pairs in two
    rankings :math:`X` and :math:`Y`, and both are normalized such that values
    close  to 1 indicate strong agreement and values close to -1 indicate
    strong disagreement. They differ in how they are normalized. To show the
    relationship, Somers' :math:`D` can be defined in terms of Kendall's
    :math:`\tau_a`:

    .. math::
        D(Y|X) = \frac{\tau_a(X, Y)}{\tau_a(X, X)}

    Suppose the first ranking :math:`X` has :math:`r` distinct ranks and the
    second ranking :math:`Y` has :math:`s` distinct ranks. These two lists of
    :math:`n` rankings can also be viewed as an :math:`r \times s` contingency
    table in which element :math:`i, j` is the number of rank pairs with rank
    :math:`i` in ranking :math:`X` and rank :math:`j` in ranking :math:`Y`.
    Accordingly, `somersd` also allows the input data to be supplied as a
    single, 2D contingency table instead of as two separate, 1D rankings.

    Note that the definition of Somers' :math:`D` is asymmetric: in general,
    :math:`D(Y|X) \neq D(X|Y)`. ``somersd(x, y)`` calculates Somers'
    :math:`D(Y|X)`: the "row" variable :math:`X` is treated as an independent
    variable, and the "column" variable :math:`Y` is dependent. For Somers'
    :math:`D(X|Y)`, swap the input lists or transpose the input table.

    Parameters
    ----------
    x : array_like
        1D array of rankings, treated as the (row) independent variable.
        Alternatively, a 2D contingency table.
    y : array_like, optional
        If `x` is a 1D array of rankings, `y` is a 1D array of rankings of the
        same length, treated as the (column) dependent variable.
        If `x` is 2D, `y` is ignored.
    alternative : {'two-sided', 'less', 'greater'}, optional
        Defines the alternative hypothesis. Default is 'two-sided'.
        The following options are available:
        * 'two-sided': the rank correlation is nonzero
        * 'less': the rank correlation is negative (less than zero)
        * 'greater':  the rank correlation is positive (greater than zero)

    Returns
    -------
    res : SomersDResult
        A `SomersDResult` object with the following fields:

            statistic : float
               The Somers' :math:`D` statistic.
            pvalue : float
               The p-value for a hypothesis test whose null
               hypothesis is an absence of association, :math:`D=0`.
               See notes for more information.
            table : 2D array
               The contingency table formed from rankings `x` and `y` (or the
               provided contingency table, if `x` is a 2D array)

    See Also
    --------
    kendalltau : Calculates Kendall's tau, another correlation measure.
    weightedtau : Computes a weighted version of Kendall's tau.
    spearmanr : Calculates a Spearman rank-order correlation coefficient.
    pearsonr : Calculates a Pearson correlation coefficient.

    Notes
    -----
    This function follows the contingency table approach of [2]_ and
    [3]_. *p*-values are computed based on an asymptotic approximation of
    the test statistic distribution under the null hypothesis :math:`D=0`.

    Theoretically, hypothesis tests based on Kendall's :math:`tau` and Somers'
    :math:`D` should be identical.
    However, the *p*-values returned by `kendalltau` are based
    on the null hypothesis of *independence* between :math:`X` and :math:`Y`
    (i.e. the population from which pairs in :math:`X` and :math:`Y` are
    sampled contains equal numbers of all possible pairs), which is more
    specific than the null hypothesis :math:`D=0` used here. If the null
    hypothesis of independence is desired, it is acceptable to use the
    *p*-value returned by `kendalltau` with the statistic returned by
    `somersd` and vice versa. For more information, see [2]_.

    Contingency tables are formatted according to the convention used by
    SAS and R: the first ranking supplied (``x``) is the "row" variable, and
    the second ranking supplied (``y``) is the "column" variable. This is
    opposite the convention of Somers' original paper [1]_.

    References
    ----------
    .. [1] Robert H. Somers, "A New Asymmetric Measure of Association for
           Ordinal Variables", *American Sociological Review*, Vol. 27, No. 6,
           pp. 799--811, 1962.

    .. [2] Morton B. Brown and Jacqueline K. Benedetti, "Sampling Behavior of
           Tests for Correlation in Two-Way Contingency Tables", *Journal of
           the American Statistical Association* Vol. 72, No. 358, pp.
           309--315, 1977.

    .. [3] SAS Institute, Inc., "The FREQ Procedure (Book Excerpt)",
           *SAS/STAT 9.2 User's Guide, Second Edition*, SAS Publishing, 2009.

    .. [4] Laerd Statistics, "Somers' d using SPSS Statistics", *SPSS
           Statistics Tutorials and Statistical Guides*,
           https://statistics.laerd.com/spss-tutorials/somers-d-using-spss-statistics.php,
           Accessed July 31, 2020.

    Examples
    --------
    We calculate Somers' D for the example given in [4]_, in which a hotel
    chain owner seeks to determine the association between hotel room
    cleanliness and customer satisfaction. The independent variable, hotel
    room cleanliness, is ranked on an ordinal scale: "below average (1)",
    "average (2)", or "above average (3)". The dependent variable, customer
    satisfaction, is ranked on a second scale: "very dissatisfied (1)",
    "moderately dissatisfied (2)", "neither dissatisfied nor satisfied (3)",
    "moderately satisfied (4)", or "very satisfied (5)". 189 customers
    respond to the survey, and the results are cast into a contingency table
    with the hotel room cleanliness as the "row" variable and customer
    satisfaction as the "column" variable.

    +-----+-----+-----+-----+-----+-----+
    |     | (1) | (2) | (3) | (4) | (5) |
    +=====+=====+=====+=====+=====+=====+
    | (1) | 27  | 25  | 14  | 7   | 0   |
    +-----+-----+-----+-----+-----+-----+
    | (2) | 7   | 14  | 18  | 35  | 12  |
    +-----+-----+-----+-----+-----+-----+
    | (3) | 1   | 3   | 2   | 7   | 17  |
    +-----+-----+-----+-----+-----+-----+

    For example, 27 customers assigned their room a cleanliness ranking of
    "below average (1)" and a corresponding satisfaction of "very
    dissatisfied (1)". We perform the analysis as follows.

    >>> from scipy.stats import somersd
    >>> table = [[27, 25, 14, 7, 0], [7, 14, 18, 35, 12], [1, 3, 2, 7, 17]]
    >>> res = somersd(table)
    >>> res.statistic
    0.6032766111513396
    >>> res.pvalue
    1.0007091191074533e-27

    The value of the Somers' D statistic is approximately 0.6, indicating
    a positive correlation between room cleanliness and customer satisfaction
    in the sample.
    The *p*-value is very small, indicating a very small probability of
    observing such an extreme value of the statistic under the null
    hypothesis that the statistic of the entire population (from which
    our sample of 189 customers is drawn) is zero. This supports the
    alternative hypothesis that the true value of Somers' D for the population
    is nonzero.

    """
    x, y = np.array(x), np.array(y)
    if x.ndim == 1:
        if x.size != y.size:
            raise ValueError("Rankings must be of equal length.")
        table = scipy.stats.contingency.crosstab(x, y)[1]
    elif x.ndim == 2:
        if np.any(x < 0):
            raise ValueError("All elements of the contingency table must be "
                             "non-negative.")
        if np.any(x != x.astype(int)):
            raise ValueError("All elements of the contingency table must be "
                             "integer.")
        if x.nonzero()[0].size < 2:
            raise ValueError("At least two elements of the contingency table "
                             "must be nonzero.")
        table = x
    else:
        raise ValueError("x must be either a 1D or 2D array")
    # The table type is converted to a float to avoid an integer overflow
    d, p = _somers_d(table.astype(float), alternative)

    # add alias for consistency with other correlation functions
    res = SomersDResult(d, p, table)
    res.correlation = d
    return res


# This could be combined with `_all_partitions` in `_resampling.py`
def _all_partitions(nx, ny):
    """
    Partition a set of indices into two fixed-length sets in all possible ways

    Partition a set of indices 0 ... nx + ny - 1 into two sets of length nx and
    ny in all possible ways (ignoring order of elements).
    """
    z = np.arange(nx+ny)
    for c in combinations(z, nx):
        x = np.array(c)
        mask = np.ones(nx+ny, bool)
        mask[x] = False
        y = z[mask]
        yield x, y


def _compute_log_combinations(n):
    """Compute all log combination of C(n, k)."""
    gammaln_arr = gammaln(np.arange(n + 1) + 1)
    return gammaln(n + 1) - gammaln_arr - gammaln_arr[::-1]


@dataclass
class BarnardExactResult:
    statistic: float
    pvalue: float


def barnard_exact(table, alternative="two-sided", pooled=True, n=32):
    r"""Perform a Barnard exact test on a 2x2 contingency table.

    Parameters
    ----------
    table : array_like of ints
        A 2x2 contingency table.  Elements should be non-negative integers.

    alternative : {'two-sided', 'less', 'greater'}, optional
        Defines the null and alternative hypotheses. Default is 'two-sided'.
        Please see explanations in the Notes section below.

    pooled : bool, optional
        Whether to compute score statistic with pooled variance (as in
        Student's t-test, for example) or unpooled variance (as in Welch's
        t-test). Default is ``True``.

    n : int, optional
        Number of sampling points used in the construction of the sampling
        method. Note that this argument will automatically be converted to
        the next higher power of 2 since `scipy.stats.qmc.Sobol` is used to
        select sample points. Default is 32. Must be positive. In most cases,
        32 points is enough to reach good precision. More points comes at
        performance cost.

    Returns
    -------
    ber : BarnardExactResult
        A result object with the following attributes.

        statistic : float
            The Wald statistic with pooled or unpooled variance, depending
            on the user choice of `pooled`.

        pvalue : float
            P-value, the probability of obtaining a distribution at least as
            extreme as the one that was actually observed, assuming that the
            null hypothesis is true.

    See Also
    --------
    chi2_contingency : Chi-square test of independence of variables in a
        contingency table.
    fisher_exact : Fisher exact test on a 2x2 contingency table.
    boschloo_exact : Boschloo's exact test on a 2x2 contingency table,
        which is an uniformly more powerful alternative to Fisher's exact test.

    Notes
    -----
    Barnard's test is an exact test used in the analysis of contingency
    tables. It examines the association of two categorical variables, and
    is a more powerful alternative than Fisher's exact test
    for 2x2 contingency tables.

    Let's define :math:`X_0` a 2x2 matrix representing the observed sample,
    where each column stores the binomial experiment, as in the example
    below. Let's also define :math:`p_1, p_2` the theoretical binomial
    probabilities for  :math:`x_{11}` and :math:`x_{12}`. When using
    Barnard exact test, we can assert three different null hypotheses :

    - :math:`H_0 : p_1 \geq p_2` versus :math:`H_1 : p_1 < p_2`,
      with `alternative` = "less"

    - :math:`H_0 : p_1 \leq p_2` versus :math:`H_1 : p_1 > p_2`,
      with `alternative` = "greater"

    - :math:`H_0 : p_1 = p_2` versus :math:`H_1 : p_1 \neq p_2`,
      with `alternative` = "two-sided" (default one)

    In order to compute Barnard's exact test, we are using the Wald
    statistic [3]_ with pooled or unpooled variance.
    Under the default assumption that both variances are equal
    (``pooled = True``), the statistic is computed as:

    .. math::

        T(X) = \frac{
            \hat{p}_1 - \hat{p}_2
        }{
            \sqrt{
                \hat{p}(1 - \hat{p})
                (\frac{1}{c_1} +
                \frac{1}{c_2})
            }
        }

    with :math:`\hat{p}_1, \hat{p}_2` and :math:`\hat{p}` the estimator of
    :math:`p_1, p_2` and :math:`p`, the latter being the combined probability,
    given the assumption that :math:`p_1 = p_2`.

    If this assumption is invalid (``pooled = False``), the statistic is:

    .. math::

        T(X) = \frac{
            \hat{p}_1 - \hat{p}_2
        }{
            \sqrt{
                \frac{\hat{p}_1 (1 - \hat{p}_1)}{c_1} +
                \frac{\hat{p}_2 (1 - \hat{p}_2)}{c_2}
            }
        }

    The p-value is then computed as:

    .. math::

        \sum
            \binom{c_1}{x_{11}}
            \binom{c_2}{x_{12}}
            \pi^{x_{11} + x_{12}}
            (1 - \pi)^{t - x_{11} - x_{12}}

    where the sum is over all  2x2 contingency tables :math:`X` such that:
    * :math:`T(X) \leq T(X_0)` when `alternative` = "less",
    * :math:`T(X) \geq T(X_0)` when `alternative` = "greater", or
    * :math:`T(X) \geq |T(X_0)|` when `alternative` = "two-sided".
    Above, :math:`c_1, c_2` are the sum of the columns 1 and 2,
    and :math:`t` the total (sum of the 4 sample's element).

    The returned p-value is the maximum p-value taken over the nuisance
    parameter :math:`\pi`, where :math:`0 \leq \pi \leq 1`.

    This function's complexity is :math:`O(n c_1 c_2)`, where `n` is the
    number of sample points.

    References
    ----------
    .. [1] Barnard, G. A. "Significance Tests for 2x2 Tables". *Biometrika*.
           34.1/2 (1947): 123-138. :doi:`dpgkg3`

    .. [2] Mehta, Cyrus R., and Pralay Senchaudhuri. "Conditional versus
           unconditional exact tests for comparing two binomials."
           *Cytel Software Corporation* 675 (2003): 1-5.

    .. [3] "Wald Test". *Wikipedia*. https://en.wikipedia.org/wiki/Wald_test

    Examples
    --------
    An example use of Barnard's test is presented in [2]_.

        Consider the following example of a vaccine efficacy study
        (Chan, 1998). In a randomized clinical trial of 30 subjects, 15 were
        inoculated with a recombinant DNA influenza vaccine and the 15 were
        inoculated with a placebo. Twelve of the 15 subjects in the placebo
        group (80%) eventually became infected with influenza whereas for the
        vaccine group, only 7 of the 15 subjects (47%) became infected. The
        data are tabulated as a 2 x 2 table::

                Vaccine  Placebo
            Yes     7        12
            No      8        3

    When working with statistical hypothesis testing, we usually use a
    threshold probability or significance level upon which we decide
    to reject the null hypothesis :math:`H_0`. Suppose we choose the common
    significance level of 5%.

    Our alternative hypothesis is that the vaccine will lower the chance of
    becoming infected with the virus; that is, the probability :math:`p_1` of
    catching the virus with the vaccine will be *less than* the probability
    :math:`p_2` of catching the virus without the vaccine.  Therefore, we call
    `barnard_exact` with the ``alternative="less"`` option:

    >>> import scipy.stats as stats
    >>> res = stats.barnard_exact([[7, 12], [8, 3]], alternative="less")
    >>> res.statistic
    -1.894...
    >>> res.pvalue
    0.03407...

    Under the null hypothesis that the vaccine will not lower the chance of
    becoming infected, the probability of obtaining test results at least as
    extreme as the observed data is approximately 3.4%. Since this p-value is
    less than our chosen significance level, we have evidence to reject
    :math:`H_0` in favor of the alternative.

    Suppose we had used Fisher's exact test instead:

    >>> _, pvalue = stats.fisher_exact([[7, 12], [8, 3]], alternative="less")
    >>> pvalue
    0.0640...

    With the same threshold significance of 5%, we would not have been able
    to reject the null hypothesis in favor of the alternative. As stated in
    [2]_, Barnard's test is uniformly more powerful than Fisher's exact test
    because Barnard's test does not condition on any margin. Fisher's test
    should only be used when both sets of marginals are fixed.

    """
    if n <= 0:
        raise ValueError(
            "Number of points `n` must be strictly positive, "
            f"found {n!r}"
        )

    table = np.asarray(table, dtype=np.int64)

    if not table.shape == (2, 2):
        raise ValueError("The input `table` must be of shape (2, 2).")

    if np.any(table < 0):
        raise ValueError("All values in `table` must be nonnegative.")

    if 0 in table.sum(axis=0):
        # If both values in column are zero, the p-value is 1 and
        # the score's statistic is NaN.
        return BarnardExactResult(np.nan, 1.0)

    total_col_1, total_col_2 = table.sum(axis=0)

    x1 = np.arange(total_col_1 + 1, dtype=np.int64).reshape(-1, 1)
    x2 = np.arange(total_col_2 + 1, dtype=np.int64).reshape(1, -1)

    # We need to calculate the wald statistics for each combination of x1 and
    # x2.
    p1, p2 = x1 / total_col_1, x2 / total_col_2

    if pooled:
        p = (x1 + x2) / (total_col_1 + total_col_2)
        variances = p * (1 - p) * (1 / total_col_1 + 1 / total_col_2)
    else:
        variances = p1 * (1 - p1) / total_col_1 + p2 * (1 - p2) / total_col_2

    # To avoid warning when dividing by 0
    with np.errstate(divide="ignore", invalid="ignore"):
        wald_statistic = np.divide((p1 - p2), np.sqrt(variances))

    wald_statistic[p1 == p2] = 0  # Removing NaN values

    wald_stat_obs = wald_statistic[table[0, 0], table[0, 1]]

    if alternative == "two-sided":
        index_arr = np.abs(wald_statistic) >= abs(wald_stat_obs)
    elif alternative == "less":
        index_arr = wald_statistic <= wald_stat_obs
    elif alternative == "greater":
        index_arr = wald_statistic >= wald_stat_obs
    else:
        msg = (
            "`alternative` should be one of {'two-sided', 'less', 'greater'},"
            f" found {alternative!r}"
        )
        raise ValueError(msg)

    x1_sum_x2 = x1 + x2

    x1_log_comb = _compute_log_combinations(total_col_1)
    x2_log_comb = _compute_log_combinations(total_col_2)
    x1_sum_x2_log_comb = x1_log_comb[x1] + x2_log_comb[x2]

    result = shgo(
        _get_binomial_log_p_value_with_nuisance_param,
        args=(x1_sum_x2, x1_sum_x2_log_comb, index_arr),
        bounds=((0, 1),),
        n=n,
        sampling_method="sobol",
    )

    # result.fun is the negative log pvalue and therefore needs to be
    # changed before return
    p_value = np.clip(np.exp(-result.fun), a_min=0, a_max=1)
    return BarnardExactResult(wald_stat_obs, p_value)


@dataclass
class BoschlooExactResult:
    statistic: float
    pvalue: float


def boschloo_exact(table, alternative="two-sided", n=32):
    r"""Perform Boschloo's exact test on a 2x2 contingency table.

    Parameters
    ----------
    table : array_like of ints
        A 2x2 contingency table.  Elements should be non-negative integers.

    alternative : {'two-sided', 'less', 'greater'}, optional
        Defines the null and alternative hypotheses. Default is 'two-sided'.
        Please see explanations in the Notes section below.

    n : int, optional
        Number of sampling points used in the construction of the sampling
        method. Note that this argument will automatically be converted to
        the next higher power of 2 since `scipy.stats.qmc.Sobol` is used to
        select sample points. Default is 32. Must be positive. In most cases,
        32 points is enough to reach good precision. More points comes at
        performance cost.

    Returns
    -------
    ber : BoschlooExactResult
        A result object with the following attributes.

        statistic : float
            The statistic used in Boschloo's test; that is, the p-value
            from Fisher's exact test.

        pvalue : float
            P-value, the probability of obtaining a distribution at least as
            extreme as the one that was actually observed, assuming that the
            null hypothesis is true.

    See Also
    --------
    chi2_contingency : Chi-square test of independence of variables in a
        contingency table.
    fisher_exact : Fisher exact test on a 2x2 contingency table.
    barnard_exact : Barnard's exact test, which is a more powerful alternative
        than Fisher's exact test for 2x2 contingency tables.

    Notes
    -----
    Boschloo's test is an exact test used in the analysis of contingency
    tables. It examines the association of two categorical variables, and
    is a uniformly more powerful alternative to Fisher's exact test
    for 2x2 contingency tables.

    Boschloo's exact test uses the p-value of Fisher's exact test as a
    statistic, and Boschloo's p-value is the probability under the null
    hypothesis of observing such an extreme value of this statistic.

    Let's define :math:`X_0` a 2x2 matrix representing the observed sample,
    where each column stores the binomial experiment, as in the example
    below. Let's also define :math:`p_1, p_2` the theoretical binomial
    probabilities for  :math:`x_{11}` and :math:`x_{12}`. When using
    Boschloo exact test, we can assert three different alternative hypotheses:

    - :math:`H_0 : p_1=p_2` versus :math:`H_1 : p_1 < p_2`,
      with `alternative` = "less"

    - :math:`H_0 : p_1=p_2` versus :math:`H_1 : p_1 > p_2`,
      with `alternative` = "greater"

    - :math:`H_0 : p_1=p_2` versus :math:`H_1 : p_1 \neq p_2`,
      with `alternative` = "two-sided" (default)

    There are multiple conventions for computing a two-sided p-value when the
    null distribution is asymmetric. Here, we apply the convention that the
    p-value of a two-sided test is twice the minimum of the p-values of the
    one-sided tests (clipped to 1.0). Note that `fisher_exact` follows a
    different convention, so for a given `table`, the statistic reported by
    `boschloo_exact` may differ from the p-value reported by `fisher_exact`
    when ``alternative='two-sided'``.

    .. versionadded:: 1.7.0

    References
    ----------
    .. [1] R.D. Boschloo. "Raised conditional level of significance for the
       2 x 2-table when testing the equality of two probabilities",
       Statistica Neerlandica, 24(1), 1970

    .. [2] "Boschloo's test", Wikipedia,
       https://en.wikipedia.org/wiki/Boschloo%27s_test

    .. [3] Lise M. Saari et al. "Employee attitudes and job satisfaction",
       Human Resource Management, 43(4), 395-407, 2004,
       :doi:`10.1002/hrm.20032`.

    Examples
    --------
    In the following example, we consider the article "Employee
    attitudes and job satisfaction" [3]_
    which reports the results of a survey from 63 scientists and 117 college
    professors. Of the 63 scientists, 31 said they were very satisfied with
    their jobs, whereas 74 of the college professors were very satisfied
    with their work. Is this significant evidence that college
    professors are happier with their work than scientists?
    The following table summarizes the data mentioned above::

                         college professors   scientists
        Very Satisfied   74                     31
        Dissatisfied     43                     32

    When working with statistical hypothesis testing, we usually use a
    threshold probability or significance level upon which we decide
    to reject the null hypothesis :math:`H_0`. Suppose we choose the common
    significance level of 5%.

    Our alternative hypothesis is that college professors are truly more
    satisfied with their work than scientists. Therefore, we expect
    :math:`p_1` the proportion of very satisfied college professors to be
    greater than :math:`p_2`, the proportion of very satisfied scientists.
    We thus call `boschloo_exact` with the ``alternative="greater"`` option:

    >>> import scipy.stats as stats
    >>> res = stats.boschloo_exact([[74, 31], [43, 32]], alternative="greater")
    >>> res.statistic
    0.0483...
    >>> res.pvalue
    0.0355...

    Under the null hypothesis that scientists are happier in their work than
    college professors, the probability of obtaining test
    results at least as extreme as the observed data is approximately 3.55%.
    Since this p-value is less than our chosen significance level, we have
    evidence to reject :math:`H_0` in favor of the alternative hypothesis.

    """
    hypergeom = distributions.hypergeom

    if n <= 0:
        raise ValueError(
            "Number of points `n` must be strictly positive,"
            f" found {n!r}"
        )

    table = np.asarray(table, dtype=np.int64)

    if not table.shape == (2, 2):
        raise ValueError("The input `table` must be of shape (2, 2).")

    if np.any(table < 0):
        raise ValueError("All values in `table` must be nonnegative.")

    if 0 in table.sum(axis=0):
        # If both values in column are zero, the p-value is 1 and
        # the score's statistic is NaN.
        return BoschlooExactResult(np.nan, np.nan)

    total_col_1, total_col_2 = table.sum(axis=0)
    total = total_col_1 + total_col_2
    x1 = np.arange(total_col_1 + 1, dtype=np.int64).reshape(1, -1)
    x2 = np.arange(total_col_2 + 1, dtype=np.int64).reshape(-1, 1)
    x1_sum_x2 = x1 + x2

    if alternative == 'less':
        pvalues = hypergeom.cdf(x1, total, x1_sum_x2, total_col_1).T
    elif alternative == 'greater':
        # Same formula as the 'less' case, but with the second column.
        pvalues = hypergeom.cdf(x2, total, x1_sum_x2, total_col_2).T
    elif alternative == 'two-sided':
        boschloo_less = boschloo_exact(table, alternative="less", n=n)
        boschloo_greater = boschloo_exact(table, alternative="greater", n=n)

        res = (
            boschloo_less if boschloo_less.pvalue < boschloo_greater.pvalue
            else boschloo_greater
        )

        # Two-sided p-value is defined as twice the minimum of the one-sided
        # p-values
        pvalue = np.clip(2 * res.pvalue, a_min=0, a_max=1)
        return BoschlooExactResult(res.statistic, pvalue)
    else:
        msg = (
            f"`alternative` should be one of {'two-sided', 'less', 'greater'},"
            f" found {alternative!r}"
        )
        raise ValueError(msg)

    fisher_stat = pvalues[table[0, 0], table[0, 1]]

    # fisher_stat * (1+1e-13) guards us from small numerical error. It is
    # equivalent to np.isclose with relative tol of 1e-13 and absolute tol of 0
    # For more throughout explanations, see gh-14178
    index_arr = pvalues <= fisher_stat * (1+1e-13)

    x1, x2, x1_sum_x2 = x1.T, x2.T, x1_sum_x2.T
    x1_log_comb = _compute_log_combinations(total_col_1)
    x2_log_comb = _compute_log_combinations(total_col_2)
    x1_sum_x2_log_comb = x1_log_comb[x1] + x2_log_comb[x2]

    result = shgo(
        _get_binomial_log_p_value_with_nuisance_param,
        args=(x1_sum_x2, x1_sum_x2_log_comb, index_arr),
        bounds=((0, 1),),
        n=n,
        sampling_method="sobol",
    )

    # result.fun is the negative log pvalue and therefore needs to be
    # changed before return
    p_value = np.clip(np.exp(-result.fun), a_min=0, a_max=1)
    return BoschlooExactResult(fisher_stat, p_value)


def _get_binomial_log_p_value_with_nuisance_param(
    nuisance_param, x1_sum_x2, x1_sum_x2_log_comb, index_arr
):
    r"""
    Compute the log pvalue in respect of a nuisance parameter considering
    a 2x2 sample space.

    Parameters
    ----------
    nuisance_param : float
        nuisance parameter used in the computation of the maximisation of
        the p-value. Must be between 0 and 1

    x1_sum_x2 : ndarray
        Sum of x1 and x2 inside barnard_exact

    x1_sum_x2_log_comb : ndarray
        sum of the log combination of x1 and x2

    index_arr : ndarray of boolean

    Returns
    -------
    p_value : float
        Return the maximum p-value considering every nuisance parameter
        between 0 and 1

    Notes
    -----

    Both Barnard's test and Boschloo's test iterate over a nuisance parameter
    :math:`\pi \in [0, 1]` to find the maximum p-value. To search this
    maxima, this function return the negative log pvalue with respect to the
    nuisance parameter passed in params. This negative log p-value is then
    used in `shgo` to find the minimum negative pvalue which is our maximum
    pvalue.

    Also, to compute the different combination used in the
    p-values' computation formula, this function uses `gammaln` which is
    more tolerant for large value than `scipy.special.comb`. `gammaln` gives
    a log combination. For the little precision loss, performances are
    improved a lot.
    """
    t1, t2 = x1_sum_x2.shape
    n = t1 + t2 - 2
    with np.errstate(divide="ignore", invalid="ignore"):
        log_nuisance = np.log(
            nuisance_param,
            out=np.zeros_like(nuisance_param),
            where=nuisance_param >= 0,
        )
        log_1_minus_nuisance = np.log(
            1 - nuisance_param,
            out=np.zeros_like(nuisance_param),
            where=1 - nuisance_param >= 0,
        )

        nuisance_power_x1_x2 = log_nuisance * x1_sum_x2
        nuisance_power_x1_x2[(x1_sum_x2 == 0)[:, :]] = 0

        nuisance_power_n_minus_x1_x2 = log_1_minus_nuisance * (n - x1_sum_x2)
        nuisance_power_n_minus_x1_x2[(x1_sum_x2 == n)[:, :]] = 0

        tmp_log_values_arr = (
            x1_sum_x2_log_comb
            + nuisance_power_x1_x2
            + nuisance_power_n_minus_x1_x2
        )

    tmp_values_from_index = tmp_log_values_arr[index_arr]

    # To avoid dividing by zero in log function and getting inf value,
    # values are centered according to the max
    max_value = tmp_values_from_index.max()

    # To have better result's precision, the log pvalue is taken here.
    # Indeed, pvalue is included inside [0, 1] interval. Passing the
    # pvalue to log makes the interval a lot bigger ([-inf, 0]), and thus
    # help us to achieve better precision
    with np.errstate(divide="ignore", invalid="ignore"):
        log_probs = np.exp(tmp_values_from_index - max_value).sum()
        log_pvalue = max_value + np.log(
            log_probs,
            out=np.full_like(log_probs, -np.inf),
            where=log_probs > 0,
        )

    # Since shgo find the minima, minus log pvalue is returned
    return -log_pvalue


def _pval_cvm_2samp_exact(s, m, n):
    """
    Compute the exact p-value of the Cramer-von Mises two-sample test
    for a given value s of the test statistic.
    m and n are the sizes of the samples.

    [1] Y. Xiao, A. Gordon, and A. Yakovlev, "A C++ Program for
        the Cramér-Von Mises Two-Sample Test", J. Stat. Soft.,
        vol. 17, no. 8, pp. 1-15, Dec. 2006.
    [2] T. W. Anderson "On the Distribution of the Two-Sample Cramer-von Mises
        Criterion," The Annals of Mathematical Statistics, Ann. Math. Statist.
        33(3), 1148-1159, (September, 1962)
    """

    # [1, p. 3]
    lcm = np.lcm(m, n)
    # [1, p. 4], below eq. 3
    a = lcm // m
    b = lcm // n
    # Combine Eq. 9 in [2] with Eq. 2 in [1] and solve for $\zeta$
    # Hint: `s` is $U$ in [2], and $T_2$ in [1] is $T$ in [2]
    mn = m * n
    zeta = lcm ** 2 * (m + n) * (6 * s - mn * (4 * mn - 1)) // (6 * mn ** 2)

    # bound maximum value that may appear in `gs` (remember both rows!)
    zeta_bound = lcm**2 * (m + n)  # bound elements in row 1
    combinations = comb(m + n, m)  # sum of row 2
    max_gs = max(zeta_bound, combinations)
    dtype = np.min_scalar_type(max_gs)

    # the frequency table of $g_{u, v}^+$ defined in [1, p. 6]
    gs = ([np.array([[0], [1]], dtype=dtype)]
          + [np.empty((2, 0), dtype=dtype) for _ in range(m)])
    for u in range(n + 1):
        next_gs = []
        tmp = np.empty((2, 0), dtype=dtype)
        for v, g in enumerate(gs):
            # Calculate g recursively with eq. 11 in [1]. Even though it
            # doesn't look like it, this also does 12/13 (all of Algorithm 1).
            vi, i0, i1 = np.intersect1d(tmp[0], g[0], return_indices=True)
            tmp = np.concatenate([
                np.stack([vi, tmp[1, i0] + g[1, i1]]),
                np.delete(tmp, i0, 1),
                np.delete(g, i1, 1)
            ], 1)
            res = (a * v - b * u) ** 2
            tmp[0] += res.astype(dtype)
            next_gs.append(tmp)
        gs = next_gs
    value, freq = gs[m]
    return np.float64(np.sum(freq[value >= zeta]) / combinations)


@_axis_nan_policy_factory(CramerVonMisesResult, n_samples=2, too_small=1,
                          result_to_tuple=_cvm_result_to_tuple)
def cramervonmises_2samp(x, y, method='auto'):
    """Perform the two-sample Cramér-von Mises test for goodness of fit.

    This is the two-sample version of the Cramér-von Mises test ([1]_):
    for two independent samples :math:`X_1, ..., X_n` and
    :math:`Y_1, ..., Y_m`, the null hypothesis is that the samples
    come from the same (unspecified) continuous distribution.

    Parameters
    ----------
    x : array_like
        A 1-D array of observed values of the random variables :math:`X_i`.
    y : array_like
        A 1-D array of observed values of the random variables :math:`Y_i`.
    method : {'auto', 'asymptotic', 'exact'}, optional
        The method used to compute the p-value, see Notes for details.
        The default is 'auto'.

    Returns
    -------
    res : object with attributes
        statistic : float
            Cramér-von Mises statistic.
        pvalue : float
            The p-value.

    See Also
    --------
    cramervonmises, anderson_ksamp, epps_singleton_2samp, ks_2samp

    Notes
    -----
    .. versionadded:: 1.7.0

    The statistic is computed according to equation 9 in [2]_. The
    calculation of the p-value depends on the keyword `method`:

    - ``asymptotic``: The p-value is approximated by using the limiting
      distribution of the test statistic.
    - ``exact``: The exact p-value is computed by enumerating all
      possible combinations of the test statistic, see [2]_.

    If ``method='auto'``, the exact approach is used
    if both samples contain equal to or less than 20 observations,
    otherwise the asymptotic distribution is used.

    If the underlying distribution is not continuous, the p-value is likely to
    be conservative (Section 6.2 in [3]_). When ranking the data to compute
    the test statistic, midranks are used if there are ties.

    References
    ----------
    .. [1] https://en.wikipedia.org/wiki/Cramer-von_Mises_criterion
    .. [2] Anderson, T.W. (1962). On the distribution of the two-sample
           Cramer-von-Mises criterion. The Annals of Mathematical
           Statistics, pp. 1148-1159.
    .. [3] Conover, W.J., Practical Nonparametric Statistics, 1971.

    Examples
    --------

    Suppose we wish to test whether two samples generated by
    ``scipy.stats.norm.rvs`` have the same distribution. We choose a
    significance level of alpha=0.05.

    >>> import numpy as np
    >>> from scipy import stats
    >>> rng = np.random.default_rng()
    >>> x = stats.norm.rvs(size=100, random_state=rng)
    >>> y = stats.norm.rvs(size=70, random_state=rng)
    >>> res = stats.cramervonmises_2samp(x, y)
    >>> res.statistic, res.pvalue
    (0.29376470588235293, 0.1412873014573014)

    The p-value exceeds our chosen significance level, so we do not
    reject the null hypothesis that the observed samples are drawn from the
    same distribution.

    For small sample sizes, one can compute the exact p-values:

    >>> x = stats.norm.rvs(size=7, random_state=rng)
    >>> y = stats.t.rvs(df=2, size=6, random_state=rng)
    >>> res = stats.cramervonmises_2samp(x, y, method='exact')
    >>> res.statistic, res.pvalue
    (0.197802197802198, 0.31643356643356646)

    The p-value based on the asymptotic distribution is a good approximation
    even though the sample size is small.

    >>> res = stats.cramervonmises_2samp(x, y, method='asymptotic')
    >>> res.statistic, res.pvalue
    (0.197802197802198, 0.2966041181527128)

    Independent of the method, one would not reject the null hypothesis at the
    chosen significance level in this example.

    """
    xa = np.sort(np.asarray(x))
    ya = np.sort(np.asarray(y))

    if xa.size <= 1 or ya.size <= 1:
        raise ValueError('x and y must contain at least two observations.')
    if method not in ['auto', 'exact', 'asymptotic']:
        raise ValueError('method must be either auto, exact or asymptotic.')

    nx = len(xa)
    ny = len(ya)

    if method == 'auto':
        if max(nx, ny) > 20:
            method = 'asymptotic'
        else:
            method = 'exact'

    # get ranks of x and y in the pooled sample
    z = np.concatenate([xa, ya])
    # in case of ties, use midrank (see [1])
    r = scipy.stats.rankdata(z, method='average')
    rx = r[:nx]
    ry = r[nx:]

    # compute U (eq. 10 in [2])
    u = nx * np.sum((rx - np.arange(1, nx+1))**2)
    u += ny * np.sum((ry - np.arange(1, ny+1))**2)

    # compute T (eq. 9 in [2])
    k, N = nx*ny, nx + ny
    t = u / (k*N) - (4*k - 1)/(6*N)

    if method == 'exact':
        p = _pval_cvm_2samp_exact(u, nx, ny)
    else:
        # compute expected value and variance of T (eq. 11 and 14 in [2])
        et = (1 + 1/N)/6
        vt = (N+1) * (4*k*N - 3*(nx**2 + ny**2) - 2*k)
        vt = vt / (45 * N**2 * 4 * k)

        # computed the normalized statistic (eq. 15 in [2])
        tn = 1/6 + (t - et) / np.sqrt(45 * vt)

        # approximate distribution of tn with limiting distribution
        # of the one-sample test statistic
        # if tn < 0.003, the _cdf_cvm_inf(tn) < 1.28*1e-18, return 1.0 directly
        if tn < 0.003:
            p = 1.0
        else:
            p = max(0, 1. - _cdf_cvm_inf(tn))

    return CramerVonMisesResult(statistic=t, pvalue=p)


class TukeyHSDResult:
    """Result of `scipy.stats.tukey_hsd`.

    Attributes
    ----------
    statistic : float ndarray
        The computed statistic of the test for each comparison. The element
        at index ``(i, j)`` is the statistic for the comparison between groups
        ``i`` and ``j``.
    pvalue : float ndarray
        The associated p-value from the studentized range distribution. The
        element at index ``(i, j)`` is the p-value for the comparison
        between groups ``i`` and ``j``.

    Notes
    -----
    The string representation of this object displays the most recently
    calculated confidence interval, and if none have been previously
    calculated, it will evaluate ``confidence_interval()``.

    References
    ----------
    .. [1] NIST/SEMATECH e-Handbook of Statistical Methods, "7.4.7.1. Tukey's
           Method."
           https://www.itl.nist.gov/div898/handbook/prc/section4/prc471.htm,
           28 November 2020.
    """

    def __init__(self, statistic, pvalue, _nobs, _ntreatments, _stand_err):
        self.statistic = statistic
        self.pvalue = pvalue
        self._ntreatments = _ntreatments
        self._nobs = _nobs
        self._stand_err = _stand_err
        self._ci = None
        self._ci_cl = None

    def __str__(self):
        # Note: `__str__` prints the confidence intervals from the most
        # recent call to `confidence_interval`. If it has not been called,
        # it will be called with the default CL of .95.
        if self._ci is None:
            self.confidence_interval(confidence_level=.95)
        s = ("Tukey's HSD Pairwise Group Comparisons"
             f" ({self._ci_cl*100:.1f}% Confidence Interval)\n")
        s += "Comparison  Statistic  p-value  Lower CI  Upper CI\n"
        for i in range(self.pvalue.shape[0]):
            for j in range(self.pvalue.shape[0]):
                if i != j:
                    s += (f" ({i} - {j}) {self.statistic[i, j]:>10.3f}"
                          f"{self.pvalue[i, j]:>10.3f}"
                          f"{self._ci.low[i, j]:>10.3f}"
                          f"{self._ci.high[i, j]:>10.3f}\n")
        return s

    def confidence_interval(self, confidence_level=.95):
        """Compute the confidence interval for the specified confidence level.

        Parameters
        ----------
        confidence_level : float, optional
            Confidence level for the computed confidence interval
            of the estimated proportion. Default is .95.

        Returns
        -------
        ci : ``ConfidenceInterval`` object
            The object has attributes ``low`` and ``high`` that hold the
            lower and upper bounds of the confidence intervals for each
            comparison. The high and low values are accessible for each
            comparison at index ``(i, j)`` between groups ``i`` and ``j``.

        References
        ----------
        .. [1] NIST/SEMATECH e-Handbook of Statistical Methods, "7.4.7.1.
               Tukey's Method."
               https://www.itl.nist.gov/div898/handbook/prc/section4/prc471.htm,
               28 November 2020.

        Examples
        --------
        >>> from scipy.stats import tukey_hsd
        >>> group0 = [24.5, 23.5, 26.4, 27.1, 29.9]
        >>> group1 = [28.4, 34.2, 29.5, 32.2, 30.1]
        >>> group2 = [26.1, 28.3, 24.3, 26.2, 27.8]
        >>> result = tukey_hsd(group0, group1, group2)
        >>> ci = result.confidence_interval()
        >>> ci.low
        array([[-3.649159, -8.249159, -3.909159],
               [ 0.950841, -3.649159,  0.690841],
               [-3.389159, -7.989159, -3.649159]])
        >>> ci.high
        array([[ 3.649159, -0.950841,  3.389159],
               [ 8.249159,  3.649159,  7.989159],
               [ 3.909159, -0.690841,  3.649159]])
        """
        # check to see if the supplied confidence level matches that of the
        # previously computed CI.
        if (self._ci is not None and self._ci_cl is not None and
                confidence_level == self._ci_cl):
            return self._ci

        if not 0 < confidence_level < 1:
            raise ValueError("Confidence level must be between 0 and 1.")
        # determine the critical value of the studentized range using the
        # appropriate confidence level, number of treatments, and degrees
        # of freedom as determined by the number of data less the number of
        # treatments. ("Confidence limits for Tukey's method")[1]. Note that
        # in the cases of unequal sample sizes there will be a criterion for
        # each group comparison.
        params = (confidence_level, self._nobs, self._ntreatments - self._nobs)
        srd = distributions.studentized_range.ppf(*params)
        # also called maximum critical value, the Tukey criterion is the
        # studentized range critical value * the square root of mean square
        # error over the sample size.
        tukey_criterion = srd * self._stand_err
        # the confidence levels are determined by the
        # `mean_differences` +- `tukey_criterion`
        upper_conf = self.statistic + tukey_criterion
        lower_conf = self.statistic - tukey_criterion
        self._ci = ConfidenceInterval(low=lower_conf, high=upper_conf)
        self._ci_cl = confidence_level
        return self._ci


def _tukey_hsd_iv(args):
    if (len(args)) < 2:
        raise ValueError("There must be more than 1 treatment.")
    args = [np.asarray(arg) for arg in args]
    for arg in args:
        if arg.ndim != 1:
            raise ValueError("Input samples must be one-dimensional.")
        if arg.size <= 1:
            raise ValueError("Input sample size must be greater than one.")
        if np.isinf(arg).any():
            raise ValueError("Input samples must be finite.")
    return args


def tukey_hsd(*args):
    """Perform Tukey's HSD test for equality of means over multiple treatments.

    Tukey's honestly significant difference (HSD) test performs pairwise
    comparison of means for a set of samples. Whereas ANOVA (e.g. `f_oneway`)
    assesses whether the true means underlying each sample are identical,
    Tukey's HSD is a post hoc test used to compare the mean of each sample
    to the mean of each other sample.

    The null hypothesis is that the distributions underlying the samples all
    have the same mean. The test statistic, which is computed for every
    possible pairing of samples, is simply the difference between the sample
    means. For each pair, the p-value is the probability under the null
    hypothesis (and other assumptions; see notes) of observing such an extreme
    value of the statistic, considering that many pairwise comparisons are
    being performed. Confidence intervals for the difference between each pair
    of means are also available.

    Parameters
    ----------
    sample1, sample2, ... : array_like
        The sample measurements for each group. There must be at least
        two arguments.

    Returns
    -------
    result : `~scipy.stats._result_classes.TukeyHSDResult` instance
        The return value is an object with the following attributes:

        statistic : float ndarray
            The computed statistic of the test for each comparison. The element
            at index ``(i, j)`` is the statistic for the comparison between
            groups ``i`` and ``j``.
        pvalue : float ndarray
            The computed p-value of the test for each comparison. The element
            at index ``(i, j)`` is the p-value for the comparison between
            groups ``i`` and ``j``.

        The object has the following methods:

        confidence_interval(confidence_level=0.95):
            Compute the confidence interval for the specified confidence level.

    See Also
    --------
    dunnett : performs comparison of means against a control group.

    Notes
    -----
    The use of this test relies on several assumptions.

    1. The observations are independent within and among groups.
    2. The observations within each group are normally distributed.
    3. The distributions from which the samples are drawn have the same finite
       variance.

    The original formulation of the test was for samples of equal size [6]_.
    In case of unequal sample sizes, the test uses the Tukey-Kramer method
    [4]_.

    References
    ----------
    .. [1] NIST/SEMATECH e-Handbook of Statistical Methods, "7.4.7.1. Tukey's
           Method."
           https://www.itl.nist.gov/div898/handbook/prc/section4/prc471.htm,
           28 November 2020.
    .. [2] Abdi, Herve & Williams, Lynne. (2021). "Tukey's Honestly Significant
           Difference (HSD) Test."
           https://personal.utdallas.edu/~herve/abdi-HSD2010-pretty.pdf
    .. [3] "One-Way ANOVA Using SAS PROC ANOVA & PROC GLM." SAS
           Tutorials, 2007, www.stattutorials.com/SAS/TUTORIAL-PROC-GLM.htm.
    .. [4] Kramer, Clyde Young. "Extension of Multiple Range Tests to Group
           Means with Unequal Numbers of Replications." Biometrics, vol. 12,
           no. 3, 1956, pp. 307-310. JSTOR, www.jstor.org/stable/3001469.
           Accessed 25 May 2021.
    .. [5] NIST/SEMATECH e-Handbook of Statistical Methods, "7.4.3.3.
           The ANOVA table and tests of hypotheses about means"
           https://www.itl.nist.gov/div898/handbook/prc/section4/prc433.htm,
           2 June 2021.
    .. [6] Tukey, John W. "Comparing Individual Means in the Analysis of
           Variance." Biometrics, vol. 5, no. 2, 1949, pp. 99-114. JSTOR,
           www.jstor.org/stable/3001913. Accessed 14 June 2021.


    Examples
    --------
    Here are some data comparing the time to relief of three brands of
    headache medicine, reported in minutes. Data adapted from [3]_.

    >>> import numpy as np
    >>> from scipy.stats import tukey_hsd
    >>> group0 = [24.5, 23.5, 26.4, 27.1, 29.9]
    >>> group1 = [28.4, 34.2, 29.5, 32.2, 30.1]
    >>> group2 = [26.1, 28.3, 24.3, 26.2, 27.8]

    We would like to see if the means between any of the groups are
    significantly different. First, visually examine a box and whisker plot.

    >>> import matplotlib.pyplot as plt
    >>> fig, ax = plt.subplots(1, 1)
    >>> ax.boxplot([group0, group1, group2])
    >>> ax.set_xticklabels(["group0", "group1", "group2"]) # doctest: +SKIP
    >>> ax.set_ylabel("mean") # doctest: +SKIP
    >>> plt.show()

    From the box and whisker plot, we can see overlap in the interquartile
    ranges group 1 to group 2 and group 3, but we can apply the ``tukey_hsd``
    test to determine if the difference between means is significant. We
    set a significance level of .05 to reject the null hypothesis.

    >>> res = tukey_hsd(group0, group1, group2)
    >>> print(res)
    Tukey's HSD Pairwise Group Comparisons (95.0% Confidence Interval)
    Comparison  Statistic  p-value   Lower CI   Upper CI
    (0 - 1)     -4.600      0.014     -8.249     -0.951
    (0 - 2)     -0.260      0.980     -3.909      3.389
    (1 - 0)      4.600      0.014      0.951      8.249
    (1 - 2)      4.340      0.020      0.691      7.989
    (2 - 0)      0.260      0.980     -3.389      3.909
    (2 - 1)     -4.340      0.020     -7.989     -0.691

    The null hypothesis is that each group has the same mean. The p-value for
    comparisons between ``group0`` and ``group1`` as well as ``group1`` and
    ``group2`` do not exceed .05, so we reject the null hypothesis that they
    have the same means. The p-value of the comparison between ``group0``
    and ``group2`` exceeds .05, so we accept the null hypothesis that there
    is not a significant difference between their means.

    We can also compute the confidence interval associated with our chosen
    confidence level.

    >>> group0 = [24.5, 23.5, 26.4, 27.1, 29.9]
    >>> group1 = [28.4, 34.2, 29.5, 32.2, 30.1]
    >>> group2 = [26.1, 28.3, 24.3, 26.2, 27.8]
    >>> result = tukey_hsd(group0, group1, group2)
    >>> conf = res.confidence_interval(confidence_level=.99)
    >>> for ((i, j), l) in np.ndenumerate(conf.low):
    ...     # filter out self comparisons
    ...     if i != j:
    ...         h = conf.high[i,j]
    ...         print(f"({i} - {j}) {l:>6.3f} {h:>6.3f}")
    (0 - 1) -9.480  0.280
    (0 - 2) -5.140  4.620
    (1 - 0) -0.280  9.480
    (1 - 2) -0.540  9.220
    (2 - 0) -4.620  5.140
    (2 - 1) -9.220  0.540
    """
    args = _tukey_hsd_iv(args)
    ntreatments = len(args)
    means = np.asarray([np.mean(arg) for arg in args])
    nsamples_treatments = np.asarray([a.size for a in args])
    nobs = np.sum(nsamples_treatments)

    # determine mean square error [5]. Note that this is sometimes called
    # mean square error within.
    mse = (np.sum([np.var(arg, ddof=1) for arg in args] *
                  (nsamples_treatments - 1)) / (nobs - ntreatments))

    # The calculation of the standard error differs when treatments differ in
    # size. See ("Unequal sample sizes")[1].
    if np.unique(nsamples_treatments).size == 1:
        # all input groups are the same length, so only one value needs to be
        # calculated [1].
        normalize = 2 / nsamples_treatments[0]
    else:
        # to compare groups of differing sizes, we must compute a variance
        # value for each individual comparison. Use broadcasting to get the
        # resulting matrix. [3], verified against [4] (page 308).
        normalize = 1 / nsamples_treatments + 1 / nsamples_treatments[None].T

    # the standard error is used in the computation of the tukey criterion and
    # finding the p-values.
    stand_err = np.sqrt(normalize * mse / 2)

    # the mean difference is the test statistic.
    mean_differences = means[None].T - means

    # Calculate the t-statistic to use within the survival function of the
    # studentized range to get the p-value.
    t_stat = np.abs(mean_differences) / stand_err

    params = t_stat, ntreatments, nobs - ntreatments
    pvalues = distributions.studentized_range.sf(*params)

    return TukeyHSDResult(mean_differences, pvalues, ntreatments,
                          nobs, stand_err)