peacock-data-public-datasets-idc-llm_eval
/
env-llmeval
/lib
/python3.10
/site-packages
/scipy
/fftpack
/_helper.py
import operator | |
import numpy as np | |
from numpy.fft import fftshift, ifftshift, fftfreq | |
import scipy.fft._pocketfft.helper as _helper | |
__all__ = ['fftshift', 'ifftshift', 'fftfreq', 'rfftfreq', 'next_fast_len'] | |
def rfftfreq(n, d=1.0): | |
"""DFT sample frequencies (for usage with rfft, irfft). | |
The returned float array contains the frequency bins in | |
cycles/unit (with zero at the start) given a window length `n` and a | |
sample spacing `d`:: | |
f = [0,1,1,2,2,...,n/2-1,n/2-1,n/2]/(d*n) if n is even | |
f = [0,1,1,2,2,...,n/2-1,n/2-1,n/2,n/2]/(d*n) if n is odd | |
Parameters | |
---------- | |
n : int | |
Window length. | |
d : scalar, optional | |
Sample spacing. Default is 1. | |
Returns | |
------- | |
out : ndarray | |
The array of length `n`, containing the sample frequencies. | |
Examples | |
-------- | |
>>> import numpy as np | |
>>> from scipy import fftpack | |
>>> sig = np.array([-2, 8, 6, 4, 1, 0, 3, 5], dtype=float) | |
>>> sig_fft = fftpack.rfft(sig) | |
>>> n = sig_fft.size | |
>>> timestep = 0.1 | |
>>> freq = fftpack.rfftfreq(n, d=timestep) | |
>>> freq | |
array([ 0. , 1.25, 1.25, 2.5 , 2.5 , 3.75, 3.75, 5. ]) | |
""" | |
n = operator.index(n) | |
if n < 0: | |
raise ValueError("n = %s is not valid. " | |
"n must be a nonnegative integer." % n) | |
return (np.arange(1, n + 1, dtype=int) // 2) / float(n * d) | |
def next_fast_len(target): | |
""" | |
Find the next fast size of input data to `fft`, for zero-padding, etc. | |
SciPy's FFTPACK has efficient functions for radix {2, 3, 4, 5}, so this | |
returns the next composite of the prime factors 2, 3, and 5 which is | |
greater than or equal to `target`. (These are also known as 5-smooth | |
numbers, regular numbers, or Hamming numbers.) | |
Parameters | |
---------- | |
target : int | |
Length to start searching from. Must be a positive integer. | |
Returns | |
------- | |
out : int | |
The first 5-smooth number greater than or equal to `target`. | |
Notes | |
----- | |
.. versionadded:: 0.18.0 | |
Examples | |
-------- | |
On a particular machine, an FFT of prime length takes 133 ms: | |
>>> from scipy import fftpack | |
>>> import numpy as np | |
>>> rng = np.random.default_rng() | |
>>> min_len = 10007 # prime length is worst case for speed | |
>>> a = rng.standard_normal(min_len) | |
>>> b = fftpack.fft(a) | |
Zero-padding to the next 5-smooth length reduces computation time to | |
211 us, a speedup of 630 times: | |
>>> fftpack.next_fast_len(min_len) | |
10125 | |
>>> b = fftpack.fft(a, 10125) | |
Rounding up to the next power of 2 is not optimal, taking 367 us to | |
compute, 1.7 times as long as the 5-smooth size: | |
>>> b = fftpack.fft(a, 16384) | |
""" | |
# Real transforms use regular sizes so this is backwards compatible | |
return _helper.good_size(target, True) | |
def _good_shape(x, shape, axes): | |
"""Ensure that shape argument is valid for scipy.fftpack | |
scipy.fftpack does not support len(shape) < x.ndim when axes is not given. | |
""" | |
if shape is not None and axes is None: | |
shape = _helper._iterable_of_int(shape, 'shape') | |
if len(shape) != np.ndim(x): | |
raise ValueError("when given, axes and shape arguments" | |
" have to be of the same length") | |
return shape | |