a BCCf @sbddlZddlZddlmZmZmZddlmm m Z gdZ d ddZ ddZd d ZdS) N)fftshift ifftshiftfftfreq)rrrrfftfreq next_fast_len?cCsBt|}|dkrtd|tjd|dtddt||S)aDFT 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. ]) rz5n = %s is not valid. n must be a nonnegative integer.)Zdtype)operatorindex ValueErrornpZarangeintfloat)ndrQ/var/www/html/django/DPS/env/lib/python3.9/site-packages/scipy/fftpack/_helper.pyr s # rcCs t|dS)a 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) T)_helperZ good_size)targetrrrr6s0rcCs:|dur6|dur6t|d}t|t|kr6td|S)zEnsure that shape argument is valid for scipy.fftpack scipy.fftpack does not support len(shape) < x.ndim when axes is not given. NshapezBwhen given, axes and shape arguments have to be of the same length)rZ_iterable_of_intlenr ndimr )xrZaxesrrr _good_shapeis  r)r)r numpyr Z numpy.fftrrrZscipy.fft._pocketfft.helperZfftZ _pocketffthelperr__all__rrrrrrrs +3