^Mh ^ddlZddlZddlmZmZmZddlmcm cm Z gdZ ddZ dZdZdS) N)fftshift ifftshiftfftfreq)rrrrfftfreq next_fast_len?ctj|}|dkrtd|dtjd|dzt dzt ||zz 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. ]) rzn = z/ is not valid. n must be a nonnegative integer.)dtype)operatorindex ValueErrornparangeintfloat)nds U/var/www/html/test/jupyter/venv/lib/python3.11/site-packages/scipy/fftpack/_helper.pyrr svF qA1uu<<<<== = IaQc * * *a /5Q<< ??c,tj|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)_helper good_size)targets rrr6s`  VT * **rc|K|Itj|d}t|tj|krt d|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)r_iterable_of_intlenrndimr)xraxess r _good_shaper#isU  T\(88 u:: # #>?? ? Lr)r)r numpyr numpy.fftrrrscipy.fft._pocketfft.helperfft _pocketffthelperr__all__rrr#rrr,s2222222222------------ K K K(@(@(@(@V0+0+0+f     r