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