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| |z � � z S )a� 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. ])
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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)
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