151 lines
4.8 KiB
Python
151 lines
4.8 KiB
Python
from __future__ import division, print_function, absolute_import
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import operator
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from numpy import arange
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from numpy.fft.helper import fftshift, ifftshift, fftfreq
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from bisect import bisect_left
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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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>>> 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 (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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>>> min_len = 10007 # prime length is worst case for speed
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>>> a = np.random.randn(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.helper.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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hams = (8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36, 40, 45, 48,
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50, 54, 60, 64, 72, 75, 80, 81, 90, 96, 100, 108, 120, 125, 128,
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135, 144, 150, 160, 162, 180, 192, 200, 216, 225, 240, 243, 250,
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256, 270, 288, 300, 320, 324, 360, 375, 384, 400, 405, 432, 450,
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480, 486, 500, 512, 540, 576, 600, 625, 640, 648, 675, 720, 729,
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750, 768, 800, 810, 864, 900, 960, 972, 1000, 1024, 1080, 1125,
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1152, 1200, 1215, 1250, 1280, 1296, 1350, 1440, 1458, 1500, 1536,
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1600, 1620, 1728, 1800, 1875, 1920, 1944, 2000, 2025, 2048, 2160,
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2187, 2250, 2304, 2400, 2430, 2500, 2560, 2592, 2700, 2880, 2916,
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3000, 3072, 3125, 3200, 3240, 3375, 3456, 3600, 3645, 3750, 3840,
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3888, 4000, 4050, 4096, 4320, 4374, 4500, 4608, 4800, 4860, 5000,
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5120, 5184, 5400, 5625, 5760, 5832, 6000, 6075, 6144, 6250, 6400,
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6480, 6561, 6750, 6912, 7200, 7290, 7500, 7680, 7776, 8000, 8100,
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8192, 8640, 8748, 9000, 9216, 9375, 9600, 9720, 10000)
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if target <= 6:
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return target
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# Quickly check if it's already a power of 2
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if not (target & (target-1)):
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return target
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# Get result quickly for small sizes, since FFT itself is similarly fast.
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if target <= hams[-1]:
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return hams[bisect_left(hams, target)]
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match = float('inf') # Anything found will be smaller
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p5 = 1
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while p5 < target:
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p35 = p5
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while p35 < target:
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# Ceiling integer division, avoiding conversion to float
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# (quotient = ceil(target / p35))
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quotient = -(-target // p35)
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# Quickly find next power of 2 >= quotient
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p2 = 2**((quotient - 1).bit_length())
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N = p2 * p35
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if N == target:
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return N
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elif N < match:
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match = N
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p35 *= 3
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if p35 == target:
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return p35
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if p35 < match:
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match = p35
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p5 *= 5
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if p5 == target:
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return p5
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if p5 < match:
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match = p5
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return match
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