numpy中多维数组的自相关 [英] Autocorrelation of a multidimensional array in numpy

问题描述

我有一个二维数组，即序列的数组，它也是数组.对于每个序列，我想计算自相关，这样对于(5,4)数组，我将获得5个结果或维数为(5,7)的数组.

I have a two dimensional array, i.e. an array of sequences which are also arrays. For each sequence I would like to calculate the autocorrelation, so that for a (5,4) array, I would get 5 results, or an array of dimension (5,7).

我知道我可以循环浏览第一个维度，但这很慢，这是我的最后选择.还有另一种方法吗?

I know I could just loop over the first dimension, but that's slow and my last resort. Is there another way?

谢谢！

基于选择的答案以及mtrw的评论，我具有以下功能:

Based on the chosen answer plus the comment from mtrw, I have the following function:

def xcorr(x):
  """FFT based autocorrelation function, which is faster than numpy.correlate"""
  # x is supposed to be an array of sequences, of shape (totalelements, length)
  fftx = fft(x, n=(length*2-1), axis=1)
  ret = ifft(fftx * np.conjugate(fftx), axis=1)
  ret = fftshift(ret, axes=1)
  return ret

请注意，在我的代码中length是一个全局变量，因此请务必声明它.我也没有将结果限制为实数，因为我还需要考虑复数.

Note that length is a global variable in my code, so be sure to declare it. I also didn't restrict the result to real numbers, since I need to take into account complex numbers as well.

推荐答案

使用我对您关于答案具有维度(5，7)的陈述感到困惑，所以也许有些我不理解的重要事情.

I'm a little confused by your statement about the answer having dimension (5, 7), so maybe there's something important I'm not understanding.

在mtrw的建议下，它没有环绕的填充版本:

At the suggestion of mtrw, a padded version that doesn't wrap around:

import numpy
from numpy.fft import fft, ifft

data = numpy.arange(5*4).reshape(5, 4)
padding = numpy.zeros((5, 3))
dataPadded = numpy.concatenate((data, padding), axis=1)
print dataPadded
##[[  0.   1.   2.   3.   0.   0.   0.   0.]
## [  4.   5.   6.   7.   0.   0.   0.   0.]
## [  8.   9.  10.  11.   0.   0.   0.   0.]
## [ 12.  13.  14.  15.   0.   0.   0.   0.]
## [ 16.  17.  18.  19.   0.   0.   0.   0.]]
dataFT = fft(dataPadded, axis=1)
dataAC = ifft(dataFT * numpy.conjugate(dataFT), axis=1).real
print numpy.round(dataAC, 10)[:, :4]
##[[   14.     8.     3.     0.     0.     3.     8.]
## [  126.    92.    59.    28.    28.    59.    92.]
## [  366.   272.   179.    88.    88.   179.   272.]
## [  734.   548.   363.   180.   180.   363.   548.]
## [ 1230.   920.   611.   304.   304.   611.   920.]]

必须有一种更有效的方法来执行此操作，尤其是因为自相关是对称的，而我没有利用它.

There must be a more efficient way to do this, especially because autocorrelation is symmetric and I don't take advantage of that.

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