numpy中多维数组的自相关

时间:2010-12-21 19:55:08

标签: python numpy

我有一个二维数组,即一个也是数组的序列数组。对于每个序列,我想计算自相关,因此对于(5,4)数组,我将得到5个结果,或维数(5,7)。

我知道我可以绕过第一个维度,但这很慢,我的最后一招。还有另一种方式吗?

谢谢!

编辑:

根据所选答案加上mtrw的评论,我有以下功能:

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,length是我的代码中的全局变量,因此请务必声明它。我也没有将结果限制为实数,因为我也需要考虑复数。

3 个答案:

答案 0 :(得分:8)

使用FFT-based autocorrelation

import numpy
from numpy.fft import fft, ifft

data = numpy.arange(5*4).reshape(5, 4)
print data
##[[ 0  1  2  3]
## [ 4  5  6  7]
## [ 8  9 10 11]
## [12 13 14 15]
## [16 17 18 19]]
dataFT = fft(data, axis=1)
dataAC = ifft(dataFT * numpy.conjugate(dataFT), axis=1).real
print dataAC
##[[   14.     8.     6.     8.]
## [  126.   120.   118.   120.]
## [  366.   360.   358.   360.]
## [  734.   728.   726.   728.]
## [ 1230.  1224.  1222.  1224.]]

我对你对维度(5,7)的答案的陈述感到有些困惑,所以也许有些重要我不理解。

编辑:根据mtrw的建议,填充版本不包围:

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做到这一点,特别是因为自相关是对称的,我没有利用它。

答案 1 :(得分:2)

对于非常大的数组,使n = 2 ** p变得很重要,其中p是整数。这将为您节省大量时间。例如:

def xcorr(x):
  l = 2 ** int(np.log2(length * 2 - 1))
  fftx = fft(x, n = l, axis = 1)
  ret = ifft(fftx * np.conjugate(fftx), axis = 1)
  ret = fftshift(ret, axes=1)
  return ret

这可能会给你环绕错误。但对于大型数组,自然相关在边缘附近应该是无关紧要的。

答案 2 :(得分:1)

也许这只是一种偏好,但我想从定义中得出结论。我个人觉得这样做更容易一些。这是我对任意nd数组的实现。

from itertools import product
from numpy import empty, roll

def autocorrelate(x): """ Compute the multidimensional autocorrelation of an nd array. input: an nd array of floats output: an nd array of autocorrelations """ # used for transposes t = roll(range(x.ndim), 1) # pairs of indexes # the first is for the autocorrelation array # the second is the shift ii = [list(enumerate(range(1, s - 1))) for s in x.shape] # initialize the resulting autocorrelation array acor = empty(shape=[len(s0) for s0 in ii]) # iterate over all combinations of directional shifts for i in product(*ii): # extract the indexes for # the autocorrelation array # and original array respectively i1, i2 = asarray(i).T x1 = x.copy() x2 = x.copy() for i0 in i2: # clip the unshifted array at the end x1 = x1[:-i0] # and the shifted array at the beginning x2 = x2[i0:] # prepare to do the same for # the next axis x1 = x1.transpose(t) x2 = x2.transpose(t) # normalize shifted and unshifted arrays x1 -= x1.mean() x1 /= x1.std() x2 -= x2.mean() x2 /= x2.std() # compute the autocorrelation directly # from the definition acor[tuple(i1)] = (x1 * x2).mean() return acor