Autocorrelation of a multidimensional array in numpy

Question

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,

Answer 1

For really large arrays it becomes important to have n = 2 ** p, where p is an integer. This will save you huge amounts of time. For example:

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

This might give you wrap-around errors. For large arrays the auto correlation should be insignificant near the edges, though.

Answer 2

Maybe it's just a preference, but I wanted to follow from the definition. I personally find it a bit easier to follow that way. This is my implementation for an arbitrary nd array.

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

Answer 3

Using 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.]]

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.

EDIT: 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.]]

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