Python：一维数组循环卷积

Question

I wonder if there's a function in numpy/scipy for 1d array circular convolution. The scipy.signal.convolve() function only provides "mode" but not "boundary", while the signal.convolve2d() function needs 2d array as input.

I need to do this to compare open vs circular convolution as part of a time series homework.

Answer 1

By convolution theorem, you can use Fourier Transform to get circular convolution.

import numpy as np
def conv_circ( signal, ker ):
    '''
        signal: real 1D array
        ker: real 1D array
        signal and ker must have same shape
    '''
    return np.real(np.fft.ifft( np.fft.fft(signal)*np.fft.fft(ker) ))

Answer 2

Since this is for homework, I'm leaving out a few details.

By the definition of convolution , if you append a signal a to itself, then the convolution between aa and b will contain inside the cyclic convolution of a and b .

Eg, consider the following:

import numpy as np
from scipy import signal

%pylab inline

a = np.array([1] * 10)
b = np.array([1] * 10)

plot(signal.convolve(a, b));

That is the standard convolution. Now this, however

plot(signal.convolve(a, np.concatenate((b, b))));

In this last figure, try to see where is the result of the circular convolution, and how to generalize this.

Answer 3

Code for this that you can copy-paste, in the spirit of StackOverflow:

n = a.shape[0]
np.convolve(np.tile(a, 2), b)[n:2 * n]

This assumes that a, b have the same shape.