Convolving a periodic image with python

Question

I want to convolve an n-dimensional image which is conceptually periodic.

What I mean is the following: if I have a 2D image

>>> image2d = [[0,0,0,0],
...            [0,0,0,1],
...            [0,0,0,0]]

and I want to convolve it with this kernel:

>>> kernel = [[ 1,1,1],
...           [ 1,1,1],
...           [ 1,1,1]]

then I want the result to be:

>>> result = [[1,0,1,1],
...           [1,0,1,1],
...           [1,0,1,1]]

How to do this in python/numpy/scipy?

Note that I am not interested in creating the kernel, but mainly the periodicity of the convolution, i.e. the three leftmost ones in the resulting image (if that makes sense).

Answer 1

This is already built in, with scipy.signal.convolve2d's optional boundary='wrap' which gives periodic boundary conditions as padding for the convolution. The mode option here is 'same' to make the output size match the input size.

In [1]: image2d = [[0,0,0,0],
    ...            [0,0,0,1],
    ...            [0,0,0,0]]

In [2]: kernel = [[ 1,1,1],
    ...           [ 1,1,1],
    ...           [ 1,1,1]]

In [3]: from scipy.signal import convolve2d

In [4]: convolve2d(image2d, kernel, mode='same', boundary='wrap')
Out[4]: 
array([[1, 0, 1, 1],
       [1, 0, 1, 1],
       [1, 0, 1, 1]])

The only disadvantage here is that you cannot use scipy.signal.fftconvolve which is often faster.