Easiest way to return sum of a matrix's neighbors in numpy

Question

I am trying to make a program that needs a matrix's neighbor(excluding itself) sum ex:

 matrix([[0, 0, 0],
        [1, 0, 1],
        [0, 1, 0]])  

would return:

matrix([[1, 2, 1],
        [1, 3, 1],
        [2, 2, 2]])

I have a working code here but its big and messy and I'm new to numpy so I need some help cleaning it up and optimizing. (I feel like there has to be a better way)

example code:

import numpy as np

def NiSum(m):
    new = []
    for x in range(m.shape[0]-1):
        row = []
        for y in range(m.shape[1]-1):
            Ni = 0
            for a in [1,1],[1,0],[1,-1],[0,1],[0,-1],[-1,1],[-1,0],[-1,-1]:
                Ni += m[x+a[0],y+a[1]]
            row.append(Ni)
        new.append(row)
    return np.matrix(new)


example = np.matrix('0 0 0 0 0 0 0 0; '*3+'0 0 0 1 1 1 0 0; '*3+'0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 ')

NiSum(example)

Thanks for any help !

Answer 1

You are summing all values in that 3x3 neighbourhood, but excluding the element itself. So, we can use Scipy's 2D convolution and subtract that input array/matrix from it for the desired output, like so -

from scipy.signal import convolve2d

convolve2d(a,np.ones((3,3),dtype=int),'same') - a

Sample run -

In [11]: a
Out[11]: 
matrix([[0, 0, 0],
        [1, 0, 1],
        [0, 1, 0]])

In [12]: convolve2d(a,np.ones((3,3),dtype=int),'same') - a
Out[12]: 
matrix([[1, 2, 1],
        [1, 3, 1],
        [2, 2, 2]])

Or simply form a kernel with all ones but zero at the center and use the same 2D convolution -

In [31]: kernel = np.array([[1,1,1],[1,0,1],[1,1,1]])

In [32]: np.asmatrix(convolve2d(a,kernel,'same'))
Out[32]: 
matrix([[1, 2, 1],
        [1, 3, 1],
        [2, 2, 2]])