Most Efficient Way to Create a Numpy Array Based on Values in Multiple Cells of Another Array

Question

I have an application where I have to process 1000's of 2D arrays. The result of the processed array is based on half of a Kings Move neighborhood in the original array. I'm trying to avoid loops if I can due to speed considerations. So, here is an example Numpy Array:

array([[ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  1.,  1.,  1.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  1.,  5.,  5.,  5.,  5.,  0.,  0.,  0.],
       [ 0.,  1.,  5.,  5.,  1.,  1.,  1.,  1.,  1.,  0.,  0.],
       [ 5.,  5.,  5.,  5.,  1.,  5.,  1.,  1.,  1.,  1.,  0.],
       [ 1.,  1.,  1.,  5.,  1.,  1.,  5.,  5.,  1.,  1.,  0.],
       [ 5.,  1.,  5.,  1.,  1.,  5.,  5.,  5.,  1.,  5.,  0.],
       [ 0.,  5.,  1.,  5.,  1.,  1.,  1.,  1.,  1.,  0.,  0.],
       [ 0.,  0.,  1.,  1.,  1.,  1.,  1.,  1.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  1.,  5.,  5.,  5.,  0.,  0.,  0.,  0.]])

At each element, I want the sum of the cell right above it, the upper right diagonal element, the cell to the immediate right and the lower diagonal. So, using the element at [6][0] I would want to sum 1 + 1 + 1 + 5.

Of course, I also have to handle the edge cases where one of the 4 cells is not there. I have started with the padded zeros on top and far right to manage some of that but I'm stuck right now. Any advice would be much appreciated!

I've suggested a couple of improvements for a similar problem in codereview.stackexchange.com/a/149176/27783; you might also find a lot of convolve2d suggestions here on SO. — hpaulj
– hpaulj, Commented Dec 14, 2016 at 23:44

littleO · Accepted Answer · 2016-12-15 01:33:59Z

1

What you're doing can be viewed as performing a convolution with a particular convolution kernel. Here's a solution using the scipy convolve2d function:

import numpy as np
import scipy as sp
import scipy.signal

x = np.random.randint(5,size=(10,10))

kernel = np.array([[0,1,1],[0,0,1],[0,0,1]])
kernel = np.fliplr(np.flipud(kernel))

check = sp.signal.convolve2d(x,kernel,mode='same')

print x
print check

answered Dec 15, 2016 at 1:33

littleO

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1 Comment

chemnteach Over a year ago

Thank you very much, this is exactly what I was looking for! Your approach to create the kernel in the form that I can see where it starts and it's mirror image applied was very helpful in understanding the process.

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