counting surrounding true's in numpy matrix (python)

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counting surrounding true's in numpy matrix (python)



I want to make something similiar to minesweeper



input matrix:


matrix = [[true, false, false],
[false, true, false],
[false, false, false]]



if a bomb is on the field I do not count it as bomb is surrounding.



I thought about doing it with numpy convolve somehow but I am struggling with how to go through the matrix and always check the left, top, right and bottom fields of the actual field (in the case of a border I check "empty" fields which is 0 for sure)



enter image description here





2d convolution with a donut-shaped seed.
– Daniel F
Aug 10 at 8:57





no it doesn't have to be numpy, but I thought that this would be the best solution, just not sure how to get it to work. @DanielF hmm how that?
– Dominik Lemberger
Aug 10 at 8:58




3 Answers
3



Here is a solution using scipy.signal.convolve2d :


scipy.signal.convolve2d


import scipy
import numpy as np

# Input matrix, can be left as boolean
matrix = np.array([[True, False, False],
[False, True, False],
[False, False, False]])

# Our dougnut filter
W = np.array([[1, 1, 1],
[1, 0, 1],
[1, 1, 1]])

# Single convolve
res = convolve2d(matrix, W, 'same')



We get the exact result:


res
array([[1, 2, 1],
[2, 1, 1],
[1, 1, 1]])





Dang, got FGITW'd. You can do all the padding and slicing using the boundary and mode parameters of convolve2d though.
– Daniel F
Aug 10 at 9:14


boundary


mode


convolve2d





thanks, exactly what I was looking for :P
– Dominik Lemberger
Aug 10 at 9:19





mode=same is enough, doesnt need the boundary tho :P - boundary wrap makes the output actually wrong as it makes 1,2,1 to 2,1,2
– Dominik Lemberger
Aug 10 at 9:24






Updated! no more boundary
– ibarrond
Aug 10 at 9:52



using only numpy, including nd_window from here


numpy


nd_window


m_pad = np.pad(matrix, ((1,1),(1,1)), 'constant', constant_values=(False, False))
filter = np.array([[1,1,1],[1,0,1],[1,1,1]], dtype = bool)
adj_matrix = np.einsum('ijkl,kl->ij', nd_window(m_pad, 3), filter)



Also, you can use scipy.signal.convolve2d


scipy.signal.convolve2d


adj_matrix = convolve2d(matrix, filter, 'same', fillvalue = False)



To keep it simple : using numpy you can simply use the sum() method (https://docs.scipy.org/doc/numpy-1.13.0/reference/generated/numpy.matrix.sum.html).


numpy


sum()



np.matrix(matrix).sum() will count all the True in your matrix.


np.matrix(matrix).sum()


True



So np.matrix(matrix).sum() - matrix[1][1] should return the number of mine around a cell.


np.matrix(matrix).sum() - matrix[1][1]



And for me the easiest way to remove the border problem is to complete your matrix with False cells all around it.


False






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