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First solution in Clear category for Moore Neighbourhood by vova-ivanyshyn
def count_neighbours(grid, row, col):
neiCount = 0;
for y in range(max(row - 1, 0), min(row + 2, len(grid))):
for x in range(max(col - 1, 0), min(col + 2, len(grid[0]))):
if y == row and x == col:
continue
elif grid[y][x] == 1:
neiCount += 1
return neiCount
if __name__ == '__main__':
#These "asserts" using only for self-checking and not necessary for auto-testing
assert count_neighbours(
((1, 0, 0, 1, 0),
(0, 1, 0, 0, 0),
(0, 0, 1, 0, 1),
(1, 0, 0, 0, 0),
(0, 0, 1, 0, 0),), 1, 2) == 3, "1st example"
assert count_neighbours(((1, 0, 0, 1, 0),
(0, 1, 0, 0, 0),
(0, 0, 1, 0, 1),
(1, 0, 0, 0, 0),
(0, 0, 1, 0, 0),), 0, 0) == 1, "2nd example"
assert count_neighbours(((1, 1, 1),
(1, 1, 1),
(1, 1, 1),), 0, 2) == 3, "Dense corner"
assert count_neighbours(((0, 0, 0),
(0, 1, 0),
(0, 0, 0),), 1, 1) == 0, "Single"
Feb. 4, 2015