获取python 2d列表中2点之间对角线上的点列表

Question

For a given square 2d list, say:

foo2d = [
[1, 1, 1, 1, 3],
[1, 3, 0, 3, 4],
[1, 1, 1, 1, 3],
[1, 3, 0, 2, 4],
[1, 3, 1, 3, 4]
]

I would like a list of the diagonals from (0, 1) ( foo2d[1][0] ) to the point diagonal from it, (2, 3) ( foo2d[3][2] ). So in the toy list above, the returned list should be: [1, 0, 1]

I have tried taking advantage of the fact that the slope of that line is 1 (or -1), so an element on the list would have to satisfy:

     pointY - startY
abs(-----------------) == 1
     pointX - startX

and be between the x minimum and x maximum. I don't have the code implementation because a) it broke everything and b) my computer crashed just as I was saving the file, forcing me to revert to a git backup that did not contain that code.

If you need it, I can try to write some pseudo code for this behavior. Thanks for any thoughts you can give me with this issue!

Answer 1

If the slop of line only can be 1 or -1, you can try this:

def get_diagonal_points(matrix, start_x, start_y, end_x, end_y):
    # make start_x <= end_x, if you don't need to check, remove this line
    if start_x > end_x:
        start_x, start_y, end_x, end_y = end_x, end_y, start_x, start_y

    result = []
    slope = (end_y - start_y) // (end_x - start_x)
    for i, j in zip(range(start_x, end_x), range(start_y, end_y, slope)):
        result.append(matrix[i][j])
    result.append(matrix[end_x][end_y])  # add end point
    return result

test and output:

foo2d = [
[1, 1, 1, 1, 3],
[1, 3, 0, 3, 4],
[1, 1, 1, 1, 3],
[1, 3, 0, 2, 4],
[1, 3, 1, 3, 4]
]
print(get_diagonal_points(foo2d, 0, 1, 2, 3))
# [1, 0, 1]
print(get_diagonal_points(foo2d, 0, 4, 3, 1))
# [3, 3, 1, 3]
print(get_diagonal_points(foo2d, 3, 1, 0, 4))
# [3, 3, 1, 3]