我有一个矩阵(由列表列表组成),看起来像:
matrix = [[0, 0, 0, 0, 5],
[0, 0, 0, 4, 0],
[2, 0, 3, 0, 0],
[3, 2, 0, 2, 0],
[1, 0, 2, 0, 1]]
[我正在努力创建的函数将把这个矩阵作为输入,以及矩阵中的一个位置(以元组表示),并返回与该点相交的两个对角线(不使用NumPy )。例如,
def getDiagonal(matrix, pos)
(row, col) = pos
# Smart diagonal finder code #
return (diag1, diag2)
diagonals = getDiagonals(matrix, (1, 1))
print(diagnonal[0])
print(diagnonal[1])
print(' ')
diagonals = getDiagonals(matrix, (1, 3))
print(diagnonal[0])
print(diagnonal[1])
预期输出:
OUT: [5, 4, 3, 2, 1]
OUT: [2, 2, 2]
OUT:
OUT: [0, 2, 2]
OUT: [0, 0, 3, 2, 1]
值得指出的是,我不介意对角线返回的元素从哪个方向(从上到下或从上到下)。它们可以轻松地以一种方式完成,并在需要时使用reverse()
进行验证。
[我看过类似的问题,例如this one,但这主要涉及获取矩阵的前导对角线,并且提供的信息很少,无法获取某个点的对角线。
非常感谢您的帮助和事先评论!
有点混乱,但是我认为是这样:
def getDiagonals(matrix, pos):
row, col = pos
nrows = len(matrix)
ncols = len(matrix[0]) if nrows > 0 else 0
# First diagonal
d1_i, d1_j = nrows - 1 - max(row - col, 0), max(col - row, 0)
d1_len = min(d1_i + 1, ncols - d1_j)
diag1 = [matrix[d1_i - k][d1_j + k] for k in range(d1_len)]
# Second diagonal
t = min(row, ncols - col - 1)
d2_i, d2_j = nrows - 1 - row + t, col + t
d2_len = min(d2_i, d2_j) + 1
diag2 = [matrix[d2_i - k][d2_j - k] for k in range(d2_len)]
return (diag1, diag2)
# Test
matrix = [[0, 0, 0, 0, 5],
[0, 0, 0, 4, 0],
[2, 0, 3, 0, 0],
[3, 2, 0, 2, 0],
[1, 0, 2, 0, 1]]
diagonals = getDiagonals(matrix, (1, 1))
print(diagonals[0])
# [1, 2, 3, 4, 5]
print(diagonals[1])
# [2, 2, 2]
diagonals = getDiagonals(matrix, (1, 3))
print(diagonals[0])
# [2, 2, 0]
print(diagonals[1])
# [1, 2, 3, 0, 0]
diagonals = getDiagonals(matrix, (2, 2))
print(diagonals[0])
# [1, 2, 3, 4, 5]
print(diagonals[1])
# [1, 2, 3, 0, 0]