让两条对角线都通过矩阵项 (i,j) 的 Pythonic 方法

获取矩阵中通过条目的对角线元素列表的Pythonic方法是什么(i,j)?

例如，给定一个矩阵：

[1  2  3   4  5]
[6  7  8   9 10]
[11 12 13 14 15]
[16 17 18 19 20]
[21 22 23 24 25]

和一个条目，比如说，(1,3)（代表元素9）如何让对角线上的元素穿过9以Python式的方式？基本上，[3,9,15] and [5,9,13,17,21] both.

Using np.diagonal https://docs.scipy.org/doc/numpy-1.10.0/reference/generated/numpy.diagonal.html有一点偏移逻辑。

import numpy as np

lst = np.array([[1,  2,  3,   4,  5],
                [6,  7,  8,   9, 10],
                [11, 12, 13, 14, 15],
                [16, 17, 18, 19, 20],
                [21, 22, 23, 24, 25]])


i, j = 1, 3
major = np.diagonal(lst, offset=(j - i))
print(major)
array([ 3,  9, 15])

minor = np.diagonal(np.rot90(lst), offset=-lst.shape[1] + (j + i) + 1)
print(minor)
array([ 5,  9, 13, 17, 21])

指数i and j是行和列。通过指定偏移量，numpy知道从哪里开始选择对角线的元素。

For the major diagonal, You want to start collecting from 3 in the first row. So you need to take the current column index and subtract it by the current row index, to figure out the correct column index at the 0^th row. Similarly for the minor diagonal, where the array is flipped (rotated by 90˚) and the process repeats.