stride_tricks 做到这一点：
>>> import numpy as np
>>>
>>> def stripe(a):
... a = np.asanyarray(a)
... *sh, i, j = a.shape
... assert i >= j
... *st, k, m = a.strides
... return np.lib.stride_tricks.as_strided(a, (*sh, i-j+1, j), (*st, k, k+m))
...
>>> a = np.arange(24).reshape(6, 4)
>>> a
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[12, 13, 14, 15],
[16, 17, 18, 19],
[20, 21, 22, 23]])
>>> stripe(a)
array([[ 0, 5, 10, 15],
[ 4, 9, 14, 19],
[ 8, 13, 18, 23]])
如果a是一个数组，则会创建一个可写的视图，这意味着，如果您觉得这样，可以执行以下操作：
>>> stripe(a)[...] *= 10
>>> a
array([[ 0, 1, 2, 3],
[ 40, 50, 6, 7],
[ 80, 90, 100, 11],
[ 12, 130, 140, 150],
[ 16, 17, 180, 190],
[ 20, 21, 22, 230]])
更新：可以相同的方式获得从左下到右上的条纹。仅很小的复杂性：它不基于与原始数组相同的地址。
>>> def reverse_stripe(a):
... a = np.asanyarray(a)
... *sh, i, j = a.shape
... assert i >= j
... *st, k, m = a.strides
... return np.lib.stride_tricks.as_strided(a[..., j-1:, :], (*sh, i-j+1, j), (*st, k, m-k))
...
>>> a = np.arange(24).reshape(6, 4)
>>> reverse_stripe(a)
array([[12, 9, 6, 3],
[16, 13, 10, 7],
[20, 17, 14, 11]])
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