如何根据第4个通道中的值设置4通道numpy阵列的前3个通道中的值？是否可以将numpy切片作为l值来实现？

给出一个3乘2像素的numpy数组，有4个通道

a = np.arange(24).reshape(3,2,4)

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]]])

我可以选择第4个通道为模3的切片。

px = np.where(0==a[:,:,3]%3)

(array([0, 1], dtype=int64), array([0, 1], dtype=int64))

a[px]

array([[ 0, 1, 2, 3],

[12, 13, 14, 15]])

现在我想将a中这些行中的前3个通道设置为0，这样结果如下所示：

a

array([[[ 0, 0, 0, 3],

[ 4, 5, 6, 7]],

[[ 8, 9, 10, 11],

[ 0, 0, 0, 15]],

[[16, 17, 18, 19],

[20, 21, 22, 23]]])

我试过了

a[px][:,0:3] = 0

但这使得数组保持不变.

解决办法：这是一种方式：

>>> px0, px1 = np.where(0==a[:,:,3]%3)

>>> a[px0, px1, :3] = 0

>>> a

array([[[ 0, 0, 0, 3],

[ 4, 5, 6, 7]],

[[ 8, 9, 10, 11],

[ 0, 0, 0, 15]],

[[16, 17, 18, 19],

[20, 21, 22, 23]]])

要么

>>> px = np.where(0==a[:,:,3]%3)

>>> a[..., :3][px] = 0

>>> a

array([[[ 0, 0, 0, 3],

[ 4, 5, 6, 7]],

[[ 8, 9, 10, 11],

[ 0, 0, 0, 15]],

[[16, 17, 18, 19],

[20, 21, 22, 23]]])

要么

>>> a[(*px, np.s_[:3])] = 0

>>> a

array([[[ 0, 0, 0, 3],

[ 4, 5, 6, 7]],

[[ 8, 9, 10, 11],

[ 0, 0, 0, 15]],

[[16, 17, 18, 19],

[20, 21, 22, 23]]])