重新排序3D数组的行

Question

我有一个像这样的3D数组：

[[[ 0  0  1  0 -1  1  1  0  0]
  [ 0  0 -1  0  1 -1 -1  0  0]
  [ 0  0  1 -2 -1  1  0  0  0]
  [ 0  0 -1  2  1 -1  0  0  0]]

 [[ 0  0  0  2  0  0  1  0  0]
  [ 0  0  0  0  0  0  0  0  0]
  [ 0  0  0 -2  0  0 -1  0  0]
  [ 0  0  0  0  0  0  0  0  0]]

 [[ 0  0  1  0 -1  1  1  0  0]
  [ 0  0 -1  2  1 -1  0  0  0]
  [ 0  0  1 -2 -1  1  0  0  0]
  [ 0  0 -1  0  1 -1 -1  0  0]]

 [[ 0  0  0  0  0  0  0  0  0]
  [ 0  0  0  0  0  0  0  0  0]
  [ 0  0  0  0  0  0  0  0  0]
  [ 0  0  0  0  0  0  0  0  0]]]

我需要对其进行整形，以便每个矩阵中的所有第一行在矩阵中组合在一起，然后是所有第二行，等等。

结果如下：

[[[ 0  0  1  0 -1  1  1  0  0]
  [ 0  0  0  2  0  0  1  0  0]
  [ 0  0  1  0 -1  1  1  0  0]
  [ 0  0  0  0  0  0  0  0  0]]

 [[ 0  0 -1  0  1 -1 -1  0  0]
  [ 0  0  0  0  0  0  0  0  0]
  [ 0  0 -1  2  1 -1  0  0  0]
  [ 0  0  0  0  0  0  0  0  0]]

 [[ 0  0  1 -2 -1  1  0  0  0]
  [ 0  0  0 -2  0  0 -1  0  0]
  [ 0  0  1 -2 -1  1  0  0  0]
  [ 0  0  0  0  0  0  0  0  0]]

 [[ 0  0 -1  2  1 -1  0  0  0]
  [ 0  0  0  0  0  0  0  0  0]
  [ 0  0 -1  0  1 -1 -1  0  0]
  [ 0  0  0  0  0  0  0  0  0]]]

Answer 1

您只需要swap the first two axes数组x。如果数组是ndarray（就像你的那样），则返回一个视图，不会复制任何数据：

>>> x.swapaxes(0,1)

例如：

>>> x = np.arange(27).reshape(3,3,3)
>>> x
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],
        [24, 25, 26]]])

>>> x.swapaxes(0,1)
array([[[ 0,  1,  2],
        [ 9, 10, 11],
        [18, 19, 20]],

       [[ 3,  4,  5],
        [12, 13, 14],
        [21, 22, 23]],

       [[ 6,  7,  8],
        [15, 16, 17],
        [24, 25, 26]]])