如何轉置 3d/2d 列表？

Question

這是我的清單：

[[[22, 13, 17, 11,  0],
[8,  2, 23,  4, 24],
[21,  9, 14, 16,  7],
[6, 10,  3, 18,  5],
[1, 12, 20, 15, 19]],

[[3, 15,  0,  2, 22],
[9, 18, 13, 17,  5],
[19,  8,  7, 25, 23],
[20, 11, 10, 24,  4],
[14, 21, 16, 12,  6]],

[[14, 21, 17, 24,  4],
[10, 16, 15,  9, 19],
[18,  8, 23, 26, 20],
[22, 11, 13,  6,  5],
[2,  0, 12,  3,  7]]]

我想要的轉置列表是這樣的：

[[[22,  8, 21,  6,  1],
[13,  2,  9,  10, 12],
[17,  23, 14,  3, 20],
[11,  4,  16, 18, 15],
[0, 24,  7,  5, 19]],

[[3, 9,  19,  20, 14],
[15, 18,  8, 11, 21],
[0,  13,  7, 10, 16],
[2, 11, 10, 24,  12],
[22, 21, 16,  4,  6]],

[[14, 10, 18, 22,  2],
[21, 16,  8,  11, 0],
[17,  15, 23, 13, 12],
[24,  9, 26,  6,  3],
[4,  19, 20,  5,  7]]]

我該怎么做？ 任何幫助將不勝感激（不確定這是 3d 列表還是 2d 列表）

每個列表項由一個空行分隔，我將其稱為一個組（5 個，因為每個組中有 5 個不同的列表）我嘗試使用 Numpy 的轉置、arange 等。但是在使用 Numpy 的轉置、arange 等; 以 3 個一組的形式返回所有內容。而且它們的順序也錯誤。

Answer 1

只需切換第一和第二軸：

import numpy as np

a = np.array(
[[[22, 13, 17, 11,  0],
[8,  2, 23,  4, 24],
[21,  9, 14, 16,  7],
[6, 10,  3, 18,  5],
[1, 12, 20, 15, 19]],
[[3, 15,  0,  2, 22],
[9, 18, 13, 17,  5],
[19,  8,  7, 25, 23],
[20, 11, 10, 24,  4],
[14, 21, 16, 12,  6]],

[[14, 21, 17, 24,  4],
[10, 16, 15,  9, 19],
[18,  8, 23, 26, 20],
[22, 11, 13,  6,  5],
[2,  0, 12,  3,  7]]])

print(np.swapaxes(a, 1, 2))

Answer 2

可以使用np.swapaxes(a, 1, 2)

https://numpy.org/doc/stable/reference/generated/numpy.swapaxes.html

這也適用於列表，而不僅僅是 numpy arrays。

Answer 3

其他答案涉及相關的 numpy 方法。 這是一個列表理解（只是為了好玩）以獲得與np.swapaxes(lst, 1, 2)相同的結果：

out = [[[row[i] for row in l] for i in range(len(l))] for l in my_list]

Output：

[[[22, 8, 21, 6, 1],
  [13, 2, 9, 10, 12],
  [17, 23, 14, 3, 20],
  [11, 4, 16, 18, 15],
  [0, 24, 7, 5, 19]],
 [[3, 9, 19, 20, 14],
  [15, 18, 8, 11, 21],
  [0, 13, 7, 10, 16],
  [2, 17, 25, 24, 12],
  [22, 5, 23, 4, 6]],
 [[14, 10, 18, 22, 2],
  [21, 16, 8, 11, 0],
  [17, 15, 23, 13, 12],
  [24, 9, 26, 6, 3],
  [4, 19, 20, 5, 7]]]