• • 此问题在此处已有答案**：

Contruct 3d array in numpy from existing 2d array（4个答案）
5天前关闭。
我有五个列表，每个列表都相当于一个(4,3)形状的numpy数组，我想用下面提到的方式组合这些列表，得到一个(3,4,5)形状的numpy数组（x）。
以下是五个清单：

import numpy as np

x1 = [[1,11,111,111.1], [1111,11111,111111,111111.1], [0.1, 0.11, 0.111, 0.1111]]
x2 = [[2,22,222,222.2], [2222,22222,222222,222222.2], [0.2, 0.22, 0.222, 0.2222]]
x3 = [[3,33,333,333.3], [3333,33333,333333,333333.3], [0.3, 0.33, 0.333, 0.3333]]
x4 = [[4,44,444,444.4], [4444,44444,444444,444444.4], [0.4, 0.44, 0.444, 0.4444]]
x5 = [[5,55,555,555.5], [5555,55555,555555,555555.5], [0.5, 0.55, 0.555, 0.5555]]

• • 我希望组合数组（x）是这样的**

x[:, :, 0] is same as x1

x[:, :, 1] is same as x2

x[:, :, 2] is same as x3 

x[:, :, 3] is same as x4

x[:, :, 4] is same as x5

• • 如何获得上述功能？**

我已经试过x = np.array([x1, x2, x3, x4, x5]).reshape([3,4,5])了，但是这给了我：

x[:, :, 0] = array([[1.000000e+00, 1.111100e+04, 1.110000e-01, 2.222000e+02],
       [2.000000e-01, 3.300000e+01, 3.333330e+05, 3.333000e-01],
       [4.444000e+03, 4.400000e-01, 5.550000e+02, 5.555555e+05]])

当我想

x[:, :, 0] = array([[1, 11, 111, 111.1],
 [1111, 11111, 111111, 111111.1],
 [0.1, 0.11, 0.111, 0.1111]])