Question

我想基于一个numpy 3d数组中的值创建一个numpy 2d数组，使用另一个numpy 2d数组确定在轴3中使用哪个元素。

import numpy as np
#--------------------------------------------------------------------
arr_3d = np.arange(2*3*4).reshape(2,3,4)
print('arr_3d shape=', arr_3d.shape, '\n', arr_3d)
arr_2d = np.array(([3,2,0], [2,3,2]))
print('\n', 'arr_2d shape=', arr_2d.shape, '\n', arr_2d)
res_2d = arr_3d[:, :, 2]
print('\n','res_2d example using element 2 of each 3rd axis...\n', res_2d)
res_2d = arr_3d[:, :, 3]
print('\n','res_2d example using element 3 of each 3rd axis...\n', res_2d)

结果...

arr_3d shape= (2, 3, 4) 
 [[[ 0  1  2  3]
  [ 4  5  6  7]
  [ 8  9 10 11]]

 [[12 13 14 15]
  [16 17 18 19]
  [20 21 22 23]]]

 arr_2d shape= (2, 3) 
 [[3 2 0]
 [2 3 2]]

 res_2d example using element 2 of each 3rd axis...
 [[ 2  6 10]
 [14 18 22]]

 res_2d example using element 3 of each 3rd axis...
 [[ 3  7 11]
 [15 19 23]]

第2个示例结果显示了如果我使用轴3的第2个元素，然后使用第3个元素，则会得到什么。但是我想从arr_2d指定的arr_3d中获得该元素。所以...

- res_2d[0,0] would use the element 3 of arr_3d axis 3
- res_2d[0,1] would use the element 2 of arr_3d axis 3
- res_2d[0,2] would use the element 0 of arr_3d axis 3
etc

所以res_2d应该看起来像这样...

[[3 6 8]
[14 19 22]]

我尝试使用此行来获取arr_2d条目，但结果为4维数组，而我需要2维数组。

res_2d = arr_3d[:, :, arr_2d[:,:]]

Answer 1

通过花式索引和广播得到的结果的形状是索引数组的形状。您需要为arr_3d

的每个轴传递2d数组

ax_0 = np.arange(arr_3d.shape[0])[:,None]
ax_1 = np.arange(arr_3d.shape[1])[None,:]

arr_3d[ax_0, ax_1, arr_2d]

Out[1127]:
array([[ 3,  6,  8],
       [14, 19, 22]])

Answer 2

In [107]: arr_3d = np.arange(2*3*4).reshape(2,3,4)                                                           
In [108]: arr_2d = np.array(([3,2,0], [2,3,2]))                                                              
In [109]: arr_2d.shape                                                                                       
Out[109]: (2, 3)
In [110]: arr_3d[[[0],[1]],[0,1,2],arr_2d]                                                                   
Out[110]: 
array([[ 3,  6,  8],
       [14, 19, 22]])

[[0],[1]]，[0,1,2]相互广播以索引与（arr_2d）大小相同的（2,3）块。

ix_可用于构造这两个索引：

In [114]: I,J = np.ix_(range(2), range(3))                                                                   
In [115]: I,J                                                                                                
Out[115]: 
(array([[0],
        [1]]), array([[0, 1, 2]]))
In [116]: arr_3d[I, J, arr_2d]                                                                               
Out[116]: 
array([[ 3,  6,  8],
       [14, 19, 22]])

用2d数组索引3d numpy数组

2 个答案: