沿每个轴将n维数组拆分为2D数组

时间:2018-10-22 05:04:04

标签: python arrays numpy

所以说我有一个任意维度的数组(目前,我们将给它三个维度)。

a=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]]])

将数组拆分为所有不同轴上的二维数组(在此例中为n-1,但为2)的最简单方法是什么。

new_array1=[[0, 1, 2, 3,], [4, 5, 6, 7]...[20, 21, 22, 23]]
new_array2=[[0, 4], [1, 5]...[19, 23]]
new_array3=[[0, 8, 16], [1, 9, 17]...[7, 15, 23]]

有没有简单的方法可以对任意维数的数组执行此操作?

5 个答案:

答案 0 :(得分:0)

尝试一下:

b = []                                                 
for i in a:
    b.append([item for sublist in i for item in sublist])

输出:

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

答案 1 :(得分:0)

您可以将numpy的swapaxes()reshape()结合使用来制作任意形状。

例如:

axis = 1
a.swapaxes(axis,2).reshape(-1,a.shape[axis])
> array([[ 0,  4],
   [ 1,  5],
   [ 2,  6],
   [ 3,  7],
   [ 8, 12],
   [ 9, 13],
   [10, 14],
   [11, 15],
   [16, 20],
   [17, 21],
   [18, 22],
   [19, 23]])

axis = 0
a.swapaxes(axis,2).reshape(-1, a.shape[axis])
>array([[ 0,  8, 16],
   [ 4, 12, 20],
   [ 1,  9, 17],
   [ 5, 13, 21],
   [ 2, 10, 18],
   [ 6, 14, 22],
   [ 3, 11, 19],
   [ 7, 15, 23]])

第一个示例可以通过重塑来完成,因为它具有相同的轴:

a.reshape(-1,4)
> 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]])

答案 2 :(得分:0)

您可以使用moveaxis和整形:

>>> from pprint import pprint
>>> import numpy as np
>>> 
>>> a = sum(np.ogrid[:2, :30:10, :400:100])
>>> a
array([[[  0, 100, 200, 300],
        [ 10, 110, 210, 310],
        [ 20, 120, 220, 320]],

       [[  1, 101, 201, 301],
        [ 11, 111, 211, 311],
        [ 21, 121, 221, 321]]])
>>> 
>>> pprint([np.moveaxis(a, j, -1).reshape(-1, d) for j, d  in enumerate(a.shape)])
[array([[  0,   1],
       [100, 101],
       [200, 201],
       [300, 301],
       [ 10,  11],
       [110, 111],
       [210, 211],
       [310, 311],
       [ 20,  21],
       [120, 121],
       [220, 221],
       [320, 321]]),
 array([[  0,  10,  20],
       [100, 110, 120],
       [200, 210, 220],
       [300, 310, 320],
       [  1,  11,  21],
       [101, 111, 121],
       [201, 211, 221],
       [301, 311, 321]]),
 array([[  0, 100, 200, 300],
       [ 10, 110, 210, 310],
       [ 20, 120, 220, 320],
       [  1, 101, 201, 301],
       [ 11, 111, 211, 311],
       [ 21, 121, 221, 321]])]

moveaxis和swapaxes的区别在于,moveaxis保留了其余尺寸的顺序。

答案 3 :(得分:0)

尝试一下:

b = list(map(lambda x:x.tolist(),np.split(a.flatten(),3)))

或者:

b = [i.tolist() for i in np.split(a.flatten(),3)]

或者:

b = list(map(lambda x:np.concatenate(x).tolist(),a))

或者:

b = [np.concatenate(i).tolist() for i in a]

所有情况:

print(b)

输出:

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

答案 4 :(得分:0)

第一个:

a1 = [j for i in range(3) for j in a[i]]

第二个:

a2 = [list(zip(i[0], i[1])) for i in a]

第三个:

a3 = [list(zip(a[0][j], a[1][j], a[2][j])) for j in range(2)]
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