重新排列 numpy 二维数组的列

Question

Is there a way to change the order of the columns in a numpy 2D array to a new and arbitrary order? For example, I have an array

array([[10, 20, 30, 40, 50],
       [ 6,  7,  8,  9, 10]])

and I want to change it into, say

array([[10, 30, 50, 40, 20],
       [ 6,  8, 10,  9,  7]])

by applying the permutation

0 -> 0
1 -> 4
2 -> 1
3 -> 3
4 -> 2

on the columns. In the new matrix, I therefore want the first column of the original to stay in place, the second to move to the last column and so on.

Is there a numpy function to do it? I have a fairly large matrix and expect to get even larger ones, so I need a solution that does this quickly and in place if possible (permutation matrices are a no-go)

Thank you.

Answer 1

This is possible in O(n) time and O(n) space using fancy indexing:

>>> import numpy as np
>>> a = np.array([[10, 20, 30, 40, 50],
...               [ 6,  7,  8,  9, 10]])
>>> permutation = [0, 4, 1, 3, 2]
>>> idx = np.empty_like(permutation)
>>> idx[permutation] = np.arange(len(permutation))
>>> a[:, idx]  # return a rearranged copy
array([[10, 30, 50, 40, 20],
       [ 6,  8, 10,  9,  7]])
>>> a[:] = a[:, idx]  # in-place modification of a

Note that a[:, idx] is returning a copy , not a view. An O(1) -space solution is not possible in the general case, due to how numpy arrays are strided in memory.

Answer 2

The easiest way in my opinion is:

a = np.array([[10, 20, 30, 40, 50],
              [6,  7,  8,  9,  10]])
print(a[:, [0, 2, 4, 3, 1]])

the result is:

[[10 30 50 40 20]
 [6  8  10 9  7 ]]

Answer 3

I have a matrix based solution for this, by post-multiplying a permutation matrix to the original one. This changes the position of the elements in original matrix

import numpy as np

a = np.array([[10, 20, 30, 40, 50],
       [ 6,  7,  8,  9, 10]])

# Create the permutation matrix by placing 1 at each row with the column to replace with
your_permutation = [0,4,1,3,2]

perm_mat = np.zeros((len(your_permutation), len(your_permutation)))

for idx, i in enumerate(your_permutation):
    perm_mat[idx, i] = 1

print np.dot(a, perm_mat)

Answer 4

如果您正在寻找任何随机排列，如果您将列转置为行，排列行，然后转回，则可以在一行中完成：

a = np.random.permutation(a.T).T