Python 重新排列 numpy 二维数组的列
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Rearrange columns of numpy 2D array
提问by 5xum
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
有没有办法将 numpy 2D 数组中列的顺序更改为新的任意顺序？例如，我有一个数组
array([[10, 20, 30, 40, 50],
[ 6, 7, 8, 9, 10]])
and I want to change it into, say
我想把它改成，说
array([[10, 30, 50, 40, 20],
[ 6, 8, 10, 9, 7]])
by applying the permutation
通过应用置换
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.
在列上。因此，在新矩阵中，我希望原始矩阵的第一列保持原位，第二列移动到最后一列，依此类推。
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)
是否有一个 numpy 函数来做到这一点？我有一个相当大的矩阵，并希望得到更大的矩阵，所以我需要一个解决方案，如果可能的话，它可以快速且适当地执行此操作（置换矩阵是不可行的）
Thank you.
谢谢你。
采纳答案by wim
This is possible in O(n)time and O(n)space using fancy indexing:
这在O(n)时间和O(n)空间中使用花式索引是可能的：
>>> 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.
请注意，a[:, idx]在返回的副本，而不是一个视图。由于 numpy 数组在内存中是如何跨步的，因此在一般情况下不可能使用O(1)空间解决方案。
回答by Parth Paritosh
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
我有一个基于矩阵的解决方案，通过将置换矩阵后乘以原始矩阵。这会改变原始矩阵中元素的位置
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)
