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],
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 ]]
If you're looking for any random permuation, you can do it in one line if you transpose columns into rows, permute the rows, then transpose back:
a = np.random.permutation(a.T).T
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)
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.