Best way to permute contents of each column in numpy

If your array is multi-dimensional, np.random.permutation permutes along the first axis (columns) by default:

>>> np.random.permutation(arr)array([[ 4,  5,  6,  7],       [ 8,  9, 10, 11],       [ 0,  1,  2,  3],       [12, 13, 14, 15]])

However, this shuffles the row indices and so each column has the same (random) ordering.

The simplest way of shuffling each column independently could be to loop over the columns and use np.random.shuffle to shuffle each one in place:

for i in range(arr.shape[1]):    np.random.shuffle(arr[:,i])

Which gives, for instance:

array([[12,  1, 14, 11],       [ 4,  9, 10,  7],       [ 8,  5,  6, 15],       [ 0, 13,  2,  3]])

This method can be useful if you have a very large array which you don't want to copy because the permutation of each column is done in place. On the other hand, even simple Python loops can be very slow and there are quicker NumPy methods such as the one provided by @jme.

Here's another way of doing this:

def permute_columns(x):    ix_i = np.random.sample(x.shape).argsort(axis=0)    ix_j = np.tile(np.arange(x.shape[1]), (x.shape[0], 1))    return x[ix_i, ix_j]

A quick test:

>>> x = np.arange(16).reshape(4,4)>>> permute_columns(x)array([[ 8,  9,  2,  3],       [ 0,  5, 10, 11],       [ 4, 13, 14,  7],       [12,  1,  6, 15]])

The idea is to generate a bunch of random numbers, then argsort them within each column independently. This produces a random permutation of each column's indices.

Note that this has sub-optimal asymptotic time complexity, since the sort takes time O(n m log m) for an array of size m x n. But since Python's for loops are pretty slow, you actually get better performance for all but very tall matrices.