[SOLVED] fastest way to sample many random permutations of a numpy array


Unlike many other numpy/random functions, numpy.random.Generator.permutation() doesn’t provide an obvious way to return multiple results in a single function call. Given a (1d) numpy array x, I want to sample n permutations of x (each of length len(x)), and have the result as a numpy array with shape (n, len(x)). A naive way of generating many permutations is np.array([rng.permutation(x) for _ in range(n)]). This is not ideal, mostly because the loop is in Python rather than inside a compiled numpy function.

import numpy as np

rng = np.random.default_rng(1234)
# x is big enough to not want to enumerate all permutations
x = rng.standard_normal(size=20)
n = 10000
perms = np.array([rng.permutation(x) for _ in range(n)])

My use case is for a brute-force search to find permutations that minimise a specific property (constituting a "good enough" search solution). I can calculate the property of interest for each permutation using numpy operations that vectorise/broadcast nicely over the resulting matrix of permutations. It turns out that naively generating the matrix of permutations is the bottleneck in my code. Is there a better way?


You can use rng.permuted instead of rng.permutation and combine it with np.tile so to repeat x multiple times and shuffle each replicates independently. Here is how:

perms = rng.permuted(np.tile(x, n).reshape(n,x.size), axis=1)

This is about 10 times faster on my machine than your initial code.

Answered By – Jérôme Richard

Answer Checked By – David Goodson (BugsFixing Volunteer)

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