Efficient Way to create a 2d array of random shuffling of a range of numbers

Question

I need to create a N x M numpy array of randomly shuffled numbers 0 to N. That is to say that each row of the matrix has exactly one of each numbers 0 to N in a random order with no repeating elements and every number is represented.

The best I've been able to come up with so far is (for N = 10 and M = 5)

import random
import numpy as np

a = np.array([random.sample(range(10), 10) for x in range(5)])

which gives me

[[5 9 1 3 8 2 6 4 0 7]
 [4 8 5 2 9 3 7 6 0 1]
 [8 4 6 7 9 2 0 5 1 3]
 [3 5 4 9 2 0 6 7 1 8]
 [6 0 4 7 3 2 1 8 5 9]]

My current solution does work, but I'd like to only have one random library if possible (I'm using numpy elsewhere so a numpy only solution would be ideal). I'm assuming numpy has an easier way to do this, but I'm currently missing it.

import numpy as np, of course.... then ... np.random.randint(0, 10, size=(5, 10)) — NaN, May 07 '18 at 15:38
@JohnColeman that was my initial thought, but I wasn't sure how to create the array so that it has the same random 0->N but in different orders — Grant Williams, May 07 '18 at 15:38
@NaN, the problem is i need it to be the range(0, N) but shuffled. I.E. no repeated elements and every element is there exactly once. — Grant Williams, May 07 '18 at 15:39
My current solution does work, but i'd like to only use one random library if possible (preferably numpy) — Grant Williams, May 07 '18 at 15:42
It doesn't reduce the number of library calls you need to make, but `a = np.array([np.random.permutation(10) for _ in range(5)])` might be cleaner. — chepner, May 07 '18 at 15:55
Thanks @Divakar, that was a perfect answer. Guess i didn't find the correct search terms when i was digging around — Grant Williams, May 07 '18 at 17:55

score 3 · Accepted Answer · answered May 07 '18 at 15:43

If the issue is simply to only use one random library (as you said in your comments), you could just use numpy.random.choice with the argument replace=False, which ensures each element is not repeated:

import numpy as np

a = np.stack([np.random.choice(range(10), 10, replace=False) for _ in range(5)])

>>> a
array([[0, 9, 3, 5, 8, 7, 1, 4, 6, 2],
       [5, 6, 8, 3, 0, 4, 7, 9, 2, 1],
       [7, 4, 9, 5, 0, 1, 6, 8, 3, 2],
       [9, 0, 3, 8, 5, 7, 6, 1, 4, 2],
       [5, 6, 0, 1, 3, 4, 9, 8, 7, 2]])

Efficient Way to create a 2d array of random shuffling of a range of numbers

1 Answers1