Generate arrays with fixed number of non-zero elements

Question

I have a question about how to generate all possible combinations of an array that satisfy the following condition:

with fixed length N
only M elements among N are 1, and the rest elements are 0.

For example, N = 4 and M = 2, we shall have

1: '0011'
2: '0101'
3: '0110'
4: '1001'
5: '1010'
6: '1100'

The basic idea is to pick up M elements from range(N) and replace the corresponding indices in np.zeros(N) with 1. However, I can only realize this idea by using several for-loops in python, which is inefficient. I'd like to ask whether there is any straightforward solution? Thanks in advance.

Seems like a task for [`itertools.permutations`](https://www.geeksforgeeks.org/permutation-and-combination-in-python/amp/) — B Remmelzwaal, Feb 05 '23 at 21:20
@mkrieger1 Yes, thanks so much. This is actually what I want ! — user8036269, Feb 05 '23 at 21:35

score 1 · Accepted Answer · answered Feb 05 '23 at 21:21

One way is to use itertools to get all possible combinations of the locations of ones and fill N-zero arrays with these ones.

import itertools
import numpy as np

N = 4
M = 2

combs = list(itertools.combinations(range(N), M))
result = [np.zeros(N) for _ in range(len(combs))]

for i, comb in enumerate(combs):
    for j in comb:
        result[i][j] = 1

print(result)
[array([1., 1., 0., 0.]), array([1., 0., 1., 0.]), array([1., 0., 0., 1.]), array([0., 1., 1., 0.]), array([0., 1., 0., 1.]), array([0., 0., 1., 1.])]

score 0 · Answer 2 · answered Feb 05 '23 at 21:24

Generate a string with as many 0s and 1s as you need, then permutate it:

from itertools import permutations 
n = 4
m = 2
item = []
for _ in range(m):
    item.append("1")
for _ in range(n-m):
    item.append("0")

perm = permutations(item)
for x in perm:
    my_str = ""
    for c in x:
        my_str += c
    print(my_str)

Generate arrays with fixed number of non-zero elements

2 Answers2