Permutation without duplicates in Python

Question

I have N positions, and each position can be either 0 or 1. I have fixed number of 1s, and I want to permutate these fixed number of 1s in these N positions.

from itertools import permutations
p = [0 for k in xrange(6)]
for k in xrange(0,3):
        p[k] = 1
print(list(permutations(p)))

But above result contains four [0,0,0,1,1,1] in the list. I only want one of them. How can I get rid of these duplicates?

Ry- · Accepted Answer · 2017-05-06T05:58:09.733

You could grab the positions of the 1s instead:

from itertools import combinations


def place_ones(size, count):
    for positions in combinations(range(size), count):
        p = [0] * size

        for i in positions:
            p[i] = 1

        yield p

In action:

>>> list(place_ones(6, 3))
[
    [1, 1, 1, 0, 0, 0],
    [1, 1, 0, 1, 0, 0],
    [1, 1, 0, 0, 1, 0],
    [1, 1, 0, 0, 0, 1],
    [1, 0, 1, 1, 0, 0],
    [1, 0, 1, 0, 1, 0],
    [1, 0, 1, 0, 0, 1],
    [1, 0, 0, 1, 1, 0],
    [1, 0, 0, 1, 0, 1],
    [1, 0, 0, 0, 1, 1],
    [0, 1, 1, 1, 0, 0],
    [0, 1, 1, 0, 1, 0],
    [0, 1, 1, 0, 0, 1],
    [0, 1, 0, 1, 1, 0],
    [0, 1, 0, 1, 0, 1],
    [0, 1, 0, 0, 1, 1],
    [0, 0, 1, 1, 1, 0],
    [0, 0, 1, 1, 0, 1],
    [0, 0, 1, 0, 1, 1],
    [0, 0, 0, 1, 1, 1],
]

Can this approach be used for 3+ kinds of values or is this only for 2 kinds of values? — Fyodor Menshikov, Jan 20 '23 at 19:29

score 6 · Answer 2 · answered May 06 '17 at 05:15

6

Set is perfect for this, as set does not not contain any duplicated element:

set(permutations(p))

answered May 06 '17 at 05:15

Peter Pei Guo

7,770
18
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54

It seems like set can not use index, then how can I grab each permutation out in the set? – JoeJackJessieJames May 06 '17 at 05:17
1

`sorted([e for e in set(permutations(p))])` convert set into list, optionaly sort to impose an ordering – f5r5e5d May 06 '17 at 05:32
7

this is very inefficient, gets hung for high numbers, and high number of duplicates for higher numbers – Sayan Dey Feb 06 '21 at 12:19
@SayanDey agree, but what is the efficient solution? – MrZH6 Dec 11 '21 at 20:46
Refer to @Ry s answer (first) – Sayan Dey Jan 30 '22 at 14:18

hiro protagonist · Answer 3 · 2023-06-24T06:27:54.367

what you are looking for is a multiset permutation.

sympy provides the function multiset_permutations:

from sympy.utilities.iterables import multiset_permutations

n_zeros, n_ones = 3, 3
for perm in multiset_permutations(n_zeros * (0,) + n_ones * (1,)):
    print(perm)

which outputs:

[0, 0, 0, 1, 1, 1]
[0, 0, 1, 0, 1, 1]
[0, 0, 1, 1, 0, 1]
[0, 0, 1, 1, 1, 0]
[0, 1, 0, 0, 1, 1]
[0, 1, 0, 1, 0, 1]
[0, 1, 0, 1, 1, 0]
[0, 1, 1, 0, 0, 1]
[0, 1, 1, 0, 1, 0]
[0, 1, 1, 1, 0, 0]
[1, 0, 0, 0, 1, 1]
[1, 0, 0, 1, 0, 1]
[1, 0, 0, 1, 1, 0]
[1, 0, 1, 0, 0, 1]
[1, 0, 1, 0, 1, 0]
[1, 0, 1, 1, 0, 0]
[1, 1, 0, 0, 0, 1]
[1, 1, 0, 0, 1, 0]
[1, 1, 0, 1, 0, 0]
[1, 1, 1, 0, 0, 0]

Permutation without duplicates in Python

3 Answers3