norbertpy norbertpy - 6 months ago 12
Python Question

Find unique permutation of a list binary values without generating all possibilities

Forgive my ignorance. I have a brain fart at the moment and unable to come up with a solution. Let's say I have a list of

[1, 1, 0, 0]
. I want to calculate all the four digits binary numbers that have exactly two 1s and two zeros, like:

['0110', '0011', '0101', '1100', '1010', '1001']


This works:

from itertools import permutations

set([''.join(x) for x in list(permutations('0011', 4))])


But this calculate the entire permutations and then discards the duplicate. Meaning, it calculates 24 times but I only need 6. It is much more significant if the collection is
[1, 1, 1, 1, 0, 0, 0, 0]
.

This should be very easy but I just can't wrap my head around it.

Answer

Use itertools.combinations() to find all possible positions for your ones, then construct the numbers using those positions:

def binary(length=4, ones=2):
    result = []
    for positions in combinations(range(length), ones):
        result.append("".join("1" if _ in positions else "0" for _ in range(length)))
    return result

Result:

In [9]: binary()
Out[9]: ['1100', '1010', '1001', '0110', '0101', '0011']

In [10]: binary(5)
Out[10]:
['11000', '10100', '10010', '10001', '01100', '01010', '01001', '00110', '00101', '00011']

In [11]: binary(4,1)
Out[11]: ['1000', '0100', '0010', '0001']

In [12]: binary(4,4)
Out[12]: ['1111']