I would like to find a clean and clever way (in python) to find all permutations of strings of 1s and 0s x chars long. Ideally this would be fast and not require doing too many iterations…

So, for x = 1 I want:
x =2


Right now I have this, which is slow and seems inelegant:

    self.nbits = n
    items = []
    for x in xrange(n+1):
        ones = x
        zeros = n-x
        item = []
        for i in xrange(ones):
        for i in xrange(zeros):
    perms = set()
    for item in items:
        for perm in itertools.permutations(item):
    perms = list(perms)
    self.to_bits = {}
    self.to_code = {}
    for x in enumerate(perms):
        self.to_bits[x[0]] = ''.join([str(y) for y in x[1]])
        self.to_code[''.join([str(y) for y in x[1]])] = x[0]

itertools.product is made for this:

>>> import itertools
>>> ["".join(seq) for seq in itertools.product("01", repeat=2)]
['00', '01', '10', '11']
>>> ["".join(seq) for seq in itertools.product("01", repeat=3)]
['000', '001', '010', '011', '100', '101', '110', '111']

There’s no need to be overly clever for something this simple:

def perms(n):
    if not n:

    for i in xrange(2**n):
        s = bin(i)[2:]
        s = "0" * (n-len(s)) + s
        yield s

print list(perms(5))

You can use itertools.product() for doing this.

import itertools
def binseq(k):
    return [''.join(x) for x in itertools.product('01', repeat=k)]

Python 2.6+:

['{0:0{width}b}'.format(v, width=x) for v in xrange(2**x)]

Kudos to all the clever solutions before me. Here is a low-level, get-you-hands-dirty way to do this:

def dec2bin(n):
    if not n:
        return ''
        return dec2bin(n/2) + str(n%2)

def pad(p, s):
    return "0"*(p-len(s))+s

def combos(n):
    for i in range(2**n):
        print pad(n, dec2bin(i))

That should do the trick