all permutations of a binary sequence x bits long


Question

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: ['0','1'] x =2 ['00','01','10','11']

etc..

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):
            item.append(1)
        for i in xrange(zeros):
            item.append(0)
        items.append(item)
    perms = set()
    for item in items:
        for perm in itertools.permutations(item):
            perms.add(perm)
    perms = list(perms)
    perms.sort()
    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]
1
24
2/8/2011 12:33:51 AM

Accepted Answer

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']
58
2/8/2011 12:31:14 AM

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

def perms(n):
    if not n:
        return

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

print list(perms(5))

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