二进制序列的所有排列x位长

Question

我想找到一个干净而聪明的方法(在python中)来查找1s和0s x chars long的字符串的所有排列.理想情况下,这将是快速的,不需要做太多的迭代......

因此,对于x = 1我想要:['0','1'] x = 2 ['00','01','10','11']

等等..

现在我有这个,这很慢,似乎不优雅:

    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]

Answer 1

itertools.product 是为此而做的:

>>> 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']

Answer 2

对于这么简单的事情,没有必要过于聪明:

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))

Answer 3

您可以itertools.product()用于执行此操作。

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

Answer 4

Python 2.6+:

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