二进制序列的所有排列x位长
我想找到一种干净巧妙的方法(在python中)来找到1s和0s x chars字符串的所有排列。理想情况下,这将是快速的,并且不需要进行太多的迭代…
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,它很慢,看起来很不雅:
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]
回答:
itertools.product
为此:
>>> 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']
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