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]
itertools.product 为此:
itertools.product
>>> 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']