好吧，不仅您有3个循环，而且如果您有3个以上的因素，这种方法将无效：)
一种可能的方式：
def genfactors(fdict): factors = set([1]) for factor, count in fdict.iteritems(): for ignore in range(count): factors.update([n*factor for n in factors]) # that line could also be: # factors.update(map(lambda e: e*factor, factors)) return factorsfactors = {2:3, 3:2, 5:1}for factor in genfactors(factors): print factor此外，您还可以避免在内部循环中重复某些工作：如果您的工作集为（1,3），并且想应用到2 ^ 3个因子，那么我们正在做：
(1,3) U (1,3)*2 = (1,2,3,6)
(1,2,3,6) U (1,2,3,6)*2 = (1,2,3,4,6,12)
(1,2,3,4,6,12) U (1,2,3,4,6,12)*2 = (1,2,3,4,6,8,12,24)
看看第二组中有多少个重复项？
但是我们可以改为：
(1,3) + (1,3)*2 = (1,2,3,6)
(1,2,3,6) + ((1,3)*2)*2 = (1,2,3,4,6,12)
(1,2,3,4,6,12) + (((1,3)*2)*2)*2 = (1,2,3,4,6,8,12,24)
没有设置，解决方案看起来更好：
def genfactors(fdict): factors = [1] for factor, count in fdict.iteritems(): newfactors = factors for ignore in range(count): newfactors = map(lambda e: e*factor, newfactors) factors += newfactors return factors
