不得不反复遍历列表并不是很好的恕我直言。
twos, threes = countmatching(xrange(1,10),
lambda a: a % 2 == 0,
lambda a: a % 3 == 0)
起点将是这样的：
def countmatching(iterable, *predicates):
v = [0] * len(predicates)
for e in iterable:
for i,p in enumerate(predicates):
if p(e):
v[i] += 1
return tuple(v)
顺便说一句，“ itertools食谱”有一个类似alt4的食谱。
def quantify(seq, pred=None):
"Count how many times the predicate is true in the sequence"
return sum(imap(pred, seq))
602392714
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