4 回答
![?](http://img1.sycdn.imooc.com/545863e80001889e02200220-100-100.jpg)
TA贡献1821条经验 获得超6个赞
import itertools
>>> for i in itertools.product([1,2,3],['a','b'],[4,5]):
... print i
...
(1, 'a', 4)
(1, 'a', 5)
(1, 'b', 4)
(1, 'b', 5)
(2, 'a', 4)
(2, 'a', 5)
(2, 'b', 4)
(2, 'b', 5)
(3, 'a', 4)
(3, 'a', 5)
(3, 'b', 4)
(3, 'b', 5)
>>>
![?](http://img1.sycdn.imooc.com/5333a0780001a6e702200220-100-100.jpg)
TA贡献1810条经验 获得超4个赞
对于Python 2.5及更早版本:
>>> [(a, b, c) for a in [1,2,3] for b in ['a','b'] for c in [4,5]]
[(1, 'a', 4), (1, 'a', 5), (1, 'b', 4), (1, 'b', 5), (2, 'a', 4),
(2, 'a', 5), (2, 'b', 4), (2, 'b', 5), (3, 'a', 4), (3, 'a', 5),
(3, 'b', 4), (3, 'b', 5)]
这是一个递归版本product()(只是一个插图):
def product(*args):
if not args:
return iter(((),)) # yield tuple()
return (items + (item,)
for items in product(*args[:-1]) for item in args[-1])
例:
>>> list(product([1,2,3], ['a','b'], [4,5]))
[(1, 'a', 4), (1, 'a', 5), (1, 'b', 4), (1, 'b', 5), (2, 'a', 4),
(2, 'a', 5), (2, 'b', 4), (2, 'b', 5), (3, 'a', 4), (3, 'a', 5),
(3, 'b', 4), (3, 'b', 5)]
>>> list(product([1,2,3]))
[(1,), (2,), (3,)]
>>> list(product([]))
[]
>>> list(product())
[()]
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