16

我想使用具有任意数量元素的 3 个(或任意数量)列表在 for 循环中进行迭代,例如:

from itertools import izip
for x in izip(["AAA", "BBB", "CCC"], ["M", "Q", "S", "K", "B"], ["00:00", "01:00", "02:00", "03:00"]):
    print x

但它给了我:

('AAA', 'M', '00:00')
('BBB', 'Q', '01:00')
('CCC', 'S', '02:00')

我想:

('AAA', 'M', '00:00')
('AAA', 'M', '01:00')
('AAA', 'M', '02:00')
.
.

('CCC', 'B', '03:00')

其实我想要这个:

for word, letter, hours in [cartesian product of 3 lists above]
    if myfunction(word,letter,hours):
       var_word_letter_hours += 1
4

3 回答 3

24

您想使用列表的产品

from itertools import product

for word, letter, hours in product(["AAA", "BBB", "CCC"], ["M", "Q", "S", "K", "B"], ["00:00", "01:00", "02:00", "03:00"]):

演示:

>>> from itertools import product
>>> for word, letter, hours in product(["AAA", "BBB", "CCC"], ["M", "Q", "S", "K", "B"], ["00:00", "01:00", "02:00", "03:00"]):
...     print word, letter, hours
... 
AAA M 00:00
AAA M 01:00
AAA M 02:00
AAA M 03:00
...
CCC B 00:00
CCC B 01:00
CCC B 02:00
CCC B 03:00
于 2012-09-02T15:39:43.457 回答
7

使用itertools.product

import itertools

for x in itertools.product(["AAA", "BBB", "CCC"],
                           ["M", "Q", "S", "K", "B"],
                           ["00:00", "01:00", "02:00", "03:00"]):
    print x

输出:

('AAA', 'M', '00:00')
('AAA', 'M', '01:00')
...
('CCC', 'B', '02:00')
('CCC', 'B', '03:00')
于 2012-09-02T15:43:17.020 回答
0

只是为了记录,另一个解决方案只是嵌套for循环:

for a in ["AAA", "BBB", "CCC"]:
    for b in ["M", "Q", "S", "K", "B"]:
       for c in ["00:00", "01:00", "02:00", "03:00"]:
           x = (a, b, c)
           # Use x ...

在我看来,这比必须找出/记住itertools.product函数的作用要清楚得多。使用它的唯一充分理由是如果您处于更抽象的情况;例如,您需要将迭代器传递给函数,而不是立即对其进行迭代,或者如果您有一个任意列表列表,您想要获取其笛卡尔积(在这种情况下,您可以使用product(*lists))。

于 2017-01-17T09:24:05.180 回答