如何在考虑两个列表时随机选择一个列表中的项？

Question

抱歉，我没能想出一个更好的标题。我的问题是我有两个列表，我想从第一个列表中随机挑选一个项目，并将其与第二个列表中的一个项目匹配，但这两个项目的索引不能相同（比如li1[3]不能和li2[3]匹配），而且一旦选中一个项目，它就必须从列表中移除。我试过这个代码，但它没有解决我的问题。谢谢大家的帮助。

from random import choice

teams1 = ["brazil", "croatia", "england", "italy", "mexico", "cameroon", "ivory coast", "japan"]
teams2 = ["spain", "germany", "belgium", "holland", "argentina", "portugal", "france", "usa"]


counter = 3
counter2 = 8 
while counter != 0:
    a = choice(range(1,counter2))
    a1 = teams1[a]
    b = choice(range(1,counter2))
    b1 = teams2[b]
    print(teams1[0], b1)
    print(teams2[0], a1)
    del teams1[a]
    del teams2[b]
    del teams1[0]
    del teams2[0]
    counter -= 1
    counter2 -= 2

print(teams1[0], teams2[0])

Answer 1

我不会感到惊讶，如果有更优雅的方法来构建成对的组合，但这个方法似乎能满足问题评论中提到的要求：

#/usr/bin/env python2.7
from random import shuffle

def main():

    teams1 = ["brazil", "croatia", "england", "italy",
              "mexico", "cameroon", "ivory coast", "japan"]
    teams2 = ["spain", "germany", "belgium", "holland",
              "argentina", "portugal", "france", "usa"]

    excluded_pairs = dict(zip(teams1, teams2))
    shuffle(teams1) # not really needed
    shuffle(teams2)

    while len(teams1):
        t1 = teams1.pop()
        if len(teams1) > 1:
            if teams2[-1] != excluded_pairs[t1]:
                t2 = teams2.pop(-1)
            else:
                t2 = teams2.pop(-2)
        elif len(teams1) == 0:
            t2 = teams2.pop(-1)
        elif len(teams1) == 1:
            if (teams2[-1] != excluded_pairs[t1]
                and teams2[-2] != excluded_pairs[teams1[0]]):
                t2 = teams2.pop(-1)
            else:
                t2 = teams2.pop(-2)

        print '{} vs. {}'.format(t1, t2)

    print len(teams1)
    print len(teams2)

if __name__ == '__main__':
    main()

Answer 2

你可以使用 random.randrange 来实现这个功能！

import random

teams1 = ["brazil", "croatia", "england", "italy", "mexico", "cameroon", "ivory coast", "japan"]
teams2 = ["spain", "germany", "belgium", "holland", "argentina", "portugal", "france", "usa"]

t1 = random.randrange(0, len(teams1))
t2 = random.randrange(0, len(teams2))

while t2 == t1:
    t2 = random.randrange(0, len(teams2))

rand_choice1 = teams1.pop(t1)
rand_choice2 = teams1.pop(t2)

这里的 while 循环是用来确保两个索引不一样，也就是说我们选出的两个队伍不会是同一个。然后，列表里的 .pop() 方法会把我们选择的队伍从列表中移除。

这种方法只会列出一场比赛，但其实可以很简单地把它做成一个函数，来随机匹配所有的队伍。

Answer 3

你的程序是对的。只是当两个列表中还有值的时候，你的程序就结束了。这是因为你的 counter 变成了 0，而这个条件：

while counter != 0:

就不再成立了。

你可以把这个条件改成检查两个列表中是否还有剩余的项目，使用：

while len(teams1) and len(teams2):

或者：

while counter > -1:

并且去掉最后一行：

print(teams1[0], teams2[0])

这样你的程序就能按预期工作了。