Python,如果它包含匹配的元组,则删除整个列表

Question

我希望以前没有问过这个问题,但我很难说出我正在努力做的事情!

如果我解释我的输入和所需的输出可能会更容易

l1 = [[(14, 'h'), (14, 'd')], [(14, 'h'), (14, 'c')], [(14, 'h'), (14, 's')], [(14, 'd'), (14, 'c')], [(14, 'd'), (14, 's')], [(14, 'c'), (14, 's')], [(13, 'h'), (13, 'd')], [(13, 'h'), (13, 'c')], [(13, 'h'), (13, 's')], [(13, 'd'), (13,'c')], [(13, 'd'), (13, 's')], [(13, 'c'), (13, 's')], [(12, 'h'), (12, 'd')], [(12, 'h'), (12, 'c')], [(12, 'h'), (12, 's')], [(12, 'd'), (12, 'c')], [(12, 'd'), (12, 's')], [(12, 'c'), (12, 's')]]

l2 = [(13,'h'),(13,'d'),(14,'c'),(14,'s'),(14,'h')]

所以这个第一个列表l1是一个列表列表,每个列表都是2张牌的扑克牌.

基本上我要做的是如果一张牌在l2中,则需要移除l1中的牌.所以条目[(14,'d'),(14,'c')]需要从l1中删除,因为(14,'c')在l2--即使(14,'d')isn'在l2.

这个例子的期望输出应该是,

[[(13, 'c'), (13, 's')], [(12, 'h'), (12, 'd')], [(12, 'h'), (12, 'c')], [(12, 'h'), (12, 's')], [(12, 'd'), (12, 'c')], [(12, 'd'), (12, 's')], [(12, 'c'), (12, 's')]]

我想过遍历L1内所有元组和删除他们,如果他们在L2,再后来回去遍历列表L1和移除任何不LEN == 2.这似乎不是最好的办法不过,去吧.

有什么想法吗？

ManyTIA!

Answer 1

l2_set = set(l2)
# or use 
#   l2_set = {(13,'h'),(13,'d'),(14,'c'),(14,'s'),(14,'h')}
# directly

l1 = [hand for hand in l1 if not (hand[0] in l2_set or hand[1] in l2_set)]

如果条目可能包含≥2个成员,则可以将过滤条件概括为

l1 = [hand for hand in l1 if all(subhand not in l2_set for subhand in hand)]