Python，刪除整個列表，如果它包含匹配的元組

Question

我希望這之前沒有被問過，但是我很難把我正在嘗試做的事情放在文字中！

如果我解釋我的輸入，以及所需的輸出

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')] 

所以這第一個列表，L1，是列出的清單可能更容易的，每個單子是2張牌撲克手。

基本上我想要做的是，如果一張卡在l2中，需要將l1中的手去掉。 （14，'c'）]需要從l1中刪除，因爲（14，'c'）在l2中 - 儘管（14，'d'）不是' t在l2。

從這個例子所需的輸出應該是，

[[(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')]] 

我想過遍歷L1內所有元組和刪除他們，如果他們在L2，再後來回去遍歷在列表l1並刪除任何沒有len == 2.這似乎不是最好的方式去，但。

有什麼想法？

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)] 

如果條目可包含≥2的成員，則可以將過濾條件

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

Answer 2

>>> [x for x in l1 if not set(x).intersection(l2)] 
[[(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')]] 

Answer 3

[x for x in l1 if not any(y in l2 for y in x)]