Python 仅将列表中的某些单词指定给变量

我正在尝试将列表

卡
中列表
playerdeck
中的单词分配给一个变量。这是我试图使用的代码，但它返回
False

playerdeck = ['Five of Spades', 'Eight of Spades',
              'Eight of Clubs', 'Four of Clubs', 'Ace of Spades',
              'Eight of Hearts', 'Four of Diamonds'] 

cards = ['King', 'Queen', 'Jack', 'Ace',
     'Two', 'Three', 'Four', 'Five',
     'Six', 'Seven', 'Eight', 'Nine',
     'Ten']

new = cards in playerdeck
print(new)

有人能帮忙吗？
课程卡：
class Card:
   def __init__(self,value):
       self.value = value
   def __eq__(self,other):          
       return str(other) in self.value
   def __str__(self):
       return self.value
   def __repr__(self):
       return "<Card:'%s'>"%self
   def __hash__(self):
       return hash(self.value.split()[0])

playerdeck = map(Card,['Five of Spades', 'Eight of Spades',
              'Eight of Clubs', 'Four of Clubs', 'Ace of Spades',
              'Eight of Hearts', 'Four of Diamonds'] )

cards = set(['King', 'Queen', 'Jack', 'Ace',
     'Two', 'Three', 'Four', 'Five',
     'Six', 'Seven', 'Eight', 'Nine',
     'Ten'])

print cards.intersection(playerdeck)

定义初始值（自身，值）：
自我价值=价值
定义（自身、其他）：
在self.value中返回str（其他）
定义（自我）：
回归自我价值
定义报告（自我）：
返回“%self”
定义散列（自我）：
返回散列（self.value.split（）[0]）
playerdeck=地图（牌，[‘黑桃五’、‘黑桃八’，
“八支梅花”、“四支梅花”、“黑桃王牌”，
“八颗红心”、“四颗钻石”]）
卡片=套装（['King'、'Queen'、'Jack'、'Ace'，
‘二’、‘三’、‘四’、‘五’，
‘六’、‘七’、‘八’、‘九’，
“十”]）
打印卡片。交叉点（playerdeck）
试试这个，它首先遍历这些卡，并检查哪些卡在玩家牌堆中。如果有匹配项，它会将其附加到新卡上并转到下一张卡

new = []

for card in cards:
    for deck in player_deck:
        if card.lower() in deck.lower():
            new.append(card)
            break

您可以尝试：

>>> playerdeck = ['Five of Spades', 'Eight of Spades',
              'Eight of Clubs', 'Four of Clubs', 'Ace of Spades',
              'Eight of Hearts', 'Four of Diamonds']
>>> cards = ['King', 'Queen', 'Jack', 'Ace',
     'Two', 'Three', 'Four', 'Five',
     'Six', 'Seven', 'Eight', 'Nine',
     'Ten']
>>> 
>>> for pd in playerdeck:
    temp = pd.split(" ")
    for data in temp:
        if data in cards:
            print data


Five
Eight
Eight
Four
Ace
Eight
Four

下面是使用两个循环和一个条件语句（仅3行）的小解决方案：


或者如果您想尝试列表理解：

>>> playerdeck = ['Five of Spades', 'Eight of Spades',
...               'Eight of Clubs', 'Four of Clubs', 'Ace of Spades',
...               'Eight of Hearts', 'Four of Diamonds']
>>> cards = ['King', 'Queen', 'Jack', 'Ace',
...      'Two', 'Three', 'Four', 'Five',
...      'Six', 'Seven', 'Eight', 'Nine',
...      'Ten']
>>> result = [card for pd in playerdeck for card in cards if card in pd]
>>> result
['Five', 'Eight', 'Eight', 'Four', 'Ace', 'Eight', 'Four']

当然，但在我的实际代码中，playerdeck是另一个列表中7个随机对象的列表。在op中，我只是简化了它。看看答案。
>>> playerdeck = ['Five of Spades', 'Eight of Spades',
...               'Eight of Clubs', 'Four of Clubs', 'Ace of Spades',
...               'Eight of Hearts', 'Four of Diamonds']
>>> cards = ['King', 'Queen', 'Jack', 'Ace',
...      'Two', 'Three', 'Four', 'Five',
...      'Six', 'Seven', 'Eight', 'Nine',
...      'Ten']
>>> result = [card for pd in playerdeck for card in cards if card in pd]
>>> result
['Five', 'Eight', 'Eight', 'Four', 'Ace', 'Eight', 'Four']