纸牌游戏模拟list.remove

Question

我跟着一本很不错的python书，它教你如何创建一些纸牌游戏的模拟。正如你在下面看到的，我已经创建了一个“卡片”类，用于索引卡片中所有可能的卡片组合。然后我创建了一个“Deck”类，它使卡片组合中的卡片成为“卡片”。我的问题是，我试图模拟某人从decl中取出一张卡片，但无法获得最后一个班级的removeCard功能，而且我不太清楚为什么。有人可以帮助我了解我的问题以及如何纠正它？谢谢。

class Card: 
def __init__(self, suit = 0, rank = 2): 
    Card.suit = suit 
    Card.rank = rank 
#Two class attributes that are helpful in determining suit/rank 
ranklist = ['narf', 'Ace', 'Two', 'Three', 'Four', 'Five', \ 
'Six', 'Seven', 'Eight', 'Nine', 'Ten', 'Jack', 'Queen', 'King'] 
suitlist = ['Clubs', 'Diamonds', 'Hearts', 'Spades'] 
def __repr__(self): 
    return (self.ranklist[self.rank] + " of " + self.suitlist[self.suit]) 


class Deck: 
def __init__(self): 
    self.cards = [] 
    for suit in range(4): 
     for rank in range(1,13): 
      x = Card() 
      x.rank = rank 
      x.suit = suit 
      self.cards.append(x) 
def printDeck(self): 
    for card in self.cards: 
     print(card) 

def removeCard(self, card): 
     if card in self.cards: 
      self.cards.remove(card) 
      return True 
     else: 
      return False 

Answer 1

首先，您的卡片的__init__函数中存在拼写错误。您应该将属性分配给self，而不是Card。

其次， 我猜你正在做的事情，如：

class Card: 
    def __init__(self, suit = 0, rank = 2): 
     self.suit = suit 
     self.rank = rank 
    #Two class attributes that are helpful in determining suit/rank 
    ranklist = ['narf', 'Ace', 'Two', 'Three', 'Four', 'Five', \ 
    'Six', 'Seven', 'Eight', 'Nine', 'Ten', 'Jack', 'Queen', 'King'] 
    suitlist = ['Clubs', 'Diamonds', 'Hearts', 'Spades'] 
    def __repr__(self): 
     return (self.ranklist[self.rank] + " of " + self.suitlist[self.suit]) 


class Deck: 
    def __init__(self): 
     self.cards = [] 
     for suit in range(4): 
      for rank in range(1,13): 
       x = Card() 
       x.rank = rank 
       x.suit = suit 
       self.cards.append(x) 
    def printDeck(self): 
     for card in self.cards: 
      print(card) 

    def removeCard(self, card): 
      if card in self.cards: 
       self.cards.remove(card) 
       return True 
      else: 
       return False 

d = Deck() 
print "The deck has {} cards.".format(len(d.cards)) 
card = Card(1,1) 
print "Removing the {}.".format(card) 
d.removeCard(card) 
print "The deck has {} cards.".format(len(d.cards)) 

你所期望的输出是：

The deck has 48 cards. 
Removing the Ace of Diamonds. 
The deck has 47 cards. 

但是你实际上：

The deck has 48 cards. 
Removing the Ace of Diamonds. 
The deck has 48 cards. 

您的removeCard方法失败，因为默认情况下，python中的对象是只有当他们的ID相等时才相等。例如：

>>> a = Card(1,1) 
>>> b = Card(1,1) 

>>> a == a 
True 
>>> id(a) == id(a) 
True 

>>> a == b 
False 
>>> id(a) == id(b) 
False 

即使a和b与Card(1,1)创建的，它们不相等。当你尝试if card in self.cards:时，Python询问自身卡是否包含一张具有这个确切ID的卡？在我的第一个代码片段中，它回复False。

您可以通过在Card类中指定自己的__eq__来覆盖此默认行为。

class Card: 
    def __init__(self, suit = 0, rank = 2): 
     self.suit = suit 
     self.rank = rank 
    #Two class attributes that are helpful in determining suit/rank 
    ranklist = ['narf', 'Ace', 'Two', 'Three', 'Four', 'Five', \ 
    'Six', 'Seven', 'Eight', 'Nine', 'Ten', 'Jack', 'Queen', 'King'] 
    suitlist = ['Clubs', 'Diamonds', 'Hearts', 'Spades'] 
    def __eq__(self, other): 
     if not isinstance(other, Card): return False 
     return self.suit == other.suit and self.rank == other.rank 
    def __repr__(self): 
     return (self.ranklist[self.rank] + " of " + self.suitlist[self.suit]) 

现在in和remove的行为不当，并根据需要删除你的卡。

The deck has 48 cards. 
Removing the Ace of Diamonds. 
The deck has 47 cards.