纸牌游戏模拟 list.remove

Question

我正在关注一本非常好的 Python 书籍，它教你如何为一些纸牌游戏创建模拟。正如您在下面看到的，我创建了一个“卡片”类，用于索引一副牌中所有可能的卡片组合。然后我创建了一个类“Deck”，它通过“Card”中的卡片组合来制作卡片组。我的问题是，我正在尝试模拟某人从 decl 中取出卡片，但无法让最后一个类函数 removeCard 工作，我不太清楚为什么。有人可以帮助我理解我的问题以及如何纠正它吗？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，您卡片的__init__ 函数中有错字。您应该将属性分配给self，而不是Card。

第二， 我猜你正在做类似的事情：

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))

你期望的输出是：

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

但你实际上得到了：

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

您的removeCard 方法失败了，因为默认情况下，python 中的对象只有在它们的 id 相等时才相等。示例：

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

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

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

尽管 a 和 b 都是用 Card(1,1) 创建的，但它们并不相等。当您尝试if card in self.cards: 时，Python 会询问 self.cards 是否包含具有此确切 ID 的卡片？对于我的第一个代码 sn-p，它会回复 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 将正常运行，并根据需要移除您的卡。

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