Question

此代码生成52张牌。在这段代码中，我不明白为什么他/她使用Deck类a方法中的b类在self.deck列表中附加参数Card和__init__？ ??

suits = ('Hearts', 'Diamonds', 'Spades', 'Clubs')
ranks = ('Two', 'Three', 'Four', 'Five', 'Six', 'Seven', 'Eight', 'Nine', 'Ten', 'Jack', 'Queen', 'King', 'Ace')

class Card():

    def __init__(self, suit, rank):
        self.suit = suit
        self.rank = rank

    def __repr__(self):
        return self.rank + " of " + self.suit


class Deck():

    def __init__(self):
        self.deck = []
        for a in suits:
            for b in ranks:
                self.deck.append(Card(a, b))

    def __repr__(self):
        deck_comp = ''
        for card in self.deck:
            deck_comp += "\n "+card.__repr__()
        return "The deck has: " + deck_comp

deck = Deck()
print(deck)

Answer 1

变量名称对您没有帮助。代码在suits和ranks上迭代以获取它们的所有组合，Card将创建一个具有2个参数的新实例：西服和等级

Hearts Two
Hearts Three
Hearts Four
...
Diamonds Two
Diamonds Three
Diamonds Four
...

使用有意义的名称，您将无法使用印刷品更好地理解：

def __init__(self):
    self.deck = []
    for suit in suits:
        print(suit) # for understanding only
        for rank in ranks:
            print(suit, rank) # for understanding only
            self.deck.append(Card(suit, rank))

这对应于两个列表之间的itertools.product，您可以映射到Card实例并保留列表

def __init__(self):
    self.deck = list(map(lambda x: Card(*x), product(suits, ranks)))

将参数附加到使用另一个类的类列表中？

1 个答案: