Python - 如何使用列表理解来填充命名元组值

Question

我有以下类来创建一副牌：

import collections


Card = collections.namedtuple('Card', ['rank', 'suit', 'value'])


class FrenchDeck:
    ranks = [str(n) for n in range(2, 11)] + list('JQKA')
    suits = 'spades clubs hearts diamonds'.split()
    card_value = [str(n + 1) for n in range(len(ranks))]

    def __init__(self):
        self._cards = [Card(rank, suit, value)
                       for suit in self.suits
                       for rank in self.ranks
                       for value in self.card_value
                       ]

    def __len__(self):
        return len(self._cards)

    def __getitem__(self, position):
        return self._cards[position]

if __name__ == '__main__':
    FrenchDeck()

我已将value值添加到卡中，以便为每张卡分配一个值，如下所示：

Card(rank='2', suit='spades', value='1')
Card(rank='2', suit='spades', value='2')
Card(rank='2', suit='spades', value='3')
Card(rank='2', suit='spades', value='4')
Card(rank='2', suit='spades', value='5')
Card(rank='2', suit='spades', value='6')
Card(rank='2', suit='spades', value='7')
Card(rank='2', suit='spades', value='8')
Card(rank='2', suit='spades', value='9')
Card(rank='2', suit='spades', value='10')
Card(rank='2', suit='spades', value='11')
Card(rank='2', suit='spades', value='12')
Card(rank='2', suit='spades', value='13')

它为每个套装的每个等级创建 13 个项目。我理解为什么会发生这种情况，但是我正在努力为套装中的每个等级添加价值，如下所示：

Card(rank='2', suit='spades', value='1')
Card(rank='3', suit='spades', value='2')
Card(rank='4', suit='spades', value='3')
Card(rank='5', suit='spades', value='4')
Card(rank='6', suit='spades', value='5')
Card(rank='7', suit='spades', value='6')
Card(rank='8', suit='spades', value='7')
Card(rank='9', suit='spades', value='8')
Card(rank='10', suit='spades', value='9')
Card(rank='J', suit='spades', value='10')
Card(rank='K', suit='spades', value='11')
Card(rank='Q', suit='spades', value='12')
Card(rank='A', suit='spades', value='13')

知道如何实现这一目标吗？

Answer 1

实际上，您并不像您认为的那样嵌套列表迭代。

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更改自

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def __init__(self):\n    self._cards = [Card(rank, suit, value)\n                   for suit in self.suits\n                   for rank in self.ranks\n                   for value in self.card_value\n                   ]\n

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到

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def __init__(self):\n    self._cards = [Card(rank, suit, value)\n                   for suit in self.suits\n                   for rank, value in zip(self.ranks,\\\n                                          self.card_value)\n                   ]\n

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实现你想要的。请注意，这相当于：

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def __init__(self):\n    self._cards = []\n    for suit in self.suits:\n        for rank, value in zip(self.ranks, self.card_value):            \n            self._cards+=[Card(rank, suit, value)]\n

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此外，正如 chepner 所说，或由 J\xc3\xa9r\xc3\xa9my 操作，

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value不是 a 的独立属性Card；它是 的函数rank。

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这意味着您可以迭代value的self.card_value每个元素self.suits，然后rank根据value您处理的内容获取相应的元素。

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更多细节

关注您的评论/问题

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您定义ranks为

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ranks = [str(n) for n in range(2, 11)] + list('JQKA')\n

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然后，基于ranks，您定义card_value如下

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card_value = [str(n + 1) for n in range(len(ranks))]\n

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这清楚地表明了两个对象之间的依赖关系：列表中的每个元素card_value都是列表中每个元素的直接变换/对应ranks。

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为了清楚起见，我会s在指针的名称上添加一个免费的符号。self.card_valueIEself.card_values

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