Question

spades = ['2S','3S','4S','5S','6S','7S','8S','9S','10S','JS','QS','KS','AS']
hearts = ['2H','3H','4H','5H','6H','7H','8H','9H','10H','JH','QH','KH','AH']
clubs = ['2C','3C','4C','5C','6C','7C','8C','9C','10C','JC','QC','KC','AC']
diamonds = ['2D','3D','4D','5D','6D','7D','8D','9D','10D','JD','QD','KD','AD']
allCards = spades + hearts + clubs + diamonds

import random
random.shuffle(allCards)

bot1 = [allCards.pop() for i in range(2)]
print(bot1)
cardVal = {'2S':1,'3S':2,'4S':3,'5S': 4,'6S':5,'7S':6,'8S':7,'9S':8,'10S':9,'JS':10,'QS':11,'KS':12,'AS':13,
    '2H':1,'3H':2,'4H':3,'5H': 4,'6H':5,'7H':6,'8H':7,'9H':8,'10H':9,'JH':10,'QH':11,'KH':12,'AH':13,
    '2C':1,'3C':2,'4C':3,'5C': 4,'6C':5,'7C':6,'8C':7,'9C':8,'10C':9,'JC':10,'QC':11,'KC':12,'AC':13,
    '2D':1,'3D':2,'4D':3,'5D': 4,'6D':5,'7D':6,'8D':7,'9D':8,'10D':9,'JD':10,'QD':11,'KD':12,'AD':13}

for i in bot1:
    print(cardVal[i])
    bot1hand = [cardVal[i]]
print(bot1hand)

我想将bot1所拥有的卡的值放在一个单独的数组中，但遇到了问题。我总是将两个值打印在单独的行上，而数组bot1hand仅存储两个值的最后一个值。

例如：

>>> 
['AC', '5C']
13
4
[4]
>>>

Answer 1

你的问题就在这里：

for i in bot1:
    print(cardVal[i])
    bot1hand = [cardVal[i]]
print(bot1hand)

特别是这一行：

bot1hand = [cardVal[i]]

您经常在写自己的价值观，因为您实际上没有正确地附加到您的列表中。实际上，您的bot1hand并未被视为列表。

首先要做的是将其初始化为循环外的列表：

bot1hand = []

然后在你的循环中，使用append方法：

bot1hand.append(cardVal[i])

所以你的最后一块代码应该是这样的：

bot1hand = []
for i in bot1:
    print(cardVal[i])
    bot1hand.append(cardVal[i])
print(bot1hand)

作为代码中的最后重构步骤，您实际上可以执行@NathanielFord建议的操作，使用理解（我看到您已在代码中使用过，因此您必须已经熟悉它）。我在这个答案中的代码块现在可以简化为：

bot1hand = [cardVal[i] for i in bot1]

Answer 2

您的for循环是个问题。您可以尝试列出comprehension：

bot1hand = [cardVal[i] for i in bot1]
print(bot1hand)

（我假设print语句用于调试目的。）

列表理解处理实际为您构建列表的责任。

如何将字典中的值拉出到数组中

2 个答案: