我有一个旨在模拟纸牌游戏21的程序。这是我的代码中非常重要的自我解释的重要元素（我已突出显示行，我将在后面提到）
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
cardVal = {'2S':2,'3S':3,'4S':4,'5S': 5,'6S':6,'7S':7,'8S':8,'9S':9,'10S':10,'JS':10,'QS':10,'KS':10,'AS':11,
'2H':2,'3H':3,'4H':4,'5H':5,'6H':6,'7H':7,'8H':8,'9H':9,'10H':10,'JH':10,'QH':10,'KH':10,'AH':11,
'2C':2,'3C':3,'4C':4,'5C':5,'6C':6,'7C':7,'8C':8,'9C':9,'10C':10,'JC':10,'QC':10,'KC':10,'AC':11,
'2D':2,'3D':3,'4D':4,'5D':5,'6D':6,'7D':7,'8D':8,'9D':9,'10D':10,'JD':10,'QD':10,'KD':10,'AD':11}
import random
random.shuffle(allCards)
playerCards = [allCards.pop() for i in range(2)]
dealerCards = [allCards.pop() for i in range(2)]
playerHand = []
dealerHand = []
playerHandVal = 0
dealerHandVal = 0
def handVal(playercards,playerhand,score):
playerhand = []
for i in playercards:
playerhand.append(cardVal[i]) ####### LINE 29 ######
score = sum(playerhand)
print(score)
handVal(playerCards,playerHand,playerHandVal)
handVal(dealerCards,dealerHand,dealerHandVal)
def twist(playercards,playerhand,score):
newCard = [allCards.pop() for i in range(1)]
playercards.append(newCard)
handVal(playercards,playerhand,score) ####### LINE 39 ########
move = input('Stick (S) or Twist (T) : ')
if move == 'T' or move == 't':
while move == 'T' or 't':
twist(playerCards,playerHand,playerHandVal) ######## LINE 45 ########
print(playerHand)
if playerHandVal > 21:
move = 's'
break
move = input('Stick (S) or Twist (T) : ')
当我运行脚本时，它将产生两个长度为2的数组，并在输入提示前为这些卡提供正确的手牌值Stick (S) or Twist (T) :。然而，当选择捻程序产生该错误，
line 45, in <module>
twist(playerCards,playerHand,playerHandVal)
line 39, in twist
handVal(playercards,playerhand,score)
line 29, in handVal
playerhand.append(cardVal[i])
TypeError: unhashable type: 'list'
我的问题是，为什么这部分代码playerhand.append(cardVal[i])最初会起作用，但是在twist函数中调用时却无法起作用
当您这样做时：
newCard = [allCards.pop() for i in range(1)]
playercards.append(newCard)
你插入list到playercards含有从弹出一个值allCards，而不是价值本身。
稍后，当您这样做时：
for i in playercards:
playerhand.append(cardVal[i]) ####### LINE 29 ######
i不是索引，也不是密钥本身，它是一个元素，list其中包含我认为是您想要的密钥；lists是可变的，因此不适合作为dicts的键，并且cardVal是dict。
它的工作一开始，因为你初始化playerhand为list包含两张牌，而不是一个list单元素list的卡片■当你这样做playerCards = [allCards.pop() for i in range(2)]，就是额外的吸引，你实现不正确。
解决方法是playercards使用卡片而非单元素卡片填充list：
newCard = allCards.pop() # Get the value, not a one-element `list` containing the value
playercards.append(newCard)
如果目标是弹出并附加几张卡片（并且您range(1)暂时只是用作占位符），则要单独附加每张卡片，而不是将list卡片的内容作为单个元素附加，您可以执行以下任一操作：
newCard = [allCards.pop() for i in range(1)]
playercards.extend(newCard) # extend appends each element from the iterable
或由于playercards和newCard均为lists，则可以使用运算符重载：
newCard = [allCards.pop() for i in range(1)]
playercards += newCard # Same as extend, but only works when both sides are same type
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