如何将程序修复到消息zip参数2必须支持迭代的位置

Question

    import random
    from Queue import Queue

    class Deck:
        def __init__(self, decks=1, shuffle=4):

使用队列系统模拟被绘制和丢弃的卡片：参数甲板：要使用的甲板数：param shuffle：洗牌的次数：返回：

            self.deck = Queue()
            self.shuffle_times = shuffle
            self.suits = ['spades', 'clubs', 'hearts', 'diamonds']
            self.cards = ['2', '3', '4', '5', '6', '7', '8', '9', '10', 'J', 'Q', 'K', 'A']
            for deck in range(decks):
                for suit in self.suits:
                    for card in self.cards:
                        self.deck.put([card, suit])
    def draw(self):

：返回：甲板上的随机卡

        card = self.deck.get()
        return card
    def shuffle(self):

洗牌n次：返回：无

        if self.shuffle_times > 0:
            temp = list(self.deck.queue)
            [random.shuffle(temp) for _ in range(self.shuffle_times)]
            self.deck = Queue()
            for card in temp:
                self.deck.put(card)
class Blackjack:
    def __init__(self, decks=1, shuffle=1):

：参数甲板：要使用的甲板数：param shuffle：洗牌的次数：返回：无

        self.num_decks = decks
        self.deck = Deck(self.num_decks, shuffle)
        self.deck.shuffle()
        self.cards_dict = {'2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8,
                       '9': 9, '10': 10, 'J': 10, 'Q': 10, 'K': 10, 'A': 11}
        self.high_low = {'2': 1, '3': 1, '4': 1, '5': 1, '6': 1, '7': 0, '8': 0, '9': 0,
                     '10': -1, 'J': -1, 'Q': -1, 'K': -1, 'A': -1}
        self.games = 0
        self.draws = 0
        self.computer_win = 0
        self.player_win = 0
        self.running_count = 0
        self.player_hand = []
        self.computer_hand = []
        self.player_total = 0
        self.computer_total = 0
    def batlle_bot(self):

使用二十一点的基本策略来决定要采取哪些措施。能够区分软手和硬手

        if 'A' in [card[0] for card in self.player_hand]:
            if self.player_total <= 17:
                return 'h'
            elif self.player_total >= 19:
                return 's'
            elif self.player_total == 18 and 2 <= self.computer_total <= 8:
                return 's'
            elif self.player_total == 18 and self.computer_total >= 9:
                return 'h'
        else:
            if self.player_total <= 11:
                return 'h'
            elif self.player_total >= 17:
                return 's'
            elif self.player_total == 12 and 2 <= self.computer_total <= 3:
                return 'h'
            elif 12 <= self.player_total <= 16 and 4 <= self.computer_total <= 6:
                return 's'
            elif 13 <= self.player_total <= 16 and 2 <= self.computer_total <= 3:
                return 's'
            elif 4 <= self.player_total <= 16 and self.computer_total >= 7:
                return 'h'
    def total(self, hand):
    """
    Helper to check for aces in the hand and calculates the value of the hand
    :param hand: list of cards in the hand
    :return: value of the hand
    """
        aces = hand.count(11)
        hand_value = sum(hand)
        if hand_value > 21 and aces > 0:
            while aces > 0 and hand_value > 21:
                hand_value -= 10
                aces -= 1
        return hand_value

    def high_low_count(self, card):

跟踪卡计数的高/低计数：参数卡：卡值：返回：根据高/低计数的卡值

        self.running_count += self.high_low[card[0]]
        return card
score = (0, 0, 0, 0)
num_cycles = 1000
num_decks = 6
for i in range(num_cycles):
    game = Blackjack(num_decks).battle_bot=True
    score = tuple([x + y for x, y in zip(score, game)])
    games, draws, player_win, computer_win = score
    print('Player win = {}'.format(player_win))
    print('Computer win = {}'.format(computer_win))
    print('Number of games = {}'.format(games))
    print('Draws = {}'.format(draws))
    try:
        print('Win/Loss = {:.2f}'.format(player_win / computer_win))
    except:
        print('Win/Loss = {}'.format(max(player_win, computer_win)))
    print('n--------------------n')

该程序应打印球员获胜的次数，经销商获胜以及多少平局。

Answer 1

您的问题在这里：

game = Blackjack(num_decks).battle_bot=True
score = tuple([x + y for x, y in zip(score, game)])

您刚分配了game和Blackjack(num_decks).battle_bot具有值True，然后尝试使用zip score（a tuple，fine）和 game（a bool， CC_8，不可能）。

我不知道您在这里的意图，即使您打电话给battle_bot，它也会返回一个字母字符串，因此您的zip只会产生一个单个输出，而不是您似乎认为您'的四个值将能够在下一行上解开包装（四个具有不同含义的值不少，而产生scores的代码似乎以对多种含义没有意义的方式对待它们）。

您需要考虑您的程序逻辑，除了发布有关的小错误之外，这里存在太多的逻辑问题。