Question

我正在创造一种通过五个玩家手牌传递五张扑克牌的方法。我已经在main()中创建了一种方法，但是我正在寻找一个创建一个函数的想法，该函数将卡片传递给另一只手，所以我不会在{{1}中重复我的代码行。 }}。只需要一些想法。

main()

Answer 1

我想出了这个：

def passAgain():
    # This line create all empty Hand objects.
    table_hands = [Hand() for _ in range(5)]

    # These 3 lines creates the 5 cards involved
    for value in ["A", "2", "3", "4", "5"]:
        for suite in ["s"]:
            table_hands[0].add(Card(value, suite))

    # Cycle the hands  
    for i, hand in enumerate(table_hands):
        # Create pairs of hands, one giving and another receiving.
        hands_passing = table_hands[i:i + 2]

        if len(hands_passing) == 2:
            # If we have two hands involved, create a list of cards that 
            #  will be given.
            giving_hand = hands_passing[0].cards[:]
            for card in giving_hand:
                hands_passing[0].give(card, hands_passing[1])
                print(table_hands)

def main():
    while True :
        ans = input("Do you want to pass cards? Press y or yes: ")
        if ans in ("y", "yes"):
            passAgain()
        else:
            break

如何创建一个函数将这些卡从手传递到另一个

1 个答案: