“虽然不在”错误 - python

Question

在本例中，我们将PH[0] 设置为'Ten of Hearts'，将guess 设置为'Ten'

列表“PH”代表Go Fish 纸牌游戏中玩家的手牌。当用户猜牌时，他们必须猜出与他们手中的牌相对应的牌。我写了这段代码，如果用户输入了一个无效的猜测（如果猜测不在PH中，则会再次提示他们输入新的猜测）

我觉得数组[guess1, guess2, guess3, guess4] 中的变量有问题，但我不太确定。

当代码执行时，循环会一直持续下去。我需要能够跳出循环以返回猜测值，以便将其输入到下一个函数中。

如果有人可以帮我解决我遇到的问题。

guess = str(raw_input("Make a guess : "))

guess11 = guess, 'of Hearts'
guess1 = " ".join(guess11)

guess22 = guess, 'of Diamonds'
guess2 = " ".join(guess22)

guess33 = guess, 'of Clubs'
guess3 = " ".join(guess33)

guess44 = guess, 'of Spades'
guess4 = " ".join(guess44)

while PH[0] not in [guess1, guess2, guess3, guess4] :
    print "You do not have a card like that in your hand."
    guess = str(raw_input("Make another guess : "))

    guess11 = guess, 'of Hearts'
    guess1 = " ".join(guess11)

    guess22 = guess, 'of Diamonds'
    guess2 = " ".join(guess22)

    guess33 = guess, 'of Clubs'
    guess3 = " ".join(guess33)

    guess44 = guess, 'of Spades'
    guess4 = " ".join(guess44)

return guess

Answer 1

你只是在测试玩家手中的第一张牌是否是他猜到的。您需要测试手中的每张牌：

while not any(guess in PH for guess in [guess1, guess2, guess3, guess4]):

这会取出每张猜出的牌，并依次测试该牌与手牌。 any() 在找到匹配项时停止循环猜测。

一个更好的主意仍然是使用集合交集：

guesses = {guess1, guess2, guess3, guess4}
while not guesses.intersection(PH):
    # ask for new guesses

您希望避免输入两次“要求猜测”代码；以while True 开始循环，并在做出正确猜测后使用break 结束循环：

suits = ('Hearts', 'Diamonds', 'Clubs', 'Spades')

while True:
    guess = raw_input("Make a guess:")
    guesses = {'{} of {}'.format(guess, suit) for suit in suits}
    if guesses.intersection(PH):
        # correct guess!
        break
    print "You do not have a card like that in your hand."

我使用集合推导在循环中构建猜测。