「雖然不是」錯誤 - 蟒蛇

Question

在這個例子中，我們將設置PH[0]到'Ten of Hearts'，並guess到'Ten'

名單「PH」表示去魚的紙牌遊戲玩家的手。當用戶猜測一張卡片時，他們必須猜出一張與他們手中的卡片相對應的卡片。我寫了這個代碼塊，以便如果用戶輸入一個無效猜測（如果猜測不在PH，他們會再次提示輸入一個新的猜測）

我覺得我有問題的變量在數組[guess1, guess2, guess3, guess4]，但我不太確定。

執行代碼時，循環會一直持續。我需要能夠退出循環以便返回猜測，以便它可以輸入到下一個函數中。

如果有人能幫我解決我遇到的問題。

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

您只是在測試如果玩家手中的第一張牌是任何的他猜測。你需要測試每張牌的牌：

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

這需要每個猜測卡和測試卡反對手。 any()在找到匹配時停止循環猜測。

一個更好的想法仍然是使用一套十字路口：

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

你想避免鍵入您的「要求猜測」代碼兩次;開始一個循環與while True和使用break結束循環時，正確的猜測已經取得：

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." 

我用了一套理解建立在猜測的循環。