Python 有没有办法只检查列表中两个值相同的值

我正在制作一个刽子手游戏，我很快就要完成了，我只有一个问题：

如果输入的单词有两个相同的字母，它将不起作用。我知道问题是什么，系统正在检查过去的字母，因为它已经找到了值，我只是不知道如何解决它

有没有办法让代码填充列表中的两个空格

代码如下：

#make player one or computer pick word
mainword = input("Enter your word\n").lower()

#varibels
builtword = []
correctLetters = list(mainword)

#establish length of build word
for x in range(len(mainword)):
    builtword.append('*')

#make player two guess letter

userGuess = input("Guess letter\n").lower()

#main code
while builtword != correctLetters:
    if userGuess in correctLetters:

        print('That was correct')
        builtword[correctLetters.index(userGuess)] = userGuess
        print(builtword)
        if builtword != correctLetters:
            userGuess = input("Guess letter").lower()

    else:
        print('sorry that\'s wrong try again')
        userGuess = input("Guess letter").lower()

因此，在这种情况下，假设我们使用HELLO并键入L，是否有任何方法可以将L（2和3）的值或us 3和4都填入

builtword

？

尝试以下方法：

#make player one or computer pick word
mainword = input("Enter your word\n").lower()

#varibels
builtword = list('*'*len(mainword))
correctLetters = list(mainword)

#make player two guess letter

userGuess = input("Guess letter\n").lower()

#main code
while builtword != correctLetters:
    if userGuess in correctLetters:
        print('That was correct')
        char_count = correctLetters.count(userGuess)
        if char_count == 1:
            builtword[correctLetters.index(userGuess)] = userGuess
            print(builtword)
        else:            
            index = -1
            for _ in range(char_count):
                index = correctLetters.index(userGuess, index+1)
                if builtword[index] != userGuess:
                    builtword[index] = userGuess
                    break
                         
        if builtword != correctLetters:
            userGuess = input("Guess letter").lower()

    else:
        print('sorry that\'s wrong try again')
        userGuess = input("Guess letter").lower()

---

检查代码缩进请在字符串中进行操作，更改

buildword

如果字符串中的字母匹配，请详细说明什么？遍历字符串？请重复您的字符串处理教程；这里应该有几个例子。是的，这解决了一些非常微小的缺陷，但我可以修复它，有一种方法可以同时填补这两个缺陷吗？@devahschaeffers这么简单，只需删除

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