Question

因此，在与字符列表进行比较后，我的代码应该理想地打印出不完整的单词。

例如，如果单词是stackoverflow并且列表包含's','a','k', 'o', 'r','l'，则输出应为

s _ a _ k o _ _ r _ l o _

我的代码是

def getGuessedWord(secretWord, lettersGuessed):
'''
secretWord: string, the word the user is guessing
lettersGuessed: list, what letters have been guessed so far
returns: string, comprised of letters and underscores that represents
  what letters in secretWord have been guessed so far.
'''
i=0
gameWord=''
while i in range(len(secretWord)):
    if(i%2!=0):
        gameWord+=' ' #to print spaces between characters
    else:
        if(secretWord[i] in lettersGuessed):
            gameWord+=secretWord[i]
        else:
            gameWord+='_'

return gameWord

但我的代码只是挂起！我无法弄清楚为什么！

Answer 1

您需要确保增加变量i。否则，你无限循环你的while循环！

循环时在循环内部的底部添加i ++。

这将解决您的无限循环问题，但不会为您提供您正在寻找的预期输出。你需要重新修改你的if / else逻辑，你就得到了它。

Answer 2

你的循环如果是无限循环因此它会挂起（因为i = 0总是在你的范围内）。

看起来你需要这样：

def getGuessedWord(secretWord, lettersGuessed):
    '''
    secretWord: string, the word the user is guessing
    lettersGuessed: list, what letters have been guessed so far
    returns: string, comprised of letters and underscores that represents
      what letters in secretWord have been guessed so far.
    '''

    gameWord=''
    for c in secretWord:
        if c in lettersGuessed:
            gameWord += c
        else:
            gameWord += '_'

    return gameWord

print(getGuessedWord('stackoverflow', ['s','a','k', 'o', 'r','l']))

输出：

s_a_ko__r_lo _

但如果你真的想要你的方式：

def getGuessedWord(secretWord, lettersGuessed):
    '''
    secretWord: string, the word the user is guessing
    lettersGuessed: list, what letters have been guessed so far
    returns: string, comprised of letters and underscores that represents
      what letters in secretWord have been guessed so far.
    '''

    i=0
    gameWord=''
    while i in range(len(secretWord)):
        gameWord += ' '
        if(secretWord[i] in lettersGuessed):
            gameWord += secretWord[i]
        else:
            gameWord += '_'
        i += 1
    return gameWord

编辑带有“_”或字符的字符串的代码挂起

2 个答案: