刽子手游戏开发：如何根据用户输入移除未猜字母集中的字符

你的代码存在两个关键问题，导致无法正确返回未被猜测的字母：

1. 循环提前终止：函数在for ch in ALL_LETTERS的第一次迭代就执行了return，不管后续字母是否被猜测，逻辑完全错误。
2. 字符替换方式错误：ALL_LETTERS.replace(letters_guessed, '')是尝试替换整个letters_guessed子串，而不是逐个移除每个已猜测的字母。

正确实现代码 #

def get_available_letters(letters_guessed):
    """
    Returns a string of letters that have not yet been guessed.

    :param letters_guessed: letters that have been guessed so far by the player
    :type letters_guessed: str
    :return: letters that have not been guessed
    :rtype: str
    """
    ALL_LETTERS = 'abcdefghijklmnopqrstuvwxyz'
    letters_guessed = letters_guessed.lower()
    available_letters = []
    for ch in ALL_LETTERS:
        if ch not in letters_guessed:
            available_letters.append(ch)
    return ''.join(available_letters)

代码说明 #

• 遍历全部26个小写字母，检查每个字母是否不在已猜测的字母集合中
• 将未被猜测的字母存入列表，最后用join方法拼接成字符串返回
• 统一将用户输入的字母转为小写，避免大小写不匹配的判断错误

简化实现（可选） #

用生成器表达式结合集合可以提升判断效率，代码更简洁：

def get_available_letters(letters_guessed):
    """
    Returns a string of letters that have not yet been guessed.

    :param letters_guessed: letters that have been guessed so far by the player
    :type letters_guessed: str
    :return: letters that have not been guessed
    :rtype: str
    """
    ALL_LETTERS = 'abcdefghijklmnopqrstuvwxyz'
    guessed_set = set(letters_guessed.lower())
    return ''.join(ch for ch in ALL_LETTERS if ch not in guessed_set)

内容的提问来源于stack exchange，提问作者Learning_How_To_Code