为什么Python中的“if-letter in-dict”工作错误？

Question

我正在尝试编写一个函数，它返回一个字符串，该字符串中的字母不是LettersGuesed

words = list('abcdefghijklmnopqrstuvwxyz')

def getAvailableLetters(lettersGuessed):
    [words.remove(letter) for letter in words if letter in lettersGuessed]
    return ''.join([str(elem) for elem in words]) 

然后我调用我的函数

getAvailableLetters(['e', 's', 'i', 'k', 'p', 'r'])

但结果与我预期的不同。 我的输出：

'abcdfghjlmnoqstuvwxyz' 

# the letter 's' in the output, but it shouldn't be there.

正确输出：

'abcdfghjlmnoqtuvwxyz' 

我做错了什么

Answer 1

原因是您在遍历列表时正在修改它-不应该这样做。 我将在这里分享@AlexMartelli在one of his answers中给出的建议：

Never alter the container you're looping on, because iterators on that container are not going to be informed of your alterations and, as you've noticed, that's quite likely to produce a very different loop and/or an incorrect one.

这里有一个替代解决方案（也使用snake_case），使用集合-假设可用字母和猜测字母都不包含重复的字母（除非错误）：

all_letters = set('abcdefghijklmnopqrstuvwxyz')

def get_available_letters(letters_guessed):
    available_letters = all_letters - set(letters_guessed)
    return ''.join([str(letter) for letter in available_letters])

这里有一个警告，上面的解决方案可能不遵守您指定字母的顺序

如果您仍想坚持列表，请分两个阶段进行：

all_letters = list('abcdefghijklmnopqrstuvwxyz')

def get_available_letters(letters_guessed):
    available_letters = [letter for letter in all_letters if letter not in letters_guessed]
    return ''.join([str(letter) for letter in available_letters])