你可以尝试下面的代码来处理变量'accepted'中的选项:

print( 'accepted: {}'.format(accepted) )
# accepted: {'yes': ['', 'Yes', 'yes', 'YES', 'y', 'Y'], 'no': ['No', 'no', 'NO', 'n', 'N']}

这是密码。

#!/usr/bin/python3

def makeChoi(yeh, neh):
    accept = {}
    # for w in words:
    accept['yes'] = [ '', yeh, yeh.lower(), yeh.upper(), yeh.lower()[0], yeh.upper()[0] ]
    accept['no'] = [ neh, neh.lower(), neh.upper(), neh.lower()[0], neh.upper()[0] ]
    return accept

accepted = makeChoi('Yes', 'No')

def doYeh():
    print('Yeh! Let\'s do it.')

def doNeh():
    print('Neh! Let\'s not do it.')

choi = None
while not choi:
    choi = input( 'Please choose: Y/n? ' )
    if choi in accepted['yes']:
        choi = True
        doYeh()
    elif choi in accepted['no']:
        choi = True
        doNeh()
    else:
        print('Your choice was "{}". Please use an accepted input value ..'.format(choi))
        print( accepted )
        choi = None

你可以尝试下面的代码来处理变量'accepted'中的选项:

print( 'accepted: {}'.format(accepted) )
# accepted: {'yes': ['', 'Yes', 'yes', 'YES', 'y', 'Y'], 'no': ['No', 'no', 'NO', 'n', 'N']}

这是密码。

#!/usr/bin/python3

def makeChoi(yeh, neh):
    accept = {}
    # for w in words:
    accept['yes'] = [ '', yeh, yeh.lower(), yeh.upper(), yeh.lower()[0], yeh.upper()[0] ]
    accept['no'] = [ neh, neh.lower(), neh.upper(), neh.lower()[0], neh.upper()[0] ]
    return accept

accepted = makeChoi('Yes', 'No')

def doYeh():
    print('Yeh! Let\'s do it.')

def doNeh():
    print('Neh! Let\'s not do it.')

choi = None
while not choi:
    choi = input( 'Please choose: Y/n? ' )
    if choi in accepted['yes']:
        choi = True
        doYeh()
    elif choi in accepted['no']:
        choi = True
        doNeh()
    else:
        print('Your choice was "{}". Please use an accepted input value ..'.format(choi))
        print( accepted )
        choi = None

对于Python 3，我使用这个函数:

def user_prompt(question: str) -> bool:
    """ Prompt the yes/no-*question* to the user. """
    from distutils.util import strtobool

    while True:
        user_input = input(question + " [y/n]: ")
        try:
            return bool(strtobool(user_input))
        except ValueError:
            print("Please use y/n or yes/no.\n")

函数的作用是:将字符串转换为bool类型。如果字符串不能被解析，它将引发ValueError。

在Python 3中，raw_input()已重命名为input()。

正如Geoff所说，strtoool实际上返回0或1，因此结果必须转换为bool类型。

这是strtobool的实现，如果你想让特殊的单词被识别为true，你可以复制代码并添加自己的case。

def strtobool (val):
    """Convert a string representation of truth to true (1) or false (0).
    True values are 'y', 'yes', 't', 'true', 'on', and '1'; false values
    are 'n', 'no', 'f', 'false', 'off', and '0'.  Raises ValueError if
    'val' is anything else.
    """
    val = val.lower()
    if val in ('y', 'yes', 't', 'true', 'on', '1'):
        return 1
    elif val in ('n', 'no', 'f', 'false', 'off', '0'):
        return 0
    else:
        raise ValueError("invalid truth value %r" % (val,))

在2.7中，这是不是太非python化了?

if raw_input('your prompt').lower()[0]=='y':
   your code here
else:
   alternate code here

它至少能捕捉到“是”的任何变化。

由于答案是“是”或“否”，在下面的例子中，第一个解决方案是使用while函数重复这个问题，第二个解决方案是使用递归-是定义事物本身的过程。

def yes_or_no(question):
    while "the answer is invalid":
        reply = str(input(question+' (y/n): ')).lower().strip()
        if reply[:1] == 'y':
            return True
        if reply[:1] == 'n':
            return False

yes_or_no("Do you know who Novak Djokovic is?")

第二个解决方案:

def yes_or_no(question):
    """Simple Yes/No Function."""
    prompt = f'{question} ? (y/n): '
    answer = input(prompt).strip().lower()
    if answer not in ['y', 'n']:
        print(f'{answer} is invalid, please try again...')
        return yes_or_no(question)
    if answer == 'y':
        return True
    return False

def main():
    """Run main function."""
    answer = yes_or_no("Do you know who Novak Djokovic is?")
    print(f'you answer was: {answer}')


if __name__ == '__main__':
    main()

我知道这已经被回答了很多方法，这可能不能回答OP的具体问题(标准列表)，但这是我为最常见的用例所做的，它比其他回答简单得多:

answer = input('Please indicate approval: [y/n]')
if not answer or answer[0].lower() != 'y':
    print('You did not indicate approval')
    exit(1)