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()

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。

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

我修改了fmark的答案，用python 2/3兼容更pythonic。

如果您对更多错误处理感兴趣，请参阅ipython的实用程序模块

# PY2/3 compatibility
from __future__ import print_function
# You could use the six package for this
try:
    input_ = raw_input
except NameError:
    input_ = input

def query_yes_no(question, default=True):
    """Ask a yes/no question via standard input and return the answer.

    If invalid input is given, the user will be asked until
    they acutally give valid input.

    Args:
        question(str):
            A question that is presented to the user.
        default(bool|None):
            The default value when enter is pressed with no value.
            When None, there is no default value and the query
            will loop.
    Returns:
        A bool indicating whether user has entered yes or no.

    Side Effects:
        Blocks program execution until valid input(y/n) is given.
    """
    yes_list = ["yes", "y"]
    no_list = ["no", "n"]

    default_dict = {  # default => prompt default string
        None: "[y/n]",
        True: "[Y/n]",
        False: "[y/N]",
    }

    default_str = default_dict[default]
    prompt_str = "%s %s " % (question, default_str)

    while True:
        choice = input_(prompt_str).lower()

        if not choice and default is not None:
            return default
        if choice in yes_list:
            return True
        if choice in no_list:
            return False

        notification_str = "Please respond with 'y' or 'n'"
        print(notification_str)

正如Alexander Artemenko提到的，这里有一个使用strtobool()的简单解决方案。

from distutils.util import strtobool

def user_yes_no_query(question):
    sys.stdout.write('%s [y/n]\n' % question)
    while True:
        try:
            return strtobool(raw_input().lower())
        except ValueError:
            sys.stdout.write('Please respond with \'y\' or \'n\'.\n')

使用

>>> user_yes_no_query('Do you like cheese?')
Do you like cheese? [y/n]
Only on tuesdays
Please respond with 'y' or 'n'.
ok
Please respond with 'y' or 'n'.
y
>>> True

def question(question, answers):
    acceptable = False
    while not acceptable:
        print(question + "specify '%s' or '%s'") % answers
        answer = raw_input()
        if answer.lower() == answers[0].lower() or answers[0].lower():
            print('Answer == %s') % answer
            acceptable = True
    return answer

raining = question("Is it raining today?", ("Y", "N"))

换做是我就会这么做。

输出

Is it raining today? Specify 'Y' or 'N'
> Y
answer = 'Y'