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)

我会这样做:

# raw_input returns the empty string for "enter"
yes = {'yes','y', 'ye', ''}
no = {'no','n'}

choice = raw_input().lower()
if choice in yes:
   return True
elif choice in no:
   return False
else:
   sys.stdout.write("Please respond with 'yes' or 'no'")

这是我所使用的:

import sys

# cs = case sensitive
# ys = whatever you want to be "yes" - string or tuple of strings

#  prompt('promptString') == 1:               # only y
#  prompt('promptString',cs = 0) == 1:        # y or Y
#  prompt('promptString','Yes') == 1:         # only Yes
#  prompt('promptString',('y','yes')) == 1:   # only y or yes
#  prompt('promptString',('Y','Yes')) == 1:   # only Y or Yes
#  prompt('promptString',('y','yes'),0) == 1: # Yes, YES, yes, y, Y etc.

def prompt(ps,ys='y',cs=1):
    sys.stdout.write(ps)
    ii = raw_input()
    if cs == 0:
        ii = ii.lower()
    if type(ys) == tuple:
        for accept in ys:
            if cs == 0:
                accept = accept.lower()
            if ii == accept:
                return True
    else:
        if ii == ys:
            return True
    return False

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'

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,))