您可以使用单击的确认方法。

import click

if click.confirm('Do you want to continue?', default=True):
    print('Do something')

这将打印:

$ Do you want to continue? [Y/n]:

应该适用于Linux, Mac或Windows上的Python 2/3。

文档:http://click.pocoo.org/5/prompts/ # confirmation-prompts

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

对python 3执行同样的操作。X, raw_input()不存在:

def ask(question, default = None):
    hasDefault = default is not None
    prompt = (question 
               + " [" + ["y", "Y"][hasDefault and default] + "/" 
               + ["n", "N"][hasDefault and not default] + "] ")

    while True:
        sys.stdout.write(prompt)
        choice = input().strip().lower()
        if choice == '':
            if default is not None:
                return default
        else:
            if "yes".startswith(choice):
                return True
            if "no".startswith(choice):
                return False

        sys.stdout.write("Please respond with 'yes' or 'no' "
                             "(or 'y' or 'n').\n")

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,))

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)

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()