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

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。

我过去常做的是……

question = 'Will the apple fall?'
print(question)
answer = int(input("Pls enter the answer: "
if answer == "y",
print('Well done')
print(answer)

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)

正如你提到的，最简单的方法是使用raw_input()(或简单的input()对于Python 3)。没有内置的方法可以做到这一点。配方577058:

import sys


def query_yes_no(question, default="yes"):
    """Ask a yes/no question via raw_input() and return their answer.

    "question" is a string that is presented to the user.
    "default" is the presumed answer if the user just hits <Enter>.
            It must be "yes" (the default), "no" or None (meaning
            an answer is required of the user).

    The "answer" return value is True for "yes" or False for "no".
    """
    valid = {"yes": True, "y": True, "ye": True, "no": False, "n": False}
    if default is None:
        prompt = " [y/n] "
    elif default == "yes":
        prompt = " [Y/n] "
    elif default == "no":
        prompt = " [y/N] "
    else:
        raise ValueError("invalid default answer: '%s'" % default)

    while True:
        sys.stdout.write(question + prompt)
        choice = input().lower()
        if default is not None and choice == "":
            return valid[default]
        elif choice in valid:
            return valid[choice]
        else:
            sys.stdout.write("Please respond with 'yes' or 'no' " "(or 'y' or 'n').\n")

(对于Python 2，使用raw_input而不是input。) 使用的例子:

>>> query_yes_no("Is cabbage yummier than cauliflower?")
Is cabbage yummier than cauliflower? [Y/n] oops
Please respond with 'yes' or 'no' (or 'y' or 'n').
Is cabbage yummier than cauliflower? [Y/n] [ENTER]
>>> True

>>> query_yes_no("Is cabbage yummier than cauliflower?", None)
Is cabbage yummier than cauliflower? [y/n] [ENTER]
Please respond with 'yes' or 'no' (or 'y' or 'n').
Is cabbage yummier than cauliflower? [y/n] y
>>> True

Python 3.8及以上版本的一行代码:

while res:= input("When correct, press enter to continue...").lower() not in {'y','yes','Y','YES',''}: pass