Dead simple python function for getting a yes or no answer.

yesOrNo.py

def yes_or_no(question):
    reply = str(raw_input(question+' (y/n): ')).lower().strip()
    if reply[0] == 'y':
        return True
    if reply[0] == 'n':
        return False
    else:
        return yes_or_no("Uhhhh... please enter ")

Correct me if I am wrong but this might be OK with the Assigment Expressions (PEP 572) in Python 3.8

while res:= input("Do you want to save? (Enter y/n)").lower() not in {"y", "n"}: pass

And if you don't like hitting ENTER for whatever reason, something like this might work:

def getch():
    """Read single character from standard input without echo."""
    import sys, tty, termios
    fd = sys.stdin.fileno()
    old_settings = termios.tcgetattr(fd)
    try:
        tty.setraw(sys.stdin.fileno())
        ch = sys.stdin.read(1)
    finally:
        termios.tcsetattr(fd, termios.TCSADRAIN, old_settings)
    return ch


def yes_or_no(question):
    c = ""
    print(question + " (Y/N): ", end = "", flush = True)
    while c not in ("y", "n"):
        c = getch().lower()
    return c == 'y'


if __name__ == "__main__":
    if not yes_or_no("Continue?"):
        print("NO") # ...or echo nothing and just provide newline
        os._exit(0)
    else:
        print("YES")

Granted, this is very far from "dead simple", but its an alternative nevertheless. (Looking at robertour's suggestion, I wish Debian 10 would have gone with 3.8 instead of 3.7...)

What about using strtobool?

def query_yes_no(question, default='no'):
    if default is None:
        prompt = " [y/n] "
    elif default == 'yes':
        prompt = " [Y/n] "
    elif default == 'no':
        prompt = " [y/N] "
    else:
        raise ValueError(f"Unknown setting '{default}' for default.")

    while True:
        try:
            resp = input(question + prompt).strip().lower()
            if default is not None and resp == '':
                return default == 'yes'
            else:
                return distutils.util.strtobool(resp)
        except ValueError:
            print("Please respond with 'yes' or 'no' (or 'y' or 'n').\n")

adapted from this SO question.

I propose renaming it to whats_it_gonna_be_boy.

I agree :)

Lots of great answers here, but prompted for something a little simpler and recursive.

#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""Simple yes/no function demo."""

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


def main():
    """Run main function."""
    ans = yesno("What'll it be")
    print(f'you answer was: {ans}')



if __name__ == '__main__':
    main()

Version without recursion:

def single_yes_or_no_question(question, default_no=True):
    choices = ' [y/N]: ' if default_no else ' [Y/n]: '
    default_answer = 'n' if default_no else 'y'
    reply = str(input(question + choices)).lower().strip() or default_answer
    if reply[0] == 'y':
        return True
    if reply[0] == 'n':
        return False
    else:
        return False if default_no else True

Correct me if I am wrong but this might be OK with the Assigment Expressions (PEP 572) in Python 3.8

while res:= input("Do you want to save? (Enter y/n)").lower() not in {"y", "n"}: pass

Great idea. A couple more brackets are needed here:

while ( res:=input("Do you want to save? (Enter y/n)").lower() ) not in {"y", "n"}: pass

and then ...

if res=='y':
  ... do something ...

Lots of great answers here, but prompted for something a little simpler and recursive.

#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""Simple yes/no function demo."""

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


def main():
    """Run main function."""
    ans = yesno("What'll it be")
    print(f'you answer was: {ans}')



if __name__ == '__main__':
    main()

Thank you so much mate, I didn't know that we can return a function that I was manipulating the code in quite different ways. It changed my many lines of stuck code to refactoring.

#!/usr/bin/python3
def confirm_prompt(question: str) -> bool:
    reply = None
    while reply not in ("y", "n"):
        reply = input(f"{question} (y/n): ").lower()
    return (reply == "y")


reply = confirm_prompt("Are you sure?")
print(reply)

Edit: added keyword argument to set either yes or no by default

#!/usr/bin/python3
def prompt(question: str, default=None) -> bool:
    choices = ("", "y", "n") if default in ("yes", "no") else ("y", "n")
    hint = "Y/n" if default == "yes" else "y/n"
    hint = "y/N" if default == "no" else hint
    reply = None

    while reply not in choices:
        reply = input(f"{question} ({hint}): ").lower()
    return (reply == "y") if default != "yes" else (reply in ("", "y"))


reply = prompt("Are you sure?")
print(reply, "\n")

reply = prompt("Are you sure?", default="yes")
print(reply, "\n")

reply = prompt("Are you sure?", default="no")
print(reply, "\n")

I agree :)

@gyoza necrothreading this just to kudos you on the fact that you quoted Meatloaf

Stop using raw_input, that's Python 2.

How would I add an --yes or -y parser arg to @icamys solution?
first
parser.add_argument("--yes", "-y", help="Always accept")
and then what? :D sorry I am really new to python

maybe something like:

def confirm(question, default_no=True):
    choices = ' [y/N]: ' if default_no else ' [Y/n]: '
    default_answer = 'n' if default_no else 'y'
    reply = str(input(question + choices)).lower().strip() or default_answer
    if reply[0] == 'y':
        return True
    if reply[0] == 'n':
        return False
    if args.yes:
        default_answer='y'
    else:
        return False if default_no else True

but that doesnt work

@willbelr, Thank you for this beautiful & elegant solution! I modified with casefold (in Python3) but am otherwise using as-is in a few scripts, and it works great!

def confirm_prompt(question: str) -> bool:
    reply = None
    while reply not in ("y", "n"):
        reply = input(f"{question} (y/n): ").casefold()
    return (reply == "y")


reply = confirm_prompt("Are you sure?")
print(reply)

My version using simple recursion:

def user_confirm(question: str) -> bool:
    reply = str(input(question + ' (y/n): ')).lower().strip()
    if reply[0] == 'y':
        return True
    elif reply[0] == 'n':
        return False
    else:
        new_question = question
        if "Please try again - " not in question:
            new_question = f"Please try again - {question}"
        return user_confirm(new_question)


if __name__ == "__main__":
    print(user_confirm("Do you love me?"))

I prefer this approach:

def yes_or_no_prompt(question:str,default_yes=True)->bool:

    inputMapping={
        'y':'y',
        "υ":'y',
        "Υ":'y',
        'n':'n',
        'ν':'n',
        'N':'n'
    }

    question = question.strip()

    if question == "":
        raise ValueError(f"Question is an empty string")

    default ='n'
    prompt_choice="(y/N)"
    if default_yes:
        default='y'
        prompt_choice="(Y/n)"

    reply = str(input(f"{question} {prompt_choice}: ")).lower().strip()
    reply = reply[0] or default
    reply = inputMapping[reply]

    if reply not in ['y','n']:
        raise ValueError(f'Invalid input for reply given {reply} instead of Y or N')

    return reply=='y'
 
 
 # Usage Example
 if __name__ == "__main__":

   while True:
     try:
        if yes_or_no_prompt("Are roses red?"):
           print("Correct")
        else:
           print("Nope")
        exit(0)
     except ValueError:
          print("I made a Boo Boo mommy")

The reason why is because I split whether I'll re-ask the question into another point whereas the value sanitization and checking is performed upon the function itself. Furthermore I set the default value as well. Furthermore I map any Greek Input given from user intop its respective English one in case keyboard layout is a non-English one.