Is there any short way to achieve what the APT (Advanced Package Tool) command line interface does in Python?
I mean, when the package manager prompts a yes/no question followed by [Yes/no], the script accepts YES/Y/yes/y or Enter (defaults to Yes as hinted by the capital letter).
The only thing I find in the official docs is input and raw_input...
I know it's not that hard to emulate, but it's annoying to rewrite :|
As you mentioned, the easiest way is to use raw_input() (or simply input() for Python 3). There is no built-in way to do this. From Recipe 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")
(For Python 2, use raw_input instead of input.)
Usage example:
>>> 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
I'd do it this way:
# 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'")
You can use click's confirm method.
import click
if click.confirm('Do you want to continue?', default=True):
print('Do something')
This will print:
$ Do you want to continue? [Y/n]:
Should work for Python 2/3 on Linux, Mac or Windows.
Docs: http://click.pocoo.org/5/prompts/#confirmation-prompts
There is a function strtobool in Python's standard library: http://docs.python.org/2/distutils/apiref.html?highlight=distutils.util#distutils.util.strtobool
You can use it to check user's input and transform it to True or False value.
A very simple (but not very sophisticated) way of doing this for a single choice would be:
msg = 'Shall I?'
shall = input("%s (y/N) " % msg).lower() == 'y'
You could also write a simple (slightly improved) function around this:
def yn_choice(message, default='y'):
choices = 'Y/n' if default.lower() in ('y', 'yes') else 'y/N'
choice = input("%s (%s) " % (message, choices))
values = ('y', 'yes', '') if choices == 'Y/n' else ('y', 'yes')
return choice.strip().lower() in values
Note: On Python 2, use raw_input instead of input.
as mentioned by @Alexander Artemenko, here's a simple solution using strtobool
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')
#usage
>>> 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
I know this has been answered a bunch of ways and this may not answer OP's specific question (with the list of criteria) but this is what I did for the most common use case and it's far simpler than the other responses:
answer = input('Please indicate approval: [y/n]')
if not answer or answer[0].lower() != 'y':
print('You did not indicate approval')
exit(1)
You can also use prompter.
Shamelessly taken from the README:
#pip install prompter
from prompter import yesno
>>> yesno('Really?')
Really? [Y/n]
True
>>> yesno('Really?')
Really? [Y/n] no
False
>>> yesno('Really?', default='no')
Really? [y/N]
True
I modified fmark's answer to by python 2/3 compatible more pythonic.
See ipython's utility module if you are interested in something with more error handling
# 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)
For Python 3, I'm using this function:
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")
The strtobool() function converts a string into a bool. If the string cant be parsed it will raise a ValueError.
In Python 3 raw_input() has been renamed to input().
As Geoff said, strtobool actually returns 0 or 1, therefore the result has to be cast to bool.
This is the implementation of strtobool, if you want special words to be recognized as true, you can copy the code and add your own cases.
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,))
on 2.7, is this too non-pythonic?
if raw_input('your prompt').lower()[0]=='y':
your code here
else:
alternate code here
it captures any variation of Yes at least.
Doing the same with python 3.x, where raw_input() doesn't exist:
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")
You could try something like the code below to be able to work with choices from the variable 'accepted' show here:
print( 'accepted: {}'.format(accepted) )
# accepted: {'yes': ['', 'Yes', 'yes', 'YES', 'y', 'Y'], 'no': ['No', 'no', 'NO', 'n', 'N']}
Here is the code ..
#!/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
As a programming noob, I found a bunch of the above answers overly complex, especially if the goal is to have a simple function that you can pass various yes/no questions to, forcing the user to select yes or no. After scouring this page and several others, and borrowing all of the various good ideas, I ended up with the following:
def yes_no(question_to_be_answered):
while True:
choice = input(question_to_be_answered).lower()
if choice[:1] == 'y':
return True
elif choice[:1] == 'n':
return False
else:
print("Please respond with 'Yes' or 'No'\n")
#See it in Practice below
musical_taste = yes_no('Do you like Pine Coladas?')
if musical_taste == True:
print('and getting caught in the rain')
elif musical_taste == False:
print('You clearly have no taste in music')
This is what I use:
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"))
This is how I'd do it.
Output
Is it raining today? Specify 'Y' or 'N'
> Y
answer = 'Y'
Here's my take on it, I simply wanted to abort if the user did not affirm the action.
import distutils
if unsafe_case:
print('Proceed with potentially unsafe thing? [y/n]')
while True:
try:
verify = distutils.util.strtobool(raw_input())
if not verify:
raise SystemExit # Abort on user reject
break
except ValueError as err:
print('Please enter \'yes\' or \'no\'')
# Try again
print('Continuing ...')
do_unsafe_thing()
How about this:
def yes(prompt = 'Please enter Yes/No: '):
while True:
try:
i = raw_input(prompt)
except KeyboardInterrupt:
return False
if i.lower() in ('yes','y'): return True
elif i.lower() in ('no','n'): return False
Since the answer is expected yes or no, in the examples below, the first solution is to repeat the question using the function while, and the second solution is to use recursion - is the process of defining something in terms of itself.
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?")
second solution:
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()
One-liner with Python 3.8 and above:
while res:= input("When correct, press enter to continue...").lower() not in {'y','yes','Y','YES',''}: pass
Python x.x
res = True
while res:
res = input("Please confirm with y/yes...").lower(); res = res not in {'y','yes','Y','YES',''}
What I used to do is...
question = 'Will the apple fall?'
print(question)
answer = int(input("Pls enter the answer: "
if answer == "y",
print('Well done')
print(answer)
