how to simplify if statement yes

Question

What's the best way to simplify this if?

#!/usr/bin/python

ans=input("choose yes or no: ")

if ans == "yes" or ans == "YES" or ans == "y" or ans == "Y":
    print("ok")
else:
        print("no")

Do you want yEs or yeS? If so, check [.lower()](https://docs.python.org/2/library/stdtypes.html#str.lower) aswell. — Lafexlos, Dec 31 '15 at 13:36

score 9 · Accepted Answer · answered Dec 31 '15 at 13:35

9

You could check it with list:

if ans in ['yes', 'YES', 'y', 'Y']:
    print('ok')
else:
    print('no')

answered Dec 31 '15 at 13:35

Anton Protopopov

27,206
10
83
90

TigerhawkT3 · Answer 2 · 2015-12-31T13:47:27.367

4

I would recommend streamlining it as follows:

if ans.lower().startswith('y'):
    print('ok')
else:
    print('no')

If lowercasing the whole thing when we're only interested in the first character feels wasteful, you can slice the first character (slices don't fail with IndexErrors like ordinary indexing can):

if ans[:1].lower() == 'y':

edited Dec 31 '15 at 13:47

answered Dec 31 '15 at 13:38

TigerhawkT3

46,954
6
53
87

1

will also catch 'yNo' as yes... but that's probably not a big deal... – pseudoDust Dec 31 '15 at 13:40
But it will also catch, say, "yse" as "yes." I'm happy with the tradeoff. – TigerhawkT3 Dec 31 '15 at 13:42
Yeah and this would have been my solution as well, just wanted to point this out... – pseudoDust Dec 31 '15 at 13:45

timgeb · Answer 3 · 2015-12-31T13:43:59.967

2

Make a set of your acceptable answers. This will also give you a lookup time of O(1) which may come in handy once you have a large number of acceptable answers.

acceptable_answers = {'yes', 'YES', 'y', 'Y'}
if ans in acceptable_answers:
     #do stuff
else:
     #do other stuff

edited Dec 31 '15 at 13:43

answered Dec 31 '15 at 13:37

timgeb

73,231
20
109
138

what is "set" keywork? – Pol Hallen Dec 31 '15 at 13:40
@PolHallen a `set` is an unordered collection of unique elements. – timgeb Dec 31 '15 at 13:41
2

You can use a literal if you want, e.g. `...= {'yes', 'YES', 'y', 'Y'}`. – TigerhawkT3 Dec 31 '15 at 13:43
@TigerhawkT3 I like it – timgeb Dec 31 '15 at 13:44

score 2 · Answer 4 · answered Dec 31 '15 at 13:44

2

One more way:

accepted = answer.lower() in ['y','yes']
print ('ok' if accepted else 'no')

answered Dec 31 '15 at 13:44

Ashalynd

12,113
2
30
36

how to simplify if statement yes

4 Answers4