Python: if answer.upper or answer.lower disregards input and always does 1st if

Asked Mar 10 '21 at 14:24

Active Mar 11 '21 at 01:37

Viewed 31 times

I want to have a loop which does something according to Y/N input. If I use .upper / .lower, the loop does not work and always goes into the first if, no matter the input. Why is that?

Here's and example of how it doesn't work:

answer = input("Do you want to begin? (Y/N): ")
if answer or answer.lower() in ("y", "yes"):        
    print("Start")
else:
    print("Bye")

If I write it like this, it works fine.

if answer in ("y","Y","yes","Yes","YES"):

This I found online as solution, but doesn't work for me:

if answer.upper() or answer.lower() in ('y', 'yes'):

Am I just missing something or did anyone else have this issue? Thanks for the help

edited Mar 11 '21 at 01:37

atin

asked Mar 10 '21 at 14:24

Val084a

2

``a or b in c`` is true when *either* ``a`` is true or ``b in c`` is true. – MisterMiyagi Mar 10 '21 at 14:30
2

you are caught with short circuit evaluation – Epsi95 Mar 10 '21 at 14:30
1

Note that if ``answer in ("y", "yes")`` then ``answer.lower() in ("y", "yes")`` as well. It's enough to just test the latter. – MisterMiyagi Mar 10 '21 at 14:32
1

@Epsi95: No, short-circuiting doesn't matter here. The results would be the same without short circuiting. – user2357112 Mar 11 '21 at 01:39
@Epsi95 What he's caught in is operator precedence, and also that `or` doesn't distribute. – Barmar Mar 11 '21 at 01:47

Python: if answer.upper or answer.lower disregards input and always does 1st if

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