+ 3

in operator python

Hello dear coders, Could you please explain why the following code outputs False: a = [1, True, 'a', 2] print('a' in a in a)

9th Jan 2023, 12:41 PM
Ruslan Guliyev
Ruslan Guliyev - avatar
5 odpowiedzi
+ 8
Ruslan Guliyev Intuitively, we would expect something like this: ('a' in a in a) --> (('a' in a) in a) --> (True in a) --> True But as you noticed, ('a' in a in a) returns False instead. This is due to the operator chaining feature of python. https://docs.python.org/3.8/reference/expressions.html#comparisons Some operators like `in`, `is`, `<`, etc are treated differently when chained together. For example, (3 < x < 5) is evaluated as ((3 < x) and (x < 5)). So this is how python actually evaluates your code: ('a' in a in a) --> (('a' in a) and (a in a)) --> (True and False) --> False Hope this helps :)
9th Jan 2023, 3:47 PM
Mozzy
Mozzy - avatar
+ 4
I could possibly be wrong, but as I understand it, chained operators go like there's a logical AND in between So ... 'a' in a in a May be equivalent to 'a' in a AND a in a True AND False With a final result of False Just a thought : )
9th Jan 2023, 3:39 PM
Ipang
+ 1
Don't know much about inner workings of python, kind of just got into it. However it seems like the in operator evaluations need precedence. I c/p your code and put print( ('a' in a) in a) and it outputs true. Not sure if this necessarily gives you the answer you are looking for, but hope it helps at least
9th Jan 2023, 12:49 PM
Bloody Halos
Bloody Halos - avatar
+ 1
So with what others said, guess it has to do less with precedence than using () to make it evaluate first in operator separately, rather than looking at the whole line As in below works as it outputs true: b = [5, 1, True, 'a', 2, False ] print(('b' in b) in b) ('b' in b) == False (False in b) == True Output: True
9th Jan 2023, 8:02 PM
Bloody Halos
Bloody Halos - avatar
+ 1
Mozzy Ipang Bloody Halos, Thanks a lot for your answers!
19th Jan 2023, 3:33 AM
Ruslan Guliyev
Ruslan Guliyev - avatar