in operator python

Question

Hello dear coders,

Could you please explain why the following code outputs False:

a = [1, True, 'a', 2]
print('a' in a in a)

Answer 1

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 :)

Answer 2

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 : )

Answer 3

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

Answer 4

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

Answer 5

Mozzy Ipang Bloody Halos, Thanks a lot for your answers!