Why is the output False?

Question

a=[1,8,8,False]
print ((True is a[2]) in a) #True
print(True is (a[2] in a)) #True

print(True is a[2] in a) #False - why??? True is not 8. So it's False. But False in a - so It must be True. It seems like 'in a' ignores at all
print(8 is a[2] in a) #True - why??? The same as above

b=[0,0]
print(8 is a[2] in b) #False . In this case we see that 'in b' wasn't be ignored. And result is totally different.
b=[True]
print(8 is a[2] in b) #False

Answer 1

The doc puts is, is not, in, not in and every comparison on one precedence level and states they are evaluated from left to right, chained up.

So True is a[2] in a should be equal to:

(True is a[2]) and (a[2] in a)

Just like 1<3<5 equals

1<3 and 3<5

Since only one of the terms is True: False.

https://docs.python.org/3/reference/expressions.html

Answer 2

print(True is a[2] in a) # False - why???
Because "python" Do not care about "in a "

Example :
print(True is a[2] in C) # False

Although "C" is not defined

Answer 3

That’s ok Aravazhi I was thinking the same 🤔

Thanks Hicham I already knew the difference in those operators but I’m missing how they’re applied in the flow of execution here?

For example True is a[2] in a

If the left/‘is’ part is evaluated first

True is a[2] = False
False in a = False (because they are two different boolean objects?)

If the right/‘in’ part is evaluated first
a[2] in a = True
True is True = False (again different boolean objects?)

If so, I then wouldn’t know how the first two statements evaluate to their boolean values

Answer 4

Sorry Эдуард for hijacking the comments

My take from Hicham‘s answer is that booleans have been evaluated as integers which makes the ‘is’ operator see them as different boolean objects/values

Someone correct me if otherwise?

Answer 5

https://code.sololearn.com/cWdzQwVT0bpC/?ref=app

Answer 6

HonFu master of Python 👏👏👏👏👏👏👏👏👏

Answer 7

Gordon explains it.

False is True in invalid_name ->
False is True and True in invalid_name

Works because 'and' short circuits the check.

Answer 8

sometimes the bool(True/False ) takes intigers values
Like : (True * 3 )== 3
I think that our bollien taking integer values, so using 'is' gives us 'False'

Answer 9

Hicham thanks so much for your reply 🙂 I understand what you said and have learned that booleans can be used that way as integers

Answer 10

https://code.sololearn.com/cUbwA585OqYl/?ref=app

I referred to the site: https://stackoverflow.com/questions/13650293/understanding-pythons-is-operator
and tried to verify the objects being referred using id() function.. still could not understand the statement in question why it returns False..

throughout the execution, true is pointing to single object and false is pointing to single object..

Answer 11

Hicham
https://code.sololearn.com/cZJtgU67wiuv/?ref=app

Answer 12

a=[1,8,8,False]

print ((True is a[2]) in a) #True
print(True is (a[2] in a)) #True

print(True is a[2] in a) #False - why???
print(True is 8) # diff values, so False

print(8 is a[2] in a) #True - why
print(8 is 8) #same values, so true

b=[0,0]
print(8 is a[2] in b) #False
b=[True]
print(8 is a[2] in b) #False

Answer 13

So without brackets - like the first print statements - the ‘in’ part of the expression is evaluated first? a[2] in a and return a[2]?

Answer 14

Jenine
Thanks, your query clarified that my response is not correct.. a[2] in a will return True, not a[2]..
In that case, I am not sure how the statement print(True is a[2] in a) returns False, as I feel it should return True whichever operator ( is or in) is evaluated first..

Answer 15

And i'm still don't understand what's going on in this code😔