+ 10
Why does the code result in such output???
Hi everyone!!! In one of the challenges I get stuck on some particular code which yields the output I can't explain to myself. So the following code yields "False": arr = [1, True, 'a', 2] print('a' in arr in arr) Why does it happen??? Furthermore the following yields "True": print(('a' in arr) in arr) and the following raises "TypeError": print('a' in (arr in arr)) How can it be???
10 odpowiedzi
+ 9
I can't explain how the first expression evaluation worked. But for the second and third, I guess the use of parentheses plays a role in changing the evaluation priority.
Second expression:
( ( 'a' in arr ) in arr ) evaluated as
( ( True ) in arr ) and this yields True because <arr> has a member whose value is True.
Third expression:
( 'a' in ( arr in arr ) ) is evaluated as
( 'a' in ( False ) ) and this is not possible because the `in` operator requires an iterable for its RHS operand. But here the RHS operand is a boolean ( False ).
I look forward to learn how the first expression is evaluated.
+ 8
I can't say this with 100% assurance
But after running some test
I think both "in" operater are executed simultaneously
Like:
'a' in arr in arr
=('a' in arr) and (arr in arr)
Because only way to get result of True in a statement like
a in b in c
there should be
(a in b) == (b in c) == True
Hope It Helps You
+ 6
The explanation given by Hacker Badshah makes sense to me. It would mean that
a in b in c
works similar to
a < b < c
as both would be treated like two expressions connected with and.
+ 4
Hacker Badshah and Simon Sauter are right!
I tried appending <arr> to itself, and the first expression evaluates to True ...
arr = [1, True, 'a', 2]
print('a' in arr in arr) # output False
arr.append( arr )
print( 'a' in arr in arr ) # output True
All this time I thought only logical and arithmetic operators are chainable. Now I see that the `in` operator is also chainable LMAO :D
+ 4
Simon Sauter Gave very good example
a < b < c
works exactly similar to this
So, Alexander Sergeev we cracked the hard nut😂
+ 3
So as I :-) The second and third expressions are explainable but the first is a hard nut!
+ 2
Alexander Sergeev Absolutely Yes 😂
+ 2
You are right, guys! Thanks a lot!!!
Indeed, I've found in the Python Docs that "in" is a sort of comparisons and therefore it's chainable. So the following two lines of code give the same result "False":
print('a' in arr and arr in arr)
print('a' in arr in arr)
Moreover the next line gives you "False" too:
print(arr in arr)
But the next one raises a TypeError:
print('a' in (arr in arr))
because, assuming that (arr in arr) = False, we have equivalence:
print('a' in (arr in arr)) = print('a' in False)
but "in" operator is only for iterables, therefore we have a TypeError.
Again, thanks a lot to everyone!!!
+ 1
The hard nut is less harder than we are 🤣