Just keep in mind the following:

3 == 3 == 3
>>> True

# Not equivalent to
(3 == 3) == 3
>>> False

# Equivalent to
3 == 3 and 3 == 3
>>> True

Nice. I've seen the chained comparators with inequalities, but not equalities. Now that I see it, it's obvious that it would work. Just never thought of it before.

Does this map avoid recursion? If so, clever and practical idea!

Indeed that is a common pitfall.
I am surprised noone mentioned it before, thank you!

You're welcome ;-)

I had the same problem but it finally worked out. good job lechevalier,thanks, I am learning new things.

same_structure_as([1,'[',']'],['[',']',1]) returns True in Python 3.6 and Python 3.8 - the versions available on CW.
No reason to return anything else in other Python versions.

Hi, the same_structure_as([1,'[',']'],['[',']',1]) results to a False when I use your solution.

Python 3.7.3

The sturcture of these two arrays is actually the same. ']' is equivalent of a symbol or any non-list structure, so basically both original and other are equiv to [1, 1, 1]

Hi there,

I know that this is not a tested case, but have you tried with these values: original = [1, '[', ']'] and other = ['[', ']', 2]. Your solution returns True, but these strings are a bit diferent, so I suppose that it should return False... Am I wrong?