Why is "myval = [(2,3)][0][-2]" 2?

Question

Can somebody explain why this line of code:
myval = [(2,3)][0][-2]

results in myval beeing a integer with the value 2?

Answer 1

myvar=[(2,3,4),(5,6,7)][1][2]
#myvar will be 7

Firstly you create a list (with two tuples inside in this case), then you indicate the position of the value you want to take;
[1] indicates the second tuple (they start from 0, remember)
[2] indicates the third number, in this case 7, so myval will be 7

myval = [(2,3)][0][-2]
#[0] represents the first list, in this case is the only one available

in your example there is also [-2], it only means that it will start counting from the end, so
[-2] is 2
[-1] is 3
[0] is 2
[1] is 3

This kind of problem is often solvable changing the numbers that you have, seeing if it gives you any error and trying to understand what they means.

Answer 2

It's simple [(2, 3)] is an array having one tuple which is (2,3). So indexing it with 0 will give you the first and only member of the array i.e. [(2, 3)] [0] == (2, 3). Again if you are indexing [(2,3)] [0] with any index, it will give you the corresponding integer inside the tuple. Simply, [(2, 3)] [0] [-2] is the same as (2, 3) [-2]. If you use positive values:

(2, 3) [0] == 2
(2, 3) [1] == 3

but if you use negative value, the counting will start from the right. i.e.
(2, 3) [-1] == 3
(2, 3) [-2] == 2

Hope it helped :)