How to parse a two-dimensional cyclically updated list into separate elements in python?

Question

Python

For example, such a list
list = . .. and it is replenished cyclically
When I have a match on the composition of the block (but not on the sequence), for example
[[30, 15, 25], ...........[15,25,30]. ...] I need to fix it.
How can I do that?
After receiving them, you need to be able to compare all the numbers, taking into account their order according to
the condition X(n)>=X(ne1)

Answer 1

It's not entirely clear what you need, but you can parse a two-dimensional list into individual elements like this:

arr = 
sum(arr, [])

[0, 0, 0, 30, 15, 25, 20, 5, 15, 10, -5, 5, 0, -15, -5]

Answer 2

To compare by the composition of the block, sort it or use set:

>>> set([1, 4, 3]) == set([4, 3, 1])
True

UPD: As rightly pointed out Umpiro set will throw out duplicate elements. So if you consider [1, 2, 1] and [1, 2] equal in composition, then use set. If not, then sort.