Comparing Two Lists to Find Bigger List

Question

I have two lists and need to compare them by their largest element, if that is tied, their 2nd largest element, and if that is tied, 3rd largest etc iterating to the entire array.

For example:

list1= [0,2,3,6,12]
list2= [1,2,3,6,12]
list3= [1,4,5,8,12]
list4= [1,4,5,9,12]

So list4 > list3 > list2 > list1.

I wrote a function that accomplishes this:

def compare(x,y):
    if sorted(x)==sorted(y):
        return "Tie"
    for index in range(len(x)-1,-1,-1):
        if sorted(x)[index]>sorted(y)[index]:
            return x
        elif sorted(x)[index]<sorted(y)[index]:
            return y

I was wondering if there was a much neater and more efficient way of writing the function because it doesn't seem very Pythonic.

Edit: comparing lists by using "<" and ">" will sort the lists from the smallest index to largest index, not largest index to smallest index. Reversed would make ">" and "<" the simplest solutions.

Answer 1

How about this?

>>> list1= [0,2,3,6,12]
>>> list2= [1,2,3,6,12]
>>> list3= [1,4,5,8,12]
>>> list4= [1,4,5,9,12]
>>> def sort_lists_by_maxes(*lists):
    return sorted(lists, key=lambda x: sorted(x, reverse=True), reverse=True)

>>> sort_lists_by_maxes(list1, list2, list3, list4)
[[1, 4, 5, 9, 12], [1, 4, 5, 8, 12], [1, 2, 3, 6, 12], [0, 2, 3, 6, 12]]

The lists are compared by their individually sorted values, and you can feed as many lists as you would like into the function as arguments.