Relative order of elements in list
list, python, sorting
Solution
This can be written as a list comprehension like this:
lst = [1, 5, 4]
s = sorted(lst)
[s.index(x) for x in lst]
=> [0, 2, 1]
And here's another test, using @frb's example:
lst = [10, 2, 3, 9]
s = sorted(lst)
[s.index(x) for x in lst]
=> [3, 0, 1, 2]
Problem
I'm writing a function that takes in a list of integers and returns a list of relative positioned elements. That is to say, if my input into said function is [1, 5, 4] the output would be [0, 2, 1], since 1 is the lowest element, 5 is the highest and 4 in the middle, all elements are unique values, or a set() But code speaks, the function i have so far is ``` def relative_order(a): rel=[] for i in a: loc = 0 for v in a: if i > v: loc += 1 rel.append(loc) return rel ``` It does work, but since i'm sending big lists into this function, and i have to compare each element to all elements in each iteration it's taking ~5sec with a list of 10.000 elements. My question is how can i improve the speed on said function and maybe be a bit more Pythonic, i tried comprehension lists, but my Python skills are lacking and i only came up with an imperative way of implementing this problem.