[FIXED] Get the index in order of values from max to min without sorting the output index list and amend another list as well

Issue

I am trying to get the indices of the values in the array from maximum to minimum without sorting the indices positions. Here is the current code I have

s = [20, 30, 45, 14, 59]
ans = sorted(range(len(s)), key=lambda k: s[k], reverse=True)
print ans

but I am getting the output as [4, 2, 1, 0, 3] which is based on sorting. I don't want the sorted list of indexes instead I want them in same position. Output should be [3, 2, 1, 4, 0]. Is there any way to achieve this?

For 2D: Now if there is another array (only with -1, 0, 1) associated with the first array sv = [(1,0), (-1,1), (-1,0), (0,-1), (1,1)]. Here each value in sv is indexed and bind to same output array [3, 2, 1, 4, 0]. Now based on certain condition say

for i in s if any val[i] < max(val[i])*25/100

value will be removed from s as well as sv and also its indexing. So in above case new s, new sv and new indexing output will be

s = [20, 30, 45, 59]
indexing = [3, 2, 1, 0]
sv = [(1,0), (-1,1), (-1,0), (1,1)]

Solution

You can use a dict that maps each list item to its sorted index:

mapping = {k: i for i, k in enumerate(sorted(s, reverse=True))}
print([mapping[i] for i in s])

This outputs:

[3, 2, 1, 4, 0]

Answered By - blhsing