按从最大值到最小值的值顺序获取索引，而无需对输出索引列表进行排序，也可以修改另一个列表

Question

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)]

Answer 1

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]

Answer 2

s = [20, 30, 45, 14, 59]
ss = sorted(s,reverse=True)


print([s.index(i) for i in ss]) ## gives [4, 2, 1, 0, 3]
print([ss.index(i) for i in s]) ## gives [3, 2, 1, 4, 0]