I have seen answers to the question:
Is it possible to arrange a numpy array (or python list) by using the indexes of the elements in decreasing order? (eg. Finding the Index of N biggest elements in Python Array / List Efficiently)
A very concise answer seems to be (from above link):
L = array([4, 1, 0, 8, 5, 2])
sorted(range(len(L)), key=lambda i:L[i])
This gives the position (in the original array) of the sorted elements.
8 --> 3
5 --> 4
4 --> 0
2 --> 5
1 --> 1
0 --> 2
So the answer is:
[3, 4, 0, 5, 1, 2]
What I am after is the position (in the sorted array) of the elements:
L = array([4, 1, 0, 8, 5, 2])
8 --> 0
5 --> 1
4 --> 2
2 --> 3
1 --> 4
0 --> 5
So I want :
[2, 4, 5, 0, 1, 3]
I realise I could take the answer from the first example, and use that to get what I want (with a bit more fiddling) but is there a short cut?
Efficiency is not a concern. I just need something that gives me the answer.
You can simply use your technique twice to get the indices in the sorted list:
A=[4, 1, 0, 8, 5, 2]
B=sorted(range(len(A)),key=lambda x:A[x],reverse=True)
C=sorted(range(len(A)),key=lambda x:B[x])
print C
prints
[2, 4, 5, 0, 1, 3]
The idea is that the first iteration produces a list:
B = [3, 4, 0, 5, 1, 2]
giving the locations in the original list of the sorted sequence.
In other words, A[3]=8 is the largest element in the original list, A[4]=5 is the next largest, etc.
The second stage then sorts these indices in B back into the order 0,1,2,3,4,5 and produces C which contains the index in the list B.
It may help to think of B as sorting the list into descending order, and C as reversing the sort back into the original unsorted order, while keeping track of the indices in the sorted list.
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