How do I get the indices of a reverse sorted list?

Question

Say I have array:

A = [3, 3, 2, 2, 4]

The indices of the sorted list (ascending) would be:

[2, 3, 0, 1, 4]

I want the reverse sorted list (descending):

[4, 0, 1, 2, 3]

Note that the reverse sorted list is not a direct reverse of the sorted list.

I would have used numpy; however, if I do this:

indices = np.argsort(A)[::-1]

# This gives: indices = [4, 1, 0, 3, 2]
# But I want: indices = [4, 0, 1, 2, 3]

Thus, how do I get the indices of the reverse sorted list that I want? Or is there some way to do this with numpy?

Thanks!

How did you end up with 0 and then 1 as the second and third output elements, considering both elements at those indices are 3? — Divakar, Aug 26 '19 at 18:45
@VickConan: it maps `3` on `-3`, etc. So that the largest element in `A` is the smallest in `-A`, etc. — Willem Van Onsem, Aug 26 '19 at 18:49
@Divakar If 2 elements are the same, like 3 and 3, I want the index of the first element. — Vick Conan, Aug 26 '19 at 18:50
@WillemVanOnsem Yup, OP simply needs to use `mergsort` to keep the order that way : `(-A).argsort(kind='mergesort')`. — Divakar, Aug 26 '19 at 18:52

Willem Van Onsem · Accepted Answer · 2019-08-26T18:53:39.087

1

You can np.argsort(..) the negative array here, like:

>>> np.argsort(-np.array([3, 3, 2, 2, 4]), kind='mergesort')
array([4, 0, 1, 2, 3])

or since numpy-1.15, you can use:

>>> np.argsort(-np.array([3, 3, 2, 2, 4]), kind='stable')
array([4, 0, 1, 2, 3])

edited Aug 26 '19 at 18:53

answered Aug 26 '19 at 18:42

Willem Van Onsem

1 Answers1