What is the most pythonic way to retrieve the indexes of values sorted by the values?

Question

I have a list a = [1, 5, 2, 8]

I need a function that will return the indexes of the entries sorted by their values, i.e. b = [3, 1, 2, 0]

I came up with a function to do this:

def indices_sorted_by_val(a, reverse=True):
    return [ y[0] for y in  sorted( [(i,a[i]) for i in range(len(a))], key=lambda x: x[1], reverse=reverse )]

But it's messy and hard to read. Is there a more pythonic way?

`[i[0] for i in sorted(enumerate(a), key=lambda x:x[1], reverse=reverse)]` or `np.argsort(a)[::-1 if reverse else None]` to use the numpy approach. — Martijn Pieters, Jun 20 '18 at 16:18

score 0 · Answer 1 · answered Jun 20 '18 at 16:19

0

First try ... enumerate will help. I vote against a one-liner, because readibility counts

a = [1, 5, 2, 8]

sorted_tuples = sorted(enumerate(a), key= lambda x: x[1], reverse=True)
print([index for index, value in sorted_tuples])

answered Jun 20 '18 at 16:19

Gelineau

2,031
4
20
30

There was a reason I closed this as a duplicate *4 whole minutes before you posted this*. – Martijn Pieters Jun 20 '18 at 16:22
I did not see it on time, sorry. – Gelineau Jun 20 '18 at 16:24

What is the most pythonic way to retrieve the indexes of values sorted by the values?

1 Answers1