# seemingly simple list indexing problem

iu2 israelu at elbit.co.il
Tue Jul 29 01:00:31 CEST 2008

```On Jul 29, 12:10 am, John Krukoff <jkruk... at ltgc.com> wrote:
> On Mon, 2008-07-28 at 16:24 -0500, Ervan Ensis wrote:
> > My programming skills are pretty rusty and I'm just learning Python so
> > this problem is giving me trouble.
>
> > I have a list like [108, 58, 68].  I want to return the sorted indices
> > of these items in the same order as the original list.  So I should
> > return [2, 0, 1]
>
> > For a list that's already in order, I'll just return the indices, i.e.
> > [56, 66, 76] should return [0, 1, 2]
>
> > Any help would be appreciated.
>
> > --
> >http://mail.python.org/mailman/listinfo/python-list
>
> If your lists aren't so large that memory is an issue, this might be a
> good place for a variation of decorate, sort, undecorate.
>
> >>> listToSort = [ 108, 58, 68 ]
> >>> decorated = [ ( data, index ) for index, data in
>
> enumerate( listToSort ) ]>>> decorated
>
> [(108, 0), (58, 1), (68, 2)]>>> result = [ None, ] * len( listToSort )
> >>> for sortedIndex, ( ignoredValue, originalIndex ) in
>
> enumerate( sorted( decorated ) ):
> ...     result[ originalIndex ] = sortedIndex
> ...>>> result
>
> [2, 0, 1]
>
> --
> John Krukoff <jkruk... at ltgc.com>
> Land Title Guarantee Company

Inspired by your idea and the above one, here is another try:

>>> a0 = [108, 58, 68, 108, 58]
>>> a1 = [(x, y) for x, y in enumerate(a0)]
>>> a1
[(0, 108), (1, 58), (2, 68), (3, 108), (4, 58)]
>>> a2 = sorted(a1, lambda x, y: cmp(x[1], y[1]))
>>> a2
[(1, 58), (4, 58), (2, 68), (0, 108), (3, 108)]
>>> a3 = [a2.index(x) for x in a1]
>>> a3
[3, 0, 2, 4, 1]

The idea is to make each item unique (by making it a tuple), and then
apply the naive solution.

```