# [Numpy-discussion] Inverted indices

Gregor Thalhammer gregor.thalhammer at gmail.com
Thu Aug 7 07:59:05 EDT 2014

```Am 07.08.2014 um 13:16 schrieb Nicolas P. Rougier <Nicolas.Rougier at inria.fr>:

>
> Hi,
>
> I've a small problem for which I cannot find a solution and I'm quite sure there is an obvious one:
>
> I've an array Z (any dtype) with some data.
> I've a (sorted) array I (of integer, same size as Z) that tells me the  index of Z[i]  (if necessary, the index can be stored in Z).
>
> Now, I have an arbitrary sequence S of indices (in the sense of I), how do I build the corresponding data ?
>
> Here is a small example:
>
> Z = [(0,0), (1,1), (2,2), (3,3), (4,4))
> I  = [0, 20, 23, 24, 37]
>
> S = [ 20,20,0,24]
> -> Result should be [(1,1), (1,1), (0,0),(3,3)]
>
> S = [15,15]
> -> Wrong (15 not in I) but ideally, I would like this to be converted to [(0,0), (0,0)]
>
>
> Any idea ?
>

If I is sorted, I would propose to use a bisection algorithm, faster than linear search:

Z = array([(0,0), (1,1), (2,2), (3,3), (4,4)])
I = array([0, 20, 23, 24, 37])
S = array([ 20,20,0,24,15,27])

a = zeros(S.shape,dtype=int)
b = a + S.shape[0]-1
for i in range(int(log2(S.shape[0]))+2):
c = (a+b)>>1
sel = I[c]<=S
a[sel] = c[sel]
b[~sel] = c[~sel]
Z[c]

If I[c] != S, then there is no corresponding index entry in I to match S.

Gregor

```