[SciPy-User] Finding Positions of Identical Entries in Arrays of Different Size

[Damian Eads]

On Thu, Aug 27, 2009 at 3:31 PM, Lorenzo Isella<lorenzo.isella at gmail.com> wrote:
> Dear All,
> A problem which should be rather easy, but which is giving me a hard time.
> (1) Consider for instance the following two arrays
>
> A=[12,78,98,654,7665,1213,765] and
> B=[98,765, 654]

np.where(np.setmember1d(A,B))

>
> Now, what I need is an efficient way of finding the positions of the
> entries of A equal to the entries of B, that is to say in the case of
> the two arrays above (counting from zero)
>
> C=[2,6,3].
> Note that in the example there is a single entry of A equal to the i-th
> entry of B.
> I have tried using scipy.where, but unsuccessfully. Any ideas about how
> to code that efficiently?
> (2) I would like to extend this to the case where A and B are m times n
> arrays, e.g. consider the case
>
> A= 233 789
>      987 734
>      890 253
>      876 321
>      534 489
>
> and
>
> B= 987 734
>      876 321
>
> and C is now the 1D vector returning the positions of the rows of A
> equal to those of B, i.e.
>
> C=[1, 3].
>

This is probably not the best solution but it works in cases where the
elements in B are not unique.

qmax=max(B.max(),A.max())
Bp=B[:,0].copy()
Ap=A[:,0].copy()
Bp+=B[:,1]*qmax
Ap+=A[:,1]*qmax
C=np.where(np.setmember1d(A,B))

I hope this helps.

Damian

> Any suggestions is really welcome.
> Kind Regards
>
> Lorenzo
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-- 
-----------------------------------------------------
Damian Eads                           Ph.D. Candidate
University of California             Computer Science
1156 High Street         Machine Learning Lab, E2-489
Santa Cruz, CA 95064    http://www.soe.ucsc.edu/~eads