M = A[..., np.newaxis] == B will give you a 40x60x20 boolean 3d-array where M[..., i] gives you a boolean mask for all the occurrences of B[i] in A.
If you wanted all the (i, j) pairs for each value in B, you could do something like import numpy as np from itertools import izip, groupby from operator import itemgetter id1, id2, id3 = np.where(A[..., np.newaxis] == B) order = np.argsort(id3) triples_iter = izip(id3[order], id1[order], id2[order]) grouped = groupby(triples_iter, itemgetter(0)) d = dict((b_value, [idx[1:] for idx in indices]) for b_value, indices in grouped) Then d[value] is a list of all the (i, j) pairs where A[i, j] == value, and the keys of d are every value in B. On Sat, Nov 24, 2012 at 3:36 PM, Siegfried Gonzi <sgo...@staffmail.ed.ac.uk>wrote: > Hi all > > This must have been answered in the past but my google search capabilities > are not the best. > > Given an array A say of dimension 40x60 and given another array/vector B > of dimension 20 (the values in B occur only once). > > What I would like to do is the following which of course does not work (by > the way doesn't work in IDL either): > > indx=where(A == B) > > I understand A and B are both of different dimensions. So my question: > what would the fastest or proper way to accomplish this (I found a solution > but think is rather awkward and not very scipy/numpy-tonic tough). > > Thanks > -- > The University of Edinburgh is a charitable body, registered in > Scotland, with registration number SC005336. > > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > http://mail.scipy.org/mailman/listinfo/numpy-discussion >
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