Yes, it is the expected result. Thanks. Maybe the set(a) & set(b) can be replaced by np.where[np.in1d(a,b)], no ?
> On 30 Dec 2015, at 18:42, Mark Miller <markperrymil...@gmail.com> wrote: > > I'm not 100% sure that I get the question, but does this help at all? > > >>> a = numpy.array([3,2,8,7]) > >>> b = numpy.array([1,3,2,4,5,7,6,8,9]) > >>> c = set(a) & set(b) > >>> c #contains elements of a that are in b (and vice versa) > set([8, 2, 3, 7]) > >>> indices = numpy.where([x in c for x in b])[0] > >>> indices #indices of b where the elements of a in b occur > array([1, 2, 5, 7], dtype=int64) > > -Mark > > > On Wed, Dec 30, 2015 at 6:45 AM, Nicolas P. Rougier > <nicolas.roug...@inria.fr> wrote: > > I’m scratching my head around a small problem but I can’t find a vectorized > solution. > I have 2 arrays A and B and I would like to get the indices (relative to B) > of elements of A that are in B: > > >>> A = np.array([2,0,1,4]) > >>> B = np.array([1,2,0]) > >>> print (some_function(A,B)) > [1,2,0] > > # A[0] == 2 is in B and 2 == B[1] -> 1 > # A[1] == 0 is in B and 0 == B[2] -> 2 > # A[2] == 1 is in B and 1 == B[0] -> 0 > > Any idea ? I tried numpy.in1d with no luck. > > > Nicolas > > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > https://mail.scipy.org/mailman/listinfo/numpy-discussion > > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > https://mail.scipy.org/mailman/listinfo/numpy-discussion _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org https://mail.scipy.org/mailman/listinfo/numpy-discussion
