[Numpy-discussion] How to find indices of values in an array (indirect in1d) ?

Nicolas P. Rougier Nicolas.Rougier at inria.fr
Wed Dec 30 13:17:16 EST 2015


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 <markperrymiller at 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.Rougier at 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 at scipy.org
> https://mail.scipy.org/mailman/listinfo/numpy-discussion
> 
> _______________________________________________
> NumPy-Discussion mailing list
> NumPy-Discussion at scipy.org
> https://mail.scipy.org/mailman/listinfo/numpy-discussion




More information about the NumPy-Discussion mailing list