Thanks for the quick answers. I think I will go with the .index and list 
comprehension.
But if someone finds with a vectorised solution for the numpy 100 exercises...


Nicolas


> On 30 Dec 2015, at 16:31, Benjamin Root <ben.v.r...@gmail.com> wrote:
> 
> Maybe use searchsorted()? I will note that I have needed to do something like 
> this once before, and I found that the list comprehension form of calling 
> .index() for each item was faster than jumping through hoops to vectorize it 
> using searchsorted (needing to sort and then map the sorted indices to the 
> original indices), and was certainly clearer, but that might depend upon the 
> problem size.
> 
> Cheers!
> Ben Root
> 
> On Wed, Dec 30, 2015 at 10:02 AM, Andy Ray Terrel <andy.ter...@gmail.com> 
> wrote:
> Using pandas one can do:
> 
> >>> A = np.array([2,0,1,4])
> >>> B = np.array([1,2,0])
> >>> s = pd.Series(range(len(B)), index=B)
> >>> s[A].values
> array([  1.,   2.,   0.,  nan])
> 
> 
> 
> On Wed, Dec 30, 2015 at 8:45 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([0,2,0])
> >>> 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
> 
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