On Mi, 2015-12-30 at 17:12 +0100, Nicolas P. Rougier wrote: > 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... >
Yeah, I doubt you can get very pretty, though maybe there is some great trick. This is one way: In [67]: A = np.array([2,0,1,4]) In [68]: B = np.array([1,2,0]) In [69]: B_sorter = np.argsort(B) In [70]: B_index = np.searchsorted(B, A, sorter=B_sorter) In [71]: invalid = B[B_sorter].take(s, mode='clip') != A In [72]: B_index[invalid] = -1 # mark invalids with -1 In [73]: B_index Out[73]: array([ 2, 0, 1, -1]) Anyway, I guess the arrays would likely have to be quite large for this to beat list comprehension. And maybe doing the searchsorted the other way around could be faster, no idea. - Sebastian > > 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 > > > > _______________________________________________ > > 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 > > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > https://mail.scipy.org/mailman/listinfo/numpy-discussion
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