Oh thanks, I would never have imagined a one-line solution...
Here are the benchmarks: In [2]: %timeit stefan(S) 100000 loops, best of 3: 10.8 µs per loop In [3]: %timeit gregor(S) 10000 loops, best of 3: 48.1 µs per loop In [4]: %timeit gregor2(S) 100000 loops, best of 3: 3.23 µs per loop Nicolas On 07 Aug 2014, at 14:04, Gregor Thalhammer <gregor.thalham...@gmail.com> wrote: > > Am 07.08.2014 um 13:59 schrieb Gregor Thalhammer > <gregor.thalham...@gmail.com>: > >> >> Am 07.08.2014 um 13:16 schrieb Nicolas P. Rougier <nicolas.roug...@inria.fr>: >> >>> >>> Hi, >>> >>> I've a small problem for which I cannot find a solution and I'm quite sure >>> there is an obvious one: >>> >>> I've an array Z (any dtype) with some data. >>> I've a (sorted) array I (of integer, same size as Z) that tells me the >>> index of Z[i] (if necessary, the index can be stored in Z). >>> >>> Now, I have an arbitrary sequence S of indices (in the sense of I), how do >>> I build the corresponding data ? >>> >>> Here is a small example: >>> >>> Z = [(0,0), (1,1), (2,2), (3,3), (4,4)) >>> I = [0, 20, 23, 24, 37] >>> >>> S = [ 20,20,0,24] >>> -> Result should be [(1,1), (1,1), (0,0),(3,3)] >>> >>> S = [15,15] >>> -> Wrong (15 not in I) but ideally, I would like this to be converted to >>> [(0,0), (0,0)] >>> >>> >>> Any idea ? >>> >> >> If I is sorted, I would propose to use a bisection algorithm, faster than >> linear search: >> >> Z = array([(0,0), (1,1), (2,2), (3,3), (4,4)]) >> I = array([0, 20, 23, 24, 37]) >> S = array([ 20,20,0,24,15,27]) >> >> a = zeros(S.shape,dtype=int) >> b = a + S.shape[0]-1 >> for i in range(int(log2(S.shape[0]))+2): >> c = (a+b)>>1 >> sel = I[c]<=S >> a[sel] = c[sel] >> b[~sel] = c[~sel] > > or even simpler: > c = searchsorted(I, S) > Z[c] > > Gregor > >> Z[c] >> >> If I[c] != S, then there is no corresponding index entry in I to match S. >> >> Gregor > > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > http://mail.scipy.org/mailman/listinfo/numpy-discussion _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
