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
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