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] 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