Dear numpy developers, The current implementation of numpy.interp(x,xp,fp) comes down to: first calculating all the slopes of the linear interpolant (these are len(xp)-1), then use a binary search to find where x is in xp (running time log(len(xp)). So we obtain a running time of
O( len(xp) + len(x)*log(len(xp) ) We could improve this to just O( len(x)*log(len(xp) ) by not caching the slopes. The point is, of course, that this is slightly slower in the common use case where x is is refinement of xp, and where you will have to compute all the slopes anyway. In my personal use case, however, I needed the value of the interp(x0,xp,fp) in order to calculate the next point x1 where I wanted to calculate interp(x1,xp,fp). The current implementation gave a severe running time penalty. I have looked at the source and I could easily produce a patch for this. Would you be interested in it? Cheers, Timo Kluck
_______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
