Dear Jake, Thanks for your response. I meant to group/count pairs in boxes (using two arrays simultaneously-hence needing 2 metrics) instead of one distance array as the binning parameter. I don't know if the algorithm supports such a thing. For now, I am proceeding with your suggestion of two ball trees at huge computational cost. I hope I am able to frame my question properly.
Thanks & Regards, Rohin. On Mon, Jul 31, 2017 at 8:16 PM, Jacob Vanderplas <jake...@cs.washington.edu > wrote: > On Sun, Jul 30, 2017 at 11:18 AM, Rohin Kumar <yrohinku...@gmail.com> > wrote: > >> *update* >> >> May be it doesn't have to be done at the tree creation level. It could be >> using loops and creating two different balltrees. Something like >> >> tree1=BallTree(X,metric='metric1') #for x-z plane >> tree2=BallTree(X,metric='metric2') #for y-z plane >> >> And then calculate correlation functions in a loop to get tpcf(X,r1,r2) >> using tree1.two_point_correlation(X,r1) and tree2.two_point_correlation( >> X,r2) >> > > Hi Rohin, > It's not exactly clear to me what you wish the tree to do with the two > different metrics, but in any case the ball tree only supports one metric > at a time. If you can construct your desired result from two ball trees > each with its own metric, then that's probably the best way to proceed, > Jake > > >> >> _______________________________________________ >> scikit-learn mailing list >> scikit-learn@python.org >> https://mail.python.org/mailman/listinfo/scikit-learn >> >> > > _______________________________________________ > scikit-learn mailing list > scikit-learn@python.org > https://mail.python.org/mailman/listinfo/scikit-learn > >
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