From hpages at fhcrc.org  Wed Oct  3 03:11:14 2012
From: hpages at fhcrc.org (=?ISO-8859-1?Q?Herv=E9_Pag=E8s?=)
Date: Tue, 02 Oct 2012 18:11:14 -0700
Subject: [Rd] Fastest non-overlapping binning mean function out there?
In-Reply-To: <CAFDcVCQ0SPAYLsgtVxyYoakWQCJavj2QSfLwtNuzwH_U-AR8_g@mail.gmail.com>
References: <CAFDcVCQ0SPAYLsgtVxyYoakWQCJavj2QSfLwtNuzwH_U-AR8_g@mail.gmail.com>
Message-ID: <506B90B2.3010602@fhcrc.org>

Hi Henrik,

On 10/02/2012 05:19 PM, Henrik Bengtsson wrote:
> Hi,
>
> I'm looking for a super-duper fast mean/sum binning implementation
> available in R, and before implementing z = binnedMeans(x y) in native
> code myself, does any one know of an existing function/package for
> this?  I'm sure it already exists.  So, given data (x,y) and B bins
> bx[1] < bx[2] < ... < bx[B] < bx[B+1], I'd like to calculate the
> binned means (or sums) 'z' such that z[1] = mean(x[bx[1] <= x & x <
> bx[2]]), z[2] = mean(x[bx[2] <= x & x < bx[3]]), .... z[B].  Let's
> assume there are no missing values and 'x' and 'bx' is already
> ordered.  The length of 'x' is in the order of 10,000-millions.  The
> number of elements in each bin vary.

You didn't say if you have a lot of bins or not. If you don't have a lot
of bins (e.g. < 10000), something like

   aggregate(x, by=list(bin=findInterval(x, bx)), FUN=mean)

might not be too bad:

   > x <- seq(0, 8, by=0.1)
   > bx <- c(2, 2.5, 4, 5.8)
   > aggregate(x, by=list(bin=findInterval(x, bx)), FUN=mean)
     bin    x
   1   0 0.95
   2   1 2.20
   3   2 3.20
   4   3 4.85
   5   4 6.90

I didn't try it on a 10,000-millions-elements vector though (and I've
no idea how I could do this).

H.


>
> Thanks,
>
> Henrik
>
> ______________________________________________
> R-devel at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-devel
>

-- 
Herv? Pag?s

Program in Computational Biology
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[Rd] Fastest non-overlapping binning mean function out there?
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