RE: [R] Getting the groupmean for each person

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From: Liaw, Andy (andy_liaw@merck.com)
Date: Mon 10 May 2004 - 21:37:29 EST


Message-id: <3A822319EB35174CA3714066D590DCD504AF7D57@usrymx25.merck.com>

Both of you might have missed my question from Friday: For very long `x'
(e.g., length=50000), indexing by names can take a long time. See that
thread for detail. (For small data, you can hardly tell the difference.)

Also, I'm trying to write the function in a way that one can pass in more
than one grouping variables in a list, much like tapply. The version I
shown is a simplified version to demonstrate the `problem' I had. I
obviously missed the fact that tapply returns 1D array...

Best,
Andy

> From: kjetil@acelerate.com
>
> On 10 May 2004 at 10:09, Christophe Pallier wrote:
>
> >
> >
> > Liaw, Andy wrote:
> >
> > >Suppose I
> > >define the function:
> > >
> > >fun <- function(x, f) {
> > > m <- tapply(x, f, mean)
> > > ans <- x - m[match(f, unique(f))]
> > > names(ans) <- names(x)
> > > ans
> > >}
> > >
> > >
> > >
> >
> > May I ask what is the purpose of match(f,unique(f)) ?
> >
> > To remove the group means, I have be using:
> >
> > x-tapply(x,f,mean)[f]
> >
> > for a while, (and I am now changing to
> > x-tapply(x,f,mean)[as.character(f)] because of the peculiarities of
>
> wouldn't
> sweep(as.array(x), 1, tapply(x,f,mean)[as.character(f)] , "-")
>
> be more natural?
>
> Kjetil Halvorsen
>
> > indexing named vectors with factors )
> >
> > The use of tapply(x,f,mean)[match(f,unique(f))] assumes a particular
> > order in the result of tapply, no? It seems a bit dangerous to me.
> >
> >
> > Christophe Pallier
> >
> > ______________________________________________
> > R-help@stat.math.ethz.ch mailing list
> > https://www.stat.math.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide!
> > http://www.R-project.org/posting-guide.html
> >
>
>
>
>

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