Re: [R] Aggregrate function

[David Winsemius]

I realized later that the which might not be necessary (and in  
addition was reminded privately). The %in% function returns a logical  
vector which works just as well with matrix or dataframe indexing as  
the numeric vector returned by which.

--
David Winsemius

On Feb 12, 2009, at 5:52 PM, David Winsemius wrote:

aggregate and by are convenience functions of tapply.

Consider this alternate solution:

xveg[which(xveg$tot %in% with(xveg, tapply(tot, loc, max))),"sp"]

It uses tapply to find the maximums by loc(ations) and then to goes
back into xveg to find the corresponding sp(ecies).  You should do
testing to see whether the handling of ties agrees with your needs.

--
David Winsemius

On Feb 12, 2:56 pm, "Christos Hatzis" <christos.hat...@nuverabio.com>
wrote:
I don't have an easy solution with aggregate, because the function in
aggregate needs to return a scalar.
But the following should work:

do.call("rbind", lapply(split(xveg, xveg$loc), function(x)
x[which.max(x$tot), ]))

   loc sp tot
L1  L1  b  60
L2  L2  e  30
L3  L3  b  68

-Christos



-----Original Message-----
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Monica Pisica
Sent: Thursday, February 12, 2009 1:58 PM
To: R help project
Subject: [R] Aggregrate function

Hi,

I have to recognize that i don't fully understand the
aggregate function, but i think it should help me with what i
want to do.

xveg is a data.frame with location, species, and total for
the species. Each location is repeated, once for every
species present at that location. For each location i want to
find out which species has the maximum total ... so i've
tried different ways to do it using aggregate.

loc <- c(rep("L1", 3), rep("L2", 5), rep("L3", 2)) sp <-
c("a", "b", "c", "a", "d", "b", "e", "c", "b", "d") tot <-
c(20, 60, 40, 15, 25, 10, 30, 20, 68, 32) xveg <-
data.frame(loc, sp, tot)

result desired:

L1   b
L2   e
L3   b

sp_maj <- aggregate(xveg[,2], list(xveg[,1], function(x)
levels(x)[which.max(table(x))])

This is wrong because it gives the first species name in each
level of location, so i get a, a, b, as species instead of b, e, b.

I've tried other few aggregate commands, all with wrong results.

I will appreciate any help,

Thanks,

Monica

_________________________________________________________________

 the go.

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.