It can be done faster and more elegant with apply and rowSums
rows <- 100000
A <- matrix(rpois(n = rows * 20, lambda = 100), nrow = rows)
A[4, c(1,3)] <- 1000
system.time({
y <- t(apply(A, 1, function(z){
1 * (z == max(z))
}))
y[rowSums(y) > 1, ] <- 0
})
system.time({
nr <- nrow(A)
nc <- ncol(A)
B <- matrix(0,nrow=nr, ncol=nc)
for(i in 1:nr){
x <- which(A[i,]==max(A[i,]))
B[i,x] <- 1
if(sum(B[i,])>1) B[i,] <- as.vector(rep(0,nc))
}
})
all.equal(y, B)
HTH,
Thierry
------------------------------------------------------------------------
----
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek
team Biometrie & Kwaliteitszorg
Gaverstraat 4
9500 Geraardsbergen
Belgium
Research Institute for Nature and Forest
team Biometrics & Quality Assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium
tel. + 32 54/436 185
thierry.onkel...@inbo.be
www.inbo.be
To call in the statistician after the experiment is done may be no more
than asking him to perform a post-mortem examination: he may be able to
say what the experiment died of.
~ Sir Ronald Aylmer Fisher
The plural of anecdote is not data.
~ Roger Brinner
The combination of some data and an aching desire for an answer does not
ensure that a reasonable answer can be extracted from a given body of
data.
~ John Tukey
> -----Oorspronkelijk bericht-----
> Van: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] Namens Owe Jessen
> Verzonden: vrijdag 9 april 2010 11:08
> Aan: r-help@r-project.org
> Onderwerp: Re: [R] How to replace all non-maximum values in a
> row with 0
>
> Am 09.04.2010 10:04, schrieb burgundy:
> > Hi,
> >
> > I would like to replace all the max values per row with "1"
> and all other
> > values with "0". If there are two max values, then "0" for
> both. Example:
> >
> > from:
> > 2 3 0 0 200
> > 30 0 0 2 50
> > 0 0 3 0 0
> > 0 0 8 8 0
> >
> > to:
> > 0 0 0 0 1
> > 0 0 0 0 1
> > 0 0 1 0 0
> > 0 0 0 0 0
> >
> > Thanks!
> >
> Nice little homework to get the day started. :-)
>
> This worked for me, but is probably not the shortest possible answer
>
> A <- matrix (c(2, 3, 0, 0, 200, 30, 0, 0, 2, 50, 0,
> 0, 3, 0,
> 0, 0, 0, 8, 8, 0), nrow = 4, byrow=T)
> nr <- nrow(A)
> nc <- ncol(A)
> B <- matrix(0,nrow=nr, ncol=nc)
> for(i in 1:nr){
> x <- which(A[i,]==max(A[i,]))
> B[i,x] <- 1
> if(sum(B[i,])>1) B[i,] <- as.vector(rep(0,nc))
> }
>
> --
> Owe Jessen
> Nettelbeckstr. 5
> 24105 Kiel
> p...@owejessen.de
> http://privat.owejessen.de
>
> ______________________________________________
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> and provide commented, minimal, self-contained, reproducible code.
>
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