[R] How to replace all non-maximum values in a row with 0

ONKELINX, Thierry Thierry.ONKELINX at inbo.be
Fri Apr 9 11:25:21 CEST 2010


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.Onkelinx op 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-bounces op r-project.org 
> [mailto:r-help-bounces op r-project.org] Namens Owe Jessen
> Verzonden: vrijdag 9 april 2010 11:08
> Aan: r-help op 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
> post op owejessen.de
> http://privat.owejessen.de
> 
> ______________________________________________
> R-help op 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.
> 

Druk dit bericht a.u.b. niet onnodig af.
Please do not print this message unnecessarily.

Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer 
en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is
door een geldig ondertekend document. The views expressed in  this message 
and any annex are purely those of the writer and may not be regarded as stating 
an official position of INBO, as long as the message is not confirmed by a duly 
signed document.



More information about the R-help mailing list