Thanks a lot to all of you! On Tue, Apr 26, 2011 at 10:34 AM, Peter Ehlers <ehl...@ucalgary.ca> wrote:
> On 2011-04-26 05:26, Dimitris Rizopoulos wrote: > >> one way is the following: >> >> a<- matrix(rnorm(9), 3 ,3) >> >> aa<- a[order(row(a), -a)] >> matrix(aa, nrow(a), byrow = TRUE)[, 2] >> > > That's clever, Dmitris. And very fast! > > Lars: my apologies; I didn't read your request > carefully. You asked for more _efficient_ and > I just thought 'shorter code'. I should know > that whenver I think 'apply', I should think > 'matrix'. > > Peter Ehlers > > > >> >> I hope it helps. >> >> Best, >> Dimitris >> >> >> On 4/26/2011 2:01 PM, Lars Bishop wrote: >> >>> Hi, >>> >>> I need to extract the second largest element from each row of a >>> matrix. Below is my solution, but I think there should be a more >>> efficient >>> way to accomplish the same, or not? >>> >>> >>> set.seed(1) >>> a<- matrix(rnorm(9), 3 ,3) >>> sec.large<- as.vector(apply(a, 1, order, decreasing=T)[2,]) >>> ans<- sapply(1:length(sec.large), function(i) a[i, sec.large[i]]) >>> ans >>> >>> Thanks in advance for your help, >>> Lars. >>> >>> [[alternative HTML version deleted]] >>> >>> ______________________________________________ >>> 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<http://www.r-project.org/posting-guide.html> >>> and provide commented, minimal, self-contained, reproducible code. >>> >>> >> > [[alternative HTML version deleted]] ______________________________________________ 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.
