HI,
I think I understand what you meant. This will output all those states where contribution for Obama is higher than all the other candidates. dat1<-read.table(text=" AL AR CA NY Doug 250 250 250 NA Jennifer 20 340 300 100 Michele 250 500 250 60 Obama 15 45 520 600 ",header=TRUE,stringsAsFactors=FALSE,sep="") res<-unlist(lapply(apply(dat1,2,function(x) x[!is.na(x)]),function(x) x[all(x["Obama"]>x[names(x)!=names(x)[grep("Obama",names(x))]])])) res[grep("Obama",names(res))] #CA.Obama NY.Obama # 520 600 A.K. ----- Original Message ----- From: noobmin <pseudov...@hotmail.com> To: r-help@r-project.org Cc: Sent: Tuesday, October 23, 2012 3:00 PM Subject: Re: [R] How to use tapply with more than one variables grouped I meant where obama has higher value compared to other candidates. Looking at the column NY, Obama has the highest. So to state that he wins. Looking for AR column, Michelle wins. I JUST want to list where obama wins. Thank you! This seems to work, just do not understand why you used a threshold? I will study your solution, thanks again! -- View this message in context: http://r.789695.n4.nabble.com/How-to-use-tapply-with-more-than-one-variables-grouped-tp4646948p4647203.html Sent from the R help mailing list archive at Nabble.com. ______________________________________________ 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. ______________________________________________ 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.