Yes, you remind me of this! Thanks!
2011/8/16 Eik Vettorazzi <e.vettora...@uke.uni-hamburg.de> > Hi Lao, > you tried to reinvent the wheel. Have a look at ?tapply > > tapply(sleep$extra,sleep$group,mean) > > Cheers > > Am 16.08.2011 09:41, schrieb Lao Meng: > > Hi all: > > My data:data(sleep) > > > > If I wanna calculate each group's extra,what I can do is: > > #method1 > > attach(sleep) > > mean(extra[group==1]) > > mean(extra[group==1]) > > > > > > #method2 > > result<-matrix(,0,2) > > g<-split(sleep,sleep$group) > > for(i in 1:length(g)) > > { > > result<-rbind(result,data.frame(unique(g[[i]]$group),mean(g[[i]]$extra))) > > } > > colnames(result)<-c("name","mean") > > > > But the above 2 method is a little bit tedious.Is there a "short cut" > manner > > to get the same result? > > > > Thanks a lot! > > > > My best > > > > [[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. > > -- > Eik Vettorazzi > Institut für Medizinische Biometrie und Epidemiologie > Universitätsklinikum Hamburg-Eppendorf > > Martinistr. 52 > 20246 Hamburg > > T ++49/40/7410-58243 > F ++49/40/7410-57790 > [[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.