On Tue, Jun 12, 2012 at 12:24 AM, Rody <rodric_seu...@hotmail.com> wrote: > I found a solution myself, but thanks for the answers. I solved it like this: > D <- matrix(1:225*100,nrow=100,ncol=225) > for(i in 1:225) > D[,i] <- rt(100,df=225) > end
but as Don said, you can do this in one step (and it is both faster and more elegant). D <- matrix(rt(100 * 225, df = 225), ncol = 225) Cheers, Josh > > > -- > View this message in context: > http://r.789695.n4.nabble.com/Storing-datasets-tp4632874p4633071.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. -- Joshua Wiley Ph.D. Student, Health Psychology Programmer Analyst II, Statistical Consulting Group University of California, Los Angeles https://joshuawiley.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.