Dear all,

Here is one more way to go though using rep() and then matrix():
> rows <- 1:3
> matrix(rep(rows,5),ncol=5)
     [,1] [,2] [,3] [,4] [,5]
[1,]    1    1    1    1    1
[2,]    2    2    2    2    2
[3,]    3    3    3    3    3

HTH,

Jorge


On Tue, Feb 24, 2009 at 1:43 PM, Gabor Grothendieck <ggrothendieck@gmail.com
> wrote:

> Suppose we want 3 rows and the ith row should have 5 columns of i.
> Create a list whose ith component is the ith row and rbind them:
>
> > rows <- 1:3
> > do.call(rbind, lapply(rows, rep, 5))
>     [,1] [,2] [,3] [,4] [,5]
> [1,]    1    1    1    1    1
> [2,]    2    2    2    2    2
> [3,]    3    3    3    3    3
>
>
> On Tue, Feb 24, 2009 at 1:01 PM, Alexy Khrabrov <deliverable@gmail.com>
> wrote:
> > I'm growing a large dataframe by composing new rows and then doing
> >
> > row <- compute.new.row.somehow(...)
> > d <- rbind(d,row)
> >
> > Is this a fast/preferred way?
> > Cheers,
> > Alexy
> >
> > ______________________________________________
> > 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.
>

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