Gabor's solution seems to optimize 'simpler'.
More efficient is to learn that in R a vector is not a matrix, but a matrix is
just an ornamented vector.
fastWolfgang <- function( v, vec ) {
matrix( c( v, rep( vec, length( v ) ) )
, now = length( v ) )
}
On July 3, 2018 6:28:45 AM PDT, "Viechtbauer, Wolfgang (SP)"
<wolfgang.viechtba...@maastrichtuniversity.nl> wrote:
>Hi All,
>
>I have one vector that I want to combine with another vector and that
>other vector should be the same for every row in the combined matrix.
>This obviously does not work:
>
>vec <- c(2,4,3)
>cbind(1:5, vec)
>
>This does, but requires me to specify the correct value for 'n' in
>replicate():
>
>cbind(1:5, t(replicate(5, vec)))
>
>Other ways that do not require this are:
>
>t(sapply(1:5, function(x) c(x, vec)))
>do.call(rbind, lapply(1:5, function(x) c(x, vec)))
>t(mapply(c, 1:5, MoreArgs=list(vec)))
>
>I wonder if there is a simpler / more efficient way of doing this.
>
>Best,
>Wolfgang
>
>______________________________________________
>R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>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.
--
Sent from my phone. Please excuse my brevity.
______________________________________________
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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.
