# [R] Efficient way for multiplying vectors with a only certain number of rows in a matrix

Ralph79 ralph.statistics at gmx.net
Wed Jan 23 12:42:59 CET 2008

```Just in case somebody is interested: I found out an efficient way for doing
the calculations mentioned below.

The code is:

n.obs = 800
n.rowsperobs = 300
n.param = 23
Designmat = matrix(rnorm(n.obs*n.rowsperobs*n.param),ncol=n.param)
Betamat = t(matrix(rnorm(n.obs*n.param),nrow=n.param))

res=rowSums(Designmat*Betamat2[rep(1:n.obs,rep(n.rowsperobs,n.obs)),])

If somebody can think of an even faster way: any comments are greatly
welcome!

Ralph79 wrote:
>
> Dear R-users,
>
> I am working on a problem that I am currently not able to solve
> efficiently. It is about multiplying one column of a matrix with only a
> certain number of rows of another matrix.
>
> Let me illustrate my problem with an example:
>
> n.obs = 800
> n.rowsperobs = 300
> n.param = 23
>
> Designmat = matrix(rnorm(n.obs*n.rowsperobs*n.param),ncol=n.param)
>
> Betamat = matrix(rnorm(n.obs*n.param),nrow=n.param)
>
> In this example, "Designmat" consists of 800*300 rows, meaning that 300
> rows belong to one of the 800 observations.
>
> Each observation has also one parameter vector, which is one column in the
> "Betamat"-matrix (i.e. Betamat contains the n.obs parameter vectors).
>
> My goal is to multiply the parameter vector of each observation (i.e. the
> respective column in Betamat) with ONLY THOSE ROWS IN DESIGNMAT that
> belong to this observation.
> Applied to the example above: The first column in Betamat has to be
> multiplied with the first 300 rows in Designmat, the second column in
> Betamat has to be multiplied with rows 301 to 600 in Betamat and so on.
> Hence, the result of this operation should be a vector of length 240000.
>
> I can think of solutions implying several loops and/or lapplys, but I
> guess that there might be a much easyer and above all faster solution.
>