This works to multiply the ith row of a by the ith value of b.
It might be what you can use
a <- matrix(1:30, 6, 5)
b <- 1:6
a
a*b
To simplify your code, I think you can do this---one multiplication
xA <- x %*% A
Now you can do the tests on xA and not have any matrix multiplications
inside the loop.
The next line is not tested but I hope gives you the idea
remove.set <- (xA[,1] < .5) || (xA[,2] < 52) || (xA[,3] > 150)
new.x <- x[remove.set,]
You now have new.x with no explicit loops at all, because this takes
advantage of vector operations.
Rich
On Fri, Jul 20, 2012 at 2:34 PM, wwreith <reith_will...@bah.com> wrote:
> That is faster than what I was doing and reducing 15% of my iterations it
> still very helpful.
>
> Next question.
>
> I need to multiply each row x[i,] of the matrix x by another matrix A.
> Specifically
>
> for(i in 1:n)
> {
> If (x[i,]%*%A[,1]<.5 || x[i,]%*%A[,2]<42 || x[i,]%*%A[,3]>150)
> {
> x<-x[-i,]
> n<-n-1
> }. #In other words remove row i from x if it does not meet criteria (>=.5,
> >=42, <=150). When multiplied to A
> }
> Is there a better way than using a for loop for this or x<-x[-i,] for that
> matter? I assume building a new matrix would be worse.
>
> Ideally I want to also exclude some x[,i] as well example if x[1,] is
> better
> than x[2,] in all three categories i.e. bigger, bigger, and smaller than
> x[2,] when multiplied to A then I want to exclude x[2,] as well. Any
> suggestions on whether it is better to do this all at once or in stages?
>
> Thanks for helping!
>
>
>
> --
> View this message in context:
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> Sent from the R help mailing list archive at Nabble.com.
>
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