I have written what I believe to be a semi-quick ols-regression function
ols32 <- function (y, x,Ridge=1.1) {
xrd<-crossprod(x)
xry<-crossprod(x, y)
diag(xrd)<-Ridge*diag(xrd)
solve(xrd,xry)
}
Now I want to apply this to the following
(vapply(1:la, function(J)
ME %*% ols32((nza[,J]),(cbind(nzaf1[,J],nzaf2[,J],nza[,-J],MOMF)))
[(la+2):(2*la+1)],FUN.VALUE=0))
Where nza,nzaf1,nzaf2 and MOMF are 500x50 matrixes and la=50 and ME is a vector of length 50. So what I actually do is I do a regression but only use the beta-coefficients from MOMF which I multiply by ME.
nza.mat<-matrix(rnorm(500*200),ncol=200)
nza<-nza.mat[,1:50]
nzaf2<-nza.mat[,101:150]
MOMF<-nza.mat[,151:200]
nzaf1<-nza.mat[,51:100]
ME<-nza.mat[500,151:200]
Is there an imediate way of making things faster or do I need to use someting like RcppEigen? Tks P
So I came up with a slightly faster way of solving this by rewriting my ols-function so that it calculates the two crossproducts only once for a whole matrix. The new function looks like this:
ols.quick <- function (y, x, ME) {
la<-dim(y)[2]
XX.cross<-crossprod(x)
XY.cross<-crossprod(x, y)
diag(XX.cross)<-Ridge*diag(XX.cross)
betas<-sapply(1:la, function(J){
idx<-c((1:la)[-J],la+J,2*la+J,(3*la+1):(4*la));
solve(XX.cross[idx,idx],XY.cross[idx,J])},simplify=T)
ME%*%betas[(la+2):(2*la+1),]
}
where y=nza (500x50) and x=cbind(nza,nzaf1,nzaf2,MOMF) (500x200) This solves the problem about 3.5 times faster.
microbenchmark(ols.quick(nza,nza.mat,ME),
vapply(1:la, function(J) ME%*%ols32(nza[,J],(cbind(nzaf1[,J],nzaf2[,J],nza[,-J],MOMF)))
[(la+2): (lb+2)],FUN.VALUE=0))
min lq median uq max neval
66.30495 67.71903 68.57001 70.17742 77.80069 100
251.59395 255.43306 258.35041 262.85742 296.51313 100
I suppose I could gain some speed with parLapply from the parallel package but I havet looked into that yet.
