HI, Tx for the fast method. dim(Rt) #b #2 dim(X) #[1] 2 50000
Is it possible to obtain the same result as X without converting X to R? According to OP "..... which will give me a (two-dimensional) array or matrix of dimension n x p with the i'th row storing the above. " A.K. ----- Original Message ----- From: "Suzen, Mehmet" <msu...@gmail.com> To: arun <smartpink...@yahoo.com> Cc: Ranjan Maitra <maitra.mbox.igno...@inbox.com>; R help <r-help@r-project.org> Sent: Saturday, December 29, 2012 1:19 PM Subject: Re: [R] efficiently multiply different matrices in 3-d array with different vectors? What I had in mind was a tensor multiplication. I think, using tensor arithmetic for multidimensional arrays looks more compacts and efficient (~ 2-3 times). I am guessing that performance will be much more pronounced for n-D arrays (n>3). # Component Wise set.seed(14) Z<-array(sample(1:1000000,800000,replace=TRUE),dim=c(50000,2,8)) set.seed(21) Y<-matrix(sample(1:400000,400000,replace=TRUE),nrow=8) system.time(X<-do.call(cbind,lapply(seq_len(dim(Z)[1]),function(i) Z[i,,]%*%Y[,i]))) # user system elapsed # 0.58 0.00 0.58 R<-cbind(sum(X[1,]), sum(X[2,])) # Tensor multiply library(tensorA) set.seed(14) Zt <-to.tensor(sample(1:1000000,800000,replace=TRUE), c(a=50000, b=2, c=8)) set.seed(21) Yt <-to.tensor(sample(1:400000,400000,replace=TRUE), c(c=8, a=50000)) system.time(Rt<-Zt %e% Yt) # user system elapsed # 0.124 0.000 0.126 # correct results? R # [,1] [,2] #[1,] 4.00595e+16 3.997387e+16 Rt #[1] 4.005950e+16 3.997387e+1 ______________________________________________ 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.
