That is about as fast as it can be done. However you may be able to avoid doing it at all if you fold V2 into a matrix instead. Did you mean to use matrix multiplication in your calculation of M3?
On September 13, 2021 11:48:48 PM PDT, nevil amos <nevil.a...@gmail.com> wrote: >Hi is there a faster way to "extract" rows of a matrix many times to for a >longer matrix based in a vector or for indices than M[ V, ] > >I need to "expand" ( rather than subset) a matrix M of 10-100,000 rows x >~50 columns to produce a matrix with a greater number (10^6-10^8) of rows >using a vector V containing the 10^6 -10^8 values that are the indices of >the rows of M. the output matrix M2 is then multiplied by another vector V2 >With the same length as V. > >Is there a faster way to achieve these calculations (which are by far the >slowest portion of a function looped 1000s of times? than the standard M2 ><- M[ V, ] and M3<-M2*V2, the two calculations are taking a similar time, >Matrix M also changes for each loop. > > >M<-matrix(runif(50*10000,0,100),nrow=10000,ncol=50) >x = 10^7 >V<-sample(1:10000,x,replace=T) >V2<-(sample(c(1,NA),x,replace=T)) >print<-(microbenchmark( >M2<-M[V,], >M3<-M2*V2, >times=5,unit = "ms")) > > > >thanks for any suggestions > >Nevil Amos > > [[alternative HTML version deleted]] > >______________________________________________ >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.