Hi Patrik, If speed is an issue, an easy way to go a bit faster using Dr. Spector's code is:
wh = which(is.na(mat), arr.ind=TRUE) mat[wh] = colMeans(mat, na.rm=TRUE)[wh[,2]] using that method, I can replace the NA values with column means on a 30000 x 10000 matrix with ~ 1% of values missing in just under 4 seconds. Any larger matrix and my system starts complaining about memory. Cheers, Josh On Wed, Dec 15, 2010 at 2:17 PM, Phil Spector <spec...@stat.berkeley.edu> wrote: > Patrik - > I don't know if it's the fastest, but, assuming your Matrix is called > mat, this seems to work fairly quickly: > > wh = which(is.na(mat),arr.ind=TRUE) > mat[wh] = apply(mat,2,mean,na.rm=TRUE)[wh[,2]] > > - Phil Spector > Statistical Computing Facility > Department of Statistics > UC Berkeley > spec...@stat.berkeley.edu > > > On Wed, 15 Dec 2010, patrik.waldm...@djingis.se wrote: > >> >> >> Dear All, >> >> ?? >> >> does anyone know which is the fastest way to replace NA values in a >> Matrix by their column mean? >> >> library(Matrix) >> mat >> >> Links: >> ------ >> [1] http://www.ownit.se >> >> >> [[alternative HTML version deleted]] >> >> > > ______________________________________________ > 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. > -- Joshua Wiley Ph.D. Student, Health Psychology University of California, Los Angeles http://www.joshuawiley.com/ ______________________________________________ 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.
