Hello Everybody, The code:
dfmed<-lapply(unique(colnames(df)), function(x) rowMedians(as.matrix(df[,colnames(df) == x]),na.rm=TRUE)) takes really long time to execute ( in hours). Is there a faster way to do this? Thanks! On Tue, May 22, 2012 at 3:46 PM, Preeti <pre...@sci.utah.edu> wrote: > Thanks Henrik! Here is the one-liner that I wrote: > > dfmed<-lapply(unique(colnames(df)), function(x) > rowMedians(as.matrix(df[,colnames(df) == x]),na.rm=TRUE)) > > Thanks again! > > > On Tue, May 22, 2012 at 3:23 PM, Henrik Bengtsson > <h...@biostat.ucsf.edu>wrote: > >> See rowMedians() of the matrixStats package for replacing apply(x, >> MARGIN=1, FUN=median). /Henrik >> >> On Tue, May 22, 2012 at 12:34 PM, Preeti <pre...@sci.utah.edu> wrote: >> > Hi, >> > >> > I have a 250,000 by 300 matrix. I am trying to calculate the median of >> > those columns (by row) with column names that are identical. I would >> like >> > this to be efficient since apply(x,1,median) where x is created by >> choosing >> > only those columns with same column name and looping on this is taking a >> > really long time. Is there an efficient way to do this? >> > >> > Thanks! >> > >> > [[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. >> > > [[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.
