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.
