Zhou Fang wrote: > Hi, > > Okay, I have a n x p matrix X, which I know is not full rank. In > particular, there may be linear dependencies amongst the columns (but > not that many). What is a fast way of finding a linearly independent > subset of the columns of X that will span the column space of X, in R? > If it helps, I have the QR decomposition of the original X 'for free'. > > I know that it's possible to do this directly by looping over the > columns and adding them, but at the very least, a solution without > horrible slow loops would be nice.
Have a look at stats:::Thin.col(), but beware that it isn't terribly robust. > Any ideas welcome. > > Zhou Fang > > ______________________________________________ > 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. -- O__ ---- Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~~~~~~~~~ - (p.dalga...@biostat.ku.dk) FAX: (+45) 35327907 ______________________________________________ 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.
