[R] Finding a basis in a set of vectors
zhou.zfang at gmail.com
Fri Feb 6 16:29:24 CET 2009
Ah ha, that does work.
What do you mean it isn't robust, though? I mean, obviously linear
dependency structures in general are not stable under small
Or is it that it's platform dependent?
On Fri, Feb 6, 2009 at 2:28 PM, Peter Dalgaard <P.Dalgaard at biostat.ku.dk> wrote:
> Zhou Fang wrote:
>> 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 at r-project.org mailing list
>> 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
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