[R] Finding a basis in a set of vectors

[Zhou Fang]

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
perturbations...?

Or is it that it's platform dependent?

Zhou

On Fri, Feb 6, 2009 at 2:28 PM, Peter Dalgaard <P.Dalgaard at biostat.ku.dk> wrote:
> 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 at 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.dalgaard at biostat.ku.dk)              FAX: (+45) 35327907
>
>