Finding a basis in a set of vectors
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
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