I'm assuming that v is an array of vectors. Does this do any better?
N=zeros(length(v[1]),max(length(v[1]),length(v))) # Because there can't be
more than this many columns
i=0
for w=v
if any(w.!=0)&&w!=N[:,1:i]*(N[:,1:i]\w)
i+=1
N[:,i]=j
end
end
N=N[:,1:i]
Note: may have minor issue for rounding errors, so some rounding during
testing may be necessary to avoid problems. Also, this is written from 0.3,
I cannot guarantee that everything will work correctly in 0.4.
On Tuesday, 6 October 2015 01:06:15 UTC+10, Matt wrote:
>
> Given a set of vectors v1, ...., vn, I'd like to construct a matrix where
> columns correspond to a linearly independent subset of these vectors.
> Currenly, I form the matrix hcat(v1, ..., vn), use qrfact!() on it, check
> the diagonal of the :R matrix and construct a new matrix as hcat(vj,...)
> I'm looking for a way that would use less memory. Is there a more memory
> efficient way ? (for instance a way to avoid the construction of an
> intermediary matrix when finding the subset of vectors, or a way to delete
> columns in place in a matrix)
>
