Re: [ublas] Filling the elements in a vector without knowing the size

From: Sourabh (sourabh_at_[hidden])
Date: 2007-03-06 05:42:08

On Tue, 6 Mar 2007, Sourabh wrote:

> Please look at the following requirement:
>  I have three vectors.
> <sparsevector>  (Sparse) contains 1 as the only non-zero entry.
> <densevector> (Dense)
> <othervector> (Dense, superset of <densevector>)
> 
> I want to fill the <densevector> with those entries of 
> <othervector> which are at the indices of non-zero 
> entries in <sparsevector>. Effectively <sparsevector> contains non-zero 
> entries as indices for the entries in <othervector> which are to be 
> filled in <densevector>.  
> 
> I have written the following code. Can it be optimized ? It is taking 
> almost 1 second, which is too much. 
> 
> densevector.resize ((sparsevector.nnz ()));
> compressed_vector<double>::const_iterator it = sparsevector.begin ();
>     for (unsigned i = 0; it != sparsevector.end (); ++it, ++i) {
> 	densevector (i) = othervector (it.index ());
>     }
> 
> 

The above operation is taking 1 second for 200000 iterations. Not for 1 
iteration. 

Just a strange thing, if I change the statement 
densevector (i) = othervector (it.index ());

to 

densevector (i) = (it.index ());

time taken is increased. What could be the possible reason ?