pysparse-users

Re: [Pysparse-users] transpose operation

[Fergus G. <fe...@go...>]

> The find/put method is surely the fastest in general. The more
> fundamental question is do you really need the explicit transpose?
> Often, algorithms that need the transpose operate directly on the
> original matrix. If what you need is matrix-vector products of the form
> A.T*x, you can always use the matvec_transp() method. If you're using
> the higher-level PysparseMatrix objects, you can do A*x and x*A. The
> latter actually computes A.T*x.
>
> I hope this helps.
>

Hi, 

thanks for all the replies.

What we want to do is generate the "outer product"  A*B^T  (actually A*A^T
in our case). 

Regards

-- 
Fergus Gallagher

"Setting a good example for children takes all the fun out of middle
age." - William Feather

Re: [Pysparse-users] transpose operation

[Dominique O. <dom...@gm...>]

On Tue, Nov 17, 2009 at 6:42 AM, Fergus Gallagher <fe...@go...>wrote:

> > The find/put method is surely the fastest in general. The more
> > fundamental question is do you really need the explicit transpose?
> > Often, algorithms that need the transpose operate directly on the
> > original matrix. If what you need is matrix-vector products of the form
> > A.T*x, you can always use the matvec_transp() method. If you're using
> > the higher-level PysparseMatrix objects, you can do A*x and x*A. The
> > latter actually computes A.T*x.
> >
> > I hope this helps.
> >
>
> Hi,
>
> thanks for all the replies.
>
> What we want to do is generate the "outer product"  A*B^T  (actually A*A^T
> in our case).
>

If you really need A*A^T explicitly, I suggest building A^T and then using
dot().  Often you don't need this matrix explicitly though (e.g., it is easy
to build matrix-vector products with A*A^T without assembling this matrix).
Similarly, if you need to factorize it to solve A*A^T x = b, you can solve
the equivalent system

  [ I  A^T ] [x]   [ 0]
  [ A   0  ] [y] = [-b]

This last system is also much sparser, and it is symmetric. It is indefinite
though.

-- 
Dominique

Re: [Pysparse-users] transpose operation

[Fergus G. <fe...@go...>]

On Tue, Nov 17, 2009 at 01:09:24PM -0500, Dominique Orban wrote:

Thanks for the reply

> 
> If you really need A*A^T explicitly, I suggest building A^T and then using
> dot().  
>

I think we do.  Our matrices are  ~ 10 Million square, so A^T not
cheap doing it either explictly as A[i,j]=A[j,i] (ouch) or as 
A^T = spmatrix.dot(A,I)

> Often you don't need this matrix explicitly though (e.g., it is easy
> to build matrix-vector products with A*A^T without assembling this matrix).
> Similarly, if you need to factorize it to solve A*A^T x = b, you can solve
> the equivalent system
> 
>   [ I  A^T ] [x]   [ 0]
>   [ A   0  ] [y] = [-b]
> 
> This last system is also much sparser, and it is symmetric. It is indefinite
> though.

Our application is more statistical (correlation) - we're not trying to solve/invert.

Regards.

-- 
Fergus Gallagher

"I take my children everywhere, but they always find their way back
home." - Robert Orben

Re: [Pysparse-users] transpose operation

[Dominique O. <dom...@gm...>]

On Tue, Nov 17, 2009 at 1:24 PM, Fergus Gallagher <fe...@go...> wrote:
> On Tue, Nov 17, 2009 at 01:09:24PM -0500, Dominique Orban wrote:

>>
>> If you really need A*A^T explicitly, I suggest building A^T and then using
>> dot().
>>
>
> I think we do.  Our matrices are  ~ 10 Million square, so A^T not
> cheap doing it either explictly as A[i,j]=A[j,i] (ouch) or as
> A^T = spmatrix.dot(A,I)

OK. What I meant was to construct B=A^T to start with, instead of A.
Then you can say dot(B,B) which will give you A*A^T.

-- 
Dominique