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From: Marios Karaoulis on 7 Jul 2010 13:40
Hi, I need some help to speed up things in a sparse matris.
I have a large sparse matrix like this

k=sparse(a_index,b_index,val_index,10000,10000);

(a_index and b_index have duplicate values).

Now another matrix have some info about some additional zeroing
elements like this
node_ebc=[1 13 155 3455 5645 3432 .....];

num_ebc=length(node_ebc);

for i=1:num_ebc
m=node_ebc(i);
k(m,m)=1;
for j=1:10000
if m~=j
fem.k(m,j)=0;
fem.k(j,m)=0;
end
end
end

This loop is very slow.
How can i speed up things?
From: Bruno Luong on 7 Jul 2010 15:22
Marios Karaoulis <marios.karaoulis(a)gmail.com> wrote in message <5dae28d8-48b8-42f1-be87-de18abf71b5e(a)q12g2000yqj.googlegroups.com>...
> Hi, I need some help to speed up things in a sparse matris.
> I have a large sparse matrix like this
>
> k=sparse(a_index,b_index,val_index,10000,10000);
>
> (a_index and b_index have duplicate values).
>
> Now another matrix have some info about some additional zeroing
> elements like this
> node_ebc=[1 13 155 3455 5645 3432 .....];
>
> num_ebc=length(node_ebc);
>
> for i=1:num_ebc
> m=node_ebc(i);
> k(m,m)=1;
> for j=1:10000
> if m~=j
> fem.k(m,j)=0;
> fem.k(j,m)=0;
> end
> end
> end
>
> This loop is very slow.
> How can i speed up things?

Well, the first obvious improvement you could try is replacing the inner loop by

k(m,:)=0;
k(:,m)=0;

then move the diagonal assignment k(m,m)=1; below it.

Bruno
From: Bruno Luong on 7 Jul 2010 15:32
The whole variable locking can be done as following:

k(node_ebc,:)=0;
k(:,node_ebc)=0;
for m=node_ebc
k(m,m) = 1;
end

% Bruno
From: Marios Karaoulis on 7 Jul 2010 15:38
Actually, what I did is the following

k(node_ebc,:)=0;
k(:,node_ebc)=0;
k(node_ebc,node_ebc)=1;

but the speed of this calculations are really slow.

If i have k matrix as full, then speed is amazing faster.
So I was hoping in zeroing only the NECESSARY non zero elements (since
there are many zero elements on those columns and rows), usinhg the
a_index, b_index, val_index.


From: Bruno Luong on 7 Jul 2010 18:35
Marios Karaoulis <marios.karaoulis(a)gmail.com> wrote in message <6265f1e1-92e6-4481-9003-31570ba53023(a)k39g2000yqb.googlegroups.com>...
> Actually, what I did is the following
>
> k(node_ebc,:)=0;
> k(:,node_ebc)=0;
> k(node_ebc,node_ebc)=1;

The last statement (sub-block assignment) is not equivalent to what you wrote earlier (subdiagonal assignment), or my simplification of it:

for m=node_ebc
% ...
k(m,m) = 1;
end

You what did you exactly do? Are you ready to settle on something for the community to work on?

Bruno
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