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KSMAT21
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Create block sparse matrix

KSMAT21
さんによって質問されました 2019 年 6 月 23 日
最新アクティビティ John D'Errico
さんによって コメントされました 2019 年 6 月 26 日
How can I create a block sparse matrix in MATLAB using the 'sparse' funcion, such as S = sparse(I,J,Z)?
Instead of Z being a vector of non-zero entries, Z contains dense matrix blocks (e.g. 100*100). The size of the full matrix S can be very large, e.g 1 million * 1 million.
Thank you.

  4 件のコメント

James Tursa
2019 年 6 月 24 日
How do you have the dense blocks currently stored? In a cell array? How do you have the desired location in the sparse matrix stored? As row&column of the upper left element of each block? Or ...? Will there be the possibility of overlap? Do the blocks always line up nicely like your picture, or could they be anywhere (e.g. half under one block and half under another block)?
KSMAT21
2019 年 6 月 24 日
Hi James,
Thanks for your response. Yes, the dense blocks are currently stored in cell arrays. There will no be overlaps among blocks.
The blocks results from a domain decomposition problem, where each subdomain has the same number of elements (100 in the figure above). The diagonal blocks represent self-interaction, and off-diagonal blocks represent mutual interaction. Thus, the blocks will always be lined up nicely, since they start at multiples of 100 (number of elements per subdomain) for both row and columns.
I know the desired locations of the block by knowing which subdomains in the problem are interacting, thus can say that I know the upper left element of each block.
Thank you.
James Tursa
2019 年 6 月 26 日
How is the subdomain information stored? I think this could be done in a mex routine without sorting or copying the data more than once as long as the subdomain information is stored in a reasonable manner. The problem I see with the methods already posted is that they will require copying the data more than once and sorting the data as well, which will take extra time.

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David Goodmanson
回答者: David Goodmanson
2019 年 6 月 24 日
編集済み: David Goodmanson
2019 年 6 月 26 日

HI ks,
Here is one way.
MODIFIED (see comment thread below)
showing two possible methods to fill the matrix by means of a for loop appoach. The second way is significantly faster, but both methods change the sparse matrix S as many times as you have blocks. For N blocks the overall time for the process appears to scale like N^2, so it's all right for a few hundred blocks. But when the numbers get large you will need to gin up some large index vectors and call sparse at most just a few times, preferably once as John has indicated.
The following may be useful.
From the [...] = ndgrid line in the code below, irow(:) is the row indices of the block itself (zero based). Then if m0vec is a column vector of N row indices for the upper left corner of the N blocks, and if you have a new enough version of Matlab
irowfull = irow(:) + m0vec';
irowfull = irowfull(:);
has the properties you want, similarly for column indices.
% create a matrix with known elements, not square
A = magic(4);
A(:,4) = [];
[Arow Acol] = size(A)
[irow icol] = ndgrid((0:Arow-1),(0:Acol-1)) % block indices
s_row = 10; s_col = 10; n_s = 100; % small example
S = spalloc(s_row, s_col,n_s)
m0 = 3; n0 = 5; % these are the row and column indices
% of the upper left corner of the block
% for loop here to index m0,n0, and A
% first way
S = S + sparse(m0+irow(:),n0+icol(:),A(:),s_row,s_col)
% second way, creating new sparse matrix not required
S(m0:m0+Arow-1,n0:n0+Acol-1) = A
% end
full(S)
A

  4 件のコメント

David Goodmanson
2019 年 6 月 24 日
Hi ks
If you only have a few thousand blocks it makes sense to just use a for loop for each one rather than something more complicated.
I modified the example code by using m0 for the row index and n0 for the column index, and by using ndgrid instead of meshgrid. Ndgrid is the more natural choice here. Compared to ndgrid, the two outputs of meshgrid are swapped which is good for plotting purposes but not as good for index calculations.
It appears that at lot of your blocks have the same shape. Clearly ndgrid only has to be done once for all of those, and you can then do a for loop on the m0 and n0 indices. (m0 and n0 could be columns in an Nx2 array, addressed by the array row index). The sample code below takes m0 and n0 out of the meshgrid (now ndgrid) calculation.
Then in the for loop you have to address the different block matrices themselves, which will of course be much easier if a lot of them are identical.
A = magic(4);
A(:,4) = [];
[Arow Acol] = size(A)
[irow icol] = ndgrid((0:Arow-1),(0:Acol-1)) % block indices
m0 = 3; n0 = 5; % these are the row and column indices
% of the upper left corner of the block
% for loop here to index m0,n0, and A
S = sparse(m0+irow(:),n0+icol(:),A(:))
% end
full(S)
A
KSMAT21
2019 年 6 月 24 日
Thanks a lot, David.
I am correct to guess that the sparse matrix must be added to the previous iterations in the loop to create multiple blocks. Editing from your above code:
n = 1e4;
S = spalloc(n,n,3*n)
%for loop here to index m0,n0, and A
S = S + sparse(m0+irow(:),n0+icol(:),A(:),n,n)
% end
Is that correct?
Because, if not, each iteration in the for loop will clear the previous entries of the sparse matrix.
Regards,
KS
David Goodmanson
2019 年 6 月 24 日
Hi KS,
That works, but gratifyingly enough, after (defining S with spalloc) so does
S(m0:m0+Arow-1,n0:n0+Acol-1) = A;
which obviates having to use ndgrid and also obviates having to create a new (albeit small) sparse matrix each time through the loop. I modified the original answer but I don't know yet how this would do speedwise.

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John D'Errico
回答者: John D'Errico
2019 年 6 月 24 日
編集済み: John D'Errico
2019 年 6 月 24 日

Are you asking how to create a block diagonal sparse matrix? Easy peasy.
tic
N = 100;
M = 10000;
Z = sparse(rand(N,N*M));
Zc = mat2cell(Z,N,repmat(N,1,M));
A = blkdiag(Zc{:});
toc
Elapsed time is 6.487005 seconds.
whos A
Name Size Bytes Class Attributes
A 1000000x1000000 1608000008 double sparse
So A is a block diagonal sparse matrix, of size 1e6x1e6, with 100x100 blocks on the diagonal, 10,000 such blocks. 6 seconds seems reasonable to build it, since almost 50% of that time was just in creating the original random matrix Z.
tic,Z = sparse(rand(N,N*M));toc
Elapsed time is 2.936146 seconds.
spy(A)
untitled.jpg

  5 件のコメント

John D'Errico
2019 年 6 月 25 日
This really is far more trivial than you may think, BUT you ABSOLUTELY need to learn to use sparse.
For example, a 3(5)x3(5) block matrix, with 3 blocks.
blocksize = 5;
nblocks = 3;
% The elements contained in the various blocks.
% If each block is the same, it is easy to build.
% I've just used random elements.
Z = rand(blocksize,blocksize,nblocks);
[subr,subc] = meshgrid(1:blocksize);
% The pattern matrix:
blockrc = [1 2;2 1;3 3];
rowind = subr + reshape((blockrc(:,1)-1)*blocksize,[1 1 nb]);
colind = subc + reshape((blockrc(:,2)-1)*blocksize,[1 1 nb]);
% create A in one call to sparse
A = sparse(rowind(:),colind(:),Z(:),blocksize*max(blockrc(:,1)),blocksize*max(blockrc(:,2)));
whos A
Name Size Bytes Class Attributes
A 15x15 1328 double sparse
spy(A)
So A is 15x15 sparse, with 3 blocks of size 5x5 each. Change the code to match your problem, in that all you need to do is set up your own pattern matrix, and assign the matrix Z with the elemnts that it contains. But see how trivial it really is to do, in ONE call to sparse. Again, learn to use sparse, and your code will be easy to write.
KSMAT21
2019 年 6 月 26 日
Thanks, this works great. Only thing, I used ndgrid instead of meshgrid as explained in one of David's comment.
The sparse function took 5.97 seconds created a sparse matrix of 60 millions non-zero entries. Is the time taken resonable for this size of matrix?
John D'Errico
2019 年 6 月 26 日
That time taken is very reasonable, to create a sparse matrix with 60 million entries. Without doing a test right now, the sparse block diagonal matrix I created in my answer had 100 million entries, taking also about 6 seconds to create. So I would not be at all surprised at your result. So 5 seconds or so seems reasonable.

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