How to generate a sparse matrix for a given array?

2021 5 24
2 个回答
回答已采纳
更新时间：2021 5 25
6 次查看（30 天）

0 个投票

Dear All,
I want to generate a sparse matrix B for a given array A. A contains two columns of indecis. For example,
A = [1 3;
1 4;
2 3;
2 5;
3 5;
4 5]
The sparse matrix B should be as follows:
B = [ 2 0 -1 -1 0;
0 2 -1 0 -1;
-1 -1 2 0 0;
-1 0 0 2 -1;
0 -1 -1 -1 3]
The characteristics of B:
  1. Sum of each row is zero.
  2. If we consider matrix A gives the information of edges of a graph, B(i,i) = sum of number of edges, B(i,j) = -1 if there is an edge between i and j.
Thanks a lot.
Benson

请先登录，再回答此问题。

 采纳的回答

Matt J
Matt J 2021-5-25
编辑：Matt J 2021-5-25

0 个投票

B=laplacian( graph(A(:,1),A(:,2)) );

5 个评论
显示 3更早的评论 隐藏 3更早的评论

Benson Gou
Benson Gou 2021-5-25
Hi, Matt,
I tried your code, but I got the following error:
"Graph has multiple edges."
Thanks a lot.
Benson
Matt J
Matt J 2021-5-25
Ran in:
Ran in:
Works fine for me:
A = [1 3;
1 4;
2 3;
2 5;
3 5;
4 5];
B=laplacian( graph(A(:,1),A(:,2)) );;
full(B)
ans = 5×5
2 0 -1 -1 0 0 2 -1 0 -1 -1 -1 3 0 -1 -1 0 0 2 -1 0 -1 -1 -1 3
Matt J
Matt J 2021-5-25
编辑：Matt J 2021-5-25
You probably need to do,
A=unique(sort(A,2),'rows')
Benson Gou
Benson Gou 2021-5-25
yes, it works well. Thanks a lot.
Benson
Matt J
Matt J 2021-5-25
You're welcome, but please Accept-click the answer that you deem best resolves your question.

请先登录，再进行评论。

更多回答（1 个）

the cyclist
the cyclist 2021-5-24
Ran in:

1 个投票

Ran in:
Here is one way:
A = [1 3;
1 4;
2 3;
2 5;
3 5;
4 5];
d = max(A(:));
B = sparse(A(:,1),A(:,2),-1,d,d) + sparse(A(:,2),A(:,1),-1,d,d);
for ii = 1:d
B(ii,ii) = -sum(B(ii,:));
end
disp(B)
(1,1) 2 (3,1) -1 (4,1) -1 (2,2) 2 (3,2) -1 (5,2) -1 (1,3) -1 (2,3) -1 (3,3) 3 (5,3) -1 (1,4) -1 (4,4) 2 (5,4) -1 (2,5) -1 (3,5) -1 (4,5) -1 (5,5) 3

7 个评论
显示 5更早的评论 隐藏 5更早的评论

Benson Gou
Benson Gou 2021-5-24
Hi, the Cyclist,
Thanks for your answer. It looks pretty good. But I am wondering if it is possible to avoid using the for iterations, because matrix A has 5000 rows.
Thanks a lot again.
Benson
the cyclist
the cyclist 2021-5-24
Ran in:
Ran in:
A = [1 3;
1 4;
2 3;
2 5;
3 5;
4 5];
d = max(A(:));
B = sparse(A(:,1),A(:,2),-1,d,d) + sparse(A(:,2),A(:,1),-1,d,d);
% for ii = 1:d
% B(ii,ii) = -sum(B(ii,:));
% end
B = B - diag(sum(B,2));
disp(B)
(1,1) 2 (3,1) -1 (4,1) -1 (2,2) 2 (3,2) -1 (5,2) -1 (1,3) -1 (2,3) -1 (3,3) 3 (5,3) -1 (1,4) -1 (4,4) 2 (5,4) -1 (2,5) -1 (3,5) -1 (4,5) -1 (5,5) 3
Benson Gou
Benson Gou 2021-5-25
Hi, Cyclist,
Thanks a lot for your prompt reply.
I tried your code on my case with 5000 rows in A. But I found there are some different results between your code and my original code which uses for iterations.
For example, row 3 in B, yours is
(1,5) 8
(1,7) -1
(1,28) -3
(1,33) -1
(1,34) -1
(1,397) -1
(1,434) -1
My result is
(1,5) 6
(1,7) -1
(1,28) -1
(1,33) -1
(1,34) -1
(1,397) -1
(1,434) -1
The locations of non-zero elemnts are correct, but the values are different. The differency may be caused by the double edges between two nodes. There are 2 edges between 5 and 28.
Thanks
Benson
Bruno Luong
Bruno Luong 2021-5-25
(1,28) -3
"There are 2 edges between 5 and 28."
Make it three.
Benson Gou
Benson Gou 2021-5-25
Hi, Bruno,
Instead of making it 3, it should be -1, as my code did. It means I would like to ignore the mulitple edges between two nodes.
Thanks.
Benson
the cyclist
the cyclist 2021-5-25
I assume that doing what @Matt J suggested in his answer:
A=unique(sort(A,2),'rows')
is what you would need to do here as well.
If you posted a small example that exhibits the problem, it would help. But I'm guessing you have your answer.
Benson Gou
Benson Gou 2021-5-25
Hi, the Cyclist,
Thanks a lot for your great help. You have a good day!
Benson

请先登录，再进行评论。

类别

在 帮助中心 和 File Exchange 中查找有关 Sparse Matrices 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by