Performing Iteration for a Matrix
2 ビュー (過去 30 日間)
古いコメントを表示
I want to perform an iterative operation on each element of a matrix according to this equation which is for a 3 x 3 matrix:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/906320/image.jpeg)
For example, by having this matrix:
a = [0 0.2 -0.6; -0.2 0 0.4; 0.6 -0.4 0]
I wrote the code for the first iteration for each element seperately but my original matrix is 1000 x 1000, so I need a code that can do this operation for the whole matrix. The code that I wrote for each element of the matrix is:
a_1(1,2) = a(1,2) + (0.25 * (a(1,3) + a(3,2))); %CF is 0.25
a_1(1,3) = a(1,3) + (0.25 * (a(1,2) + a(2,3)));
a_1(2,1) = a(2,1) + (0.25 * (a(2,3) + a(3,1)));
a_1(2,3) = a(2,3) + (0.25 * (a(2,1) + a(1,3)));
a_1(3,1) = a(3,1) + (0.25 * (a(3,2) + a(2,1)));
a_1(3,2) = a(3,2) + (0.25 * (a(3,1) + a(1,2)))
Which will give the result as shown above. The term that starts with CF on the RH is essentially saying no index is chosen twice in the same place. I'd appreciate it if someone can please help me. Thank you!
0 件のコメント
採用された回答
Torsten
2022 年 2 月 25 日
編集済み: Torsten
2022 年 2 月 25 日
L = 2;
CF = 0.25;
a = [0 0.2 -0.6; -0.2 0 0.4; 0.6 -0.4 0];
n = size(a,1);
A = zeros(L,n,n);
A(1,:,:) = a;
for l = 2:L
for i = 1:n
for j = 1:n
A(l,i,j) = A(l-1,i,j) - CF*(A(l-1,i,i) + 2*A(l-1,i,j) + A(l-1,j,j));
for k = 1:n
A(l,i,j) = A(l,i,j) + CF*(A(l-1,i,k) + A(l-1,k,j));
end
end
end
end
0 件のコメント
その他の回答 (0 件)
参考
カテゴリ
Help Center および File Exchange で Matrix Indexing についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!