Is there an easier way to index diagonal elements of a matrix?
50 次查看(过去 30 天)
显示 更早的评论
Let's say I have a 10 x 10 diagonal matrix of random integers between 0 and 100:
A = diag(randi(100,10,1));
I want to replace some of the diagonals with different values. In particular, I want to replace the 3rd, 6th, 7th, and 9th diagonal element with the value 1000.
One intuitive way to do this would be:
A([3 6 7 9],[3 6 7 9]) = 1000;
But this doesn't work because MATLAB reads this as replacing matrix entries (3,3), (3,6), (3,7), (3,9), (6,3), (6,6), (6,7), (6,9), and so on.
One way that does work is to go:
v = diag(A);
v([3 6 7 9]) = 1000;
A = diag(v);
But this seems kind of clunky with double calls to "diag" and the additional variable "v" needing to be stored in memory. Is there a more elegant way to do it using matrix indexing?
Thanks
0 个评论
采纳的回答
John D'Errico
2022-7-27
编辑:John D'Errico
2022-7-27
A = diag(randi(100,10,1));
n = size(A,1);
A(sub2ind([n,n],[3 6 7 9],[3 6 7 9])) = 1000;
A
If you understand how matrix elements are stored in memory, it is not that hard either, even if we avoid sub2ind. Next, I'll change them to 999.
ind = [3 6 7 9];
A(ind + (ind - 1)*n) = 999;
A
更多回答(1 个)
David Hill
2022-7-27
Or linear indexing
n=20;%size of matrix
A = diag(randi(100,n,1));
c=[4 7 8 12];%places on the diagonal wanting to replace
A((c-1)*(n+1)+1)=1000;
0 个评论
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 Creating and Concatenating Matrices 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!