# Is there an easier way to index diagonal elements of a matrix?

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Darcy Cordell am 27 Jul. 2022
Kommentiert: Darcy Cordell am 28 Jul. 2022
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
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### Akzeptierte Antwort

John D'Errico am 27 Jul. 2022
Bearbeitet: John D'Errico am 27 Jul. 2022
A = diag(randi(100,10,1));
n = size(A,1);
A(sub2ind([n,n],[3 6 7 9],[3 6 7 9])) = 1000;
A
A = 10×10
56 0 0 0 0 0 0 0 0 0 0 52 0 0 0 0 0 0 0 0 0 0 1000 0 0 0 0 0 0 0 0 0 0 23 0 0 0 0 0 0 0 0 0 0 84 0 0 0 0 0 0 0 0 0 0 1000 0 0 0 0 0 0 0 0 0 0 1000 0 0 0 0 0 0 0 0 0 0 25 0 0 0 0 0 0 0 0 0 0 1000 0 0 0 0 0 0 0 0 0 0 33
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
A = 10×10
56 0 0 0 0 0 0 0 0 0 0 52 0 0 0 0 0 0 0 0 0 0 999 0 0 0 0 0 0 0 0 0 0 23 0 0 0 0 0 0 0 0 0 0 84 0 0 0 0 0 0 0 0 0 0 999 0 0 0 0 0 0 0 0 0 0 999 0 0 0 0 0 0 0 0 0 0 25 0 0 0 0 0 0 0 0 0 0 999 0 0 0 0 0 0 0 0 0 0 33
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Darcy Cordell am 28 Jul. 2022
Thanks! The linear indexing method is exactly the kind of nice solution I was looking for.

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### Weitere Antworten (1)

David Hill am 27 Jul. 2022
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;
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