Matrix manipulation and using the sum command

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Ataman Ugur 2017-1-23

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编辑： Walter Roberson 2017-1-30

Hey guys I'm trying to use the "sum" command to replace the central value in a 3x3 matrix by the 8 surrounding values, but I need to exclude the central value from the summation. A=[8 8 8;8 7 8;8 8 8], I need to replace the 7 with the sum of the surrounding 8's using the "sum" command. Can anyone show me how to exclude the 7 from the command A(2,2)=sum(A(1:3,1:3))?

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Answer 1

John Chilleri 2017-1-23

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编辑：John Chilleri 2017-1-29

在 MATLAB Online 中打开

Hello,

Would,

 A(2,2) = 0;
 A(2,2) = sum(A(:));

or

A(2,2) = sum(A(:)) - A(2,2);

suffice?

You could also generalize this to any matrix that has a center value (i.e. n x n where n is odd):

n = size(A,1);
A((n+1)/2,(n+1)/2) = sum(A(:)) - A((n+1)/2,(n+1)/2);

Hope this helps!

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Jan 2017-1-29

编辑：Jan 2017-1-29

+1, Or sum(A(:)).

John Chilleri 2017-1-29

Thanks for all the feedback on many of my posts! It's very helpful and much appreciated!

I decided to test if sum(A(:)) is faster than sum(sum(A))!

Other than being more elegant, it's also faster; performing 1,000,000 sum(A(:)) took about ~1.4 seconds versus ~2.2 seconds for sum(sum(A)). Will switch to A(:) from here on out!

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