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Asked by I Made on 12 Mar 2013

So i have e.g data < 3x3 int 16 >

4 5 3

5 4 1

2 1 2

i wanted to got

78 60 23

and the process is like :

Column 1->(4*4*1)+(5*5*2)+(2*2*3) then Column 2->(5*5*1)+(4*4*2)+(1*1*3) then Column 3->(3*3*1+1*1*2+2*2*3)

i tried this :

[g,h]=size(data); m = int16(1:g); t=sum(data(:,m).*data(:,m).*m);

but i have matrix dimension problem, How can i solve,achieve this?

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Answer by Andrei Bobrov on 12 Mar 2013

Accepted answer

out = cast((1:size(A,1))*double(A.*A),class(A));

Show 1 older comment

Cedric Wannaz on 12 Mar 2013

If you are not interested in the output to be the same type as `A`, you can do:

out = (1:size(A,1)) * double(A).^2 ;

so you are not limited by int16 upper limit.

I Made on 12 Mar 2013

Works like magic.. thanks andrei tough i don't understand yet the principle of the precision using double or using class like you mention on your first answer.

Cedric Wannaz on 13 Mar 2013

Andrei type-casts `A.*A` to double so it can be multiplied by the vector `(1:size(..))` avoiding the error that you got when you tried my solution (that I wrote without paying enough attention to the fact that `A` was `int16`). He then type-casts back to the original type afterwards (which is the type/class of `A`).

In my comment above, I type-cast `A` to double before squaring it, so the square is not truncated by int16 limits. I don't type-cast back to the type of `A`, because I don't think that you need that (and there would be the truncation issue as the matrix multiplication performs a weighted sum of squares with weights >=1).

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