I have the following code (recognizing individuals pixels of a matrix). For a high number of images, this is very time consuming. Therefore, ¿Could the loop be avoided ? Thanks!

V=zeros(x,y,num_images,'uint16')
for k=1:num_images
for i= 1:x
for j= 1:y
    if B(i,j,k)==0
        V(i,j,k)= A(i,j,k)/3
    else
        V(i,j,k)= A(i,j,k)/3+(2^15)
    end
end
end
end

[KSSV]

First you terminate the outputs with ;. Else it will take hell lot of time, because it will print result on the screen.

V=zeros(x,y,num_images,'uint16') ;
for k=1:num_images
    for i= 1:x
        for j= 1:y
            if B(i,j,k)==0
                V(i,j,k)= A(i,j,k)/3 ;
            else
                V(i,j,k)= A(i,j,k)/3+(2^15) ;
            end
        end
    end
end

Answer 1

[David Barry]

It's difficult to know the relationship between A, B and V from your example code but I am assuming they are of equal dimensions. In which case I think you probably want something like this instead of the loop.

V = zeros(size(B));
V(B == 0) = A(B == 0)/3;
V(B ~= 0) = A(B ~=0)/3 + 2^15;

Answer 2

[Santi]

Very nice and simple answer, worked! Thanks a lot

[David Barry]

Great. Please accept the answer then.

Answer 3

[John BG]

there is no need to initialise the void V, go straight to the indices returned from functions

• find( zeros())
• find( nonzeros())

A=A/3
V(find(zeros(B)))=A(find(zeros(B)))+2^15
V(find(nonzeros(B)))=A(find(nonzeros(B)))

if you find my answer useful would you please mark it as Accepted Answer by clicking on the ACCEPT ANSWER button?

thanks in advance for time and attention

John BG

<mailto:[e-mail] [e-mail]>

[Jan]

Both lines do not work:

• nonzeros() replies the vector of the non-zero elements and the information about the position inside B is lost.
• zeros() creates an array of zeros with the elements of B as dimensions. Therefore find(zeros(B)) will reply the empty matrix.