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!

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Santi
Santi 2016-12-22
编辑： Jan 2017-2-20
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
  1 个评论
KSSV
KSSV 2016-12-22
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

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采纳的回答

David Barry
David Barry 2016-12-22
编辑：David Barry 2016-12-22
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;

更多回答（2 个）

Santi
Santi 2016-12-22
Very nice and simple answer, worked! Thanks a lot
John BG
John BG 2016-12-22
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:jgb2012@sky.com jgb2012@sky.com>
  1 个评论
Jan
Jan 2017-2-19
编辑：Jan 2017-2-20
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.

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