how can i simplify this for loop?

for i=1:2 %for-loop1
  for j=1:5 %for-loop2
      x(j)=i;
      y(j)=j+1;
      t=[x; y]';
  end
  for j=1:5 %for-loop3
      t(j,1)=t(j,1)+j;
      t(j,2)=t(j,2)+2*j;
  end
  if i==1 %if
      a=t;
  elseif i>1
      a=[a; t];
  end
end
disp(a);

thanks for your advice and help.

---------------------------

OHHHH.. I'm sorry for my stupid question.

upper code is simplification, and

for i=1:10 %for-loop1
  for j=1:11 %for-loop2
  ca=Gimg(round((i-8/10)*H):round((i-2/10)*H),round((j-8/10)*W):round((j-2/10)*W));
  if mean(mean(ca))<60
      ca(ca>mean(mean(ca))-3)=0;
  else
      ca(ca<mean(mean(ca))+3)=0;
  end
  [x(j) y(j)]=pjm_centroid(ca);
  array=[x; y]';
  end
  for j=1:11 %for-loop3
      array(j,1)=array(j,1)+round((j-8/10)*W);
      array(j,2)=array(j,2)+round((i-8/10)*H);
  end
  if i==1 %if
      p=array;
  elseif i>1
      p=[p; array];
  end
end
disp(p);

---------------------

this is an originally code.

variables i and j are meaningful I think.

When you compare simplification and of original code,

how can I simplify structure of for-loop in original code?

Answer 1

[kjetil87]

a=[];
t=zeros(5,2);
for i=1:2
    t(:,1)=(1:5)+i;
    t(:,2)=(2:6)+2*(1:5);
    a=[a;t];
end
disp(a)

You can also get rid of the for i=1:2 but i included it to show the process.

a=zeros(10,2);
a(:,1)=[(2:6),(3:7)];  
a(:,2)=repmat((2:6)+2*(1:5),1,2); %or just hardcode 
                                  %t(:,2)=[(2:6)+2*(1:5),(2:6)+2*(1:5)]
disp(a)

[kjetil87]

After your edit: I do not know how the Gimg funciton works, but if it can take a vectorized inputs you may try to use the same strategy (i.e remove the for loop and replace j with (1:11) . If so the mean test can probably be performed on all outpus at once.

It will atleast work for for-loop3, just remember to pre allocate array and index it with array(:,1) etc.

then

array(:,1)=array(:,1)+round( ( (1:11)-8/1) *W)

will be the same as your third loop etc.

Also your if i==1 test is not needed if you declare p as empty before the loop

p=[];

Then in p=[p;array] will work even when i=1.

In general if you have a for loop that uses the index on each element in an array with the "same operation every time" you can replace the entire loop with a vector such as 1:11 instead of j.

[kjetil87]

Also from a closer look at your code you may have intended to move

array=[x,y]';

below the end line of for loop 2. Now you are doing 10 meaningless operations before you overwrite it again with the correct result of x and y.

[PJM]

Thank you for your help!! :D

Answer 2

[Evan]

Here's how you can get rid of the nested loops:

a = [];
for i = 1:2
   x(1:5) = i;
   y(1:5) = (1:5) + 1;
   t = [x; y]';

   t(1:5,1) = t(1:5,1) + (1:5)';
   t(1:5,2) = t(1:5,2) + 2*(1:5)';

   a = [a; t];
end
disp(a)