## How to vectorize this code to eliminate nested For loops?

Asked by Jae Min Lee

### Jae Min Lee (view profile)

on 6 Sep 2018
Latest activity Commented on by Greg

### Greg (view profile)

on 6 Sep 2018
Accepted Answer by Greg

### Greg (view profile)

Would like to know how this code is vectorized:
for r=1:rows
for c=1:cols
if ((Lin <= Limg(r,c)) && (Limg(r,c) <= Hin))
enhanceImg(r,c) = Lout + (Hout - Lout) * ((Limg(r,c) - Lin) / (Hin - Lin))^gamma;
elseif (Limg(r,c) > Lin)
enhanceImg(r,c) = Hout;
else
enhanceImg(r,c) = Lout;
end
end
end
Lin, Hin, Lout, Hout, gamma are scalar values.

Greg

### Greg (view profile)

on 6 Sep 2018
+1 for a very well formed question. Clear and concise, showed an attempt, and described relevant variable sizes.
That all makes it much easier to answer.

on 6 Sep 2018
Edited by Greg

### Greg (view profile)

on 6 Sep 2018

Most of that is straightforward, just a couple elementwise periods for safety. Then, the hardest part is getting the logic to put the right values into your output matrix.
%%%Pre-allocate to Lout saves one round of comparison + assignment
enhancedImg = Lout.*ones(rows,cols);
%%%Build a mask for your first if condition
blnOne = (Lin <= Limg) & (Limg <= Hin);
%%%Assign the arithmetic result to elements behind that mask
enhancedImg(blnOne) = Lout+(Hout-Lout).*((Limg(blnOne)-Lin)./(Hin-Lin)).^gamma;
%%%Another mask for the elseif condition
blnTwo = Limg > Lin;
%%%Assign Hout to those elements
enhancedImg(blnTwo) = Hout;
%%%As mentioned above, all remaining elements are already Lout, so no "else" condition is needed
EDIT: Fixed Limg(blnOne) per first comment below.

Show 1 older comment
Greg

### Greg (view profile)

on 6 Sep 2018
Sorry, see updated answer.
Jae Min Lee

### Jae Min Lee (view profile)

on 6 Sep 2018
Thank you very much!
Greg

on 6 Sep 2018
Happy to help.