Getting rid of loops

Hi, I am trying to clean matrix A (which is a 1064 x 4 matrix) while relying in the cleaning process on Matrices B and C that have the same size as A (Mat file attached). I developed a loop to clean A but I feel that a loop is unnecessary and that I can do it without relying on loops. Could the cleaning process be done much more efficiently? Your help is appreciated.

for i=1:size(A,1)
	for j=1:size(A,2)
		if A(i,j)<1 && A(i,j)>0 % this is the first issue in the cleaning process. A values between 0 and 1 should be equal to the respective value in matrix X
			A (i,j)=B(i,j);
		end
		if A(i,j)<C(i,j) % this is the 2nd issue I am facing A values lower than their respective C values are erroneous and should be replaced by NaN
			A (i,j)=nan;
		end
		if B(i,j)>100 %this is the third issue:  When B is greater than 100, A values should be equal to B values
			A (i,j)=B(i,j);
		elseif B(i,j)<0 %This is the fourth issue: negative percentages are missing observations
			A(i,j)=-999;
		end
		if C(i,j)==0%This is the 5th issue: when C is zero, A should be zero as well
			A(i,j)=0;
		end
	end
end

[Stephen Cobeldick]

http://web.archive.org/web/20210109204538/https://www.mathworks.com/matlabcentral/answers/712828-getting-rid-of-loops

Answer 1

[Stephen Cobeldick]

Logical indexing is much simpler than using loops:

idx    = A<1 & A>0;
A(idx) = B(idx);
A(A<C) = NaN;
idy    = B>100 | B<0;
A(idy) = B(idy);
A(C==0) = 0

[Wesso]

It is not working perfectly. Look at A(1,3) it is 2. C(1,3) is 1. When A<C then A should be NaN

[Wesso]

similar to my comments to image analyst the codes are not working well. A(2,1)=0 and C(2,1)=30. The codes replaced A(2,1) with NaN

[Stephen Cobeldick]

"A(2,1)=0 and C(2,1)=30. The codes replaced A(2,1) with NaN"

Because that is what you requested. Your original code includes this comment: "A values lower than their respective C values are erroneous and should be replaced by NaN". The value 0 is clearly less than 30, and so that erroneous value is set to NaN. Just as you wrote. If you actually want something else to occur (different to what you specified in your question), then please explain the new rule so that we can help you implement it.

[Wesso]

my bad I was looking at the wrong matrix. Thanks a lot

Answer 2

[Image Analyst]

Did you consider just using a mask and no for loops?

mask = A<1 & A>0; % this is the first issue in the cleaning process. A values between 0 and 1 should be equal to the respective value in matrix X
A(mask) = B(mask);
mask = A<C; % this is the 2nd issue I am facing A values lower than their respective C values are erroneous and should be replaced by NaN
A(mask) = nan;
mask = B>100; %this is the third issue:  When B is greater than 100, A values should be equal to B values
A(mask) = B(mask);
mask = B<0; %This is the fourth issue: negative percentages are missing observations
A(mask) = -999;
mask = C==0%This is the 5th issue: when C is zero, A should be zero as well
A(mask) = 0;

[Wesso]

I am receiving errors

for example:

A(mask) = B is yielding the following error: Unable to perform assignment because the left and right sides have a different number of elements.

Error in table1 (line 10)

A(mask) = B;

[Stephen Cobeldick]

The answer has a small bug: the logical index should be used on both sides.

[Image Analyst]

Right. Thanks for catching that Steven - I've corrected it. I got it right later but missed it on line 2.

[Wesso]

Also it is not working perfectly. Look at A(1,3) it is 2. C(1,3) is 1. When A<C then A should be NaN

[Wesso]

I don't think the codes are working well. A(2,1)=0 and C(2,1)=30. The codes replaced A(2,1) with NaN

[Image Analyst]

"Look at A(1,3) it is 2. C(1,3) is 1. When A<C then A should be NaN" <== Yes, so A is not less than 1 there so NaN is not assigned there, as requested. Are you implying that it SHOULD be NaN even though 2 not less than 1????

"A(2,1)=0 and C(2,1)=30. The codes replaced A(2,1) with NaN". Yes, it does get replaced by NaN just like you asked for when A is less than C. Why is that not working well?

Finally, you may have competing conditions. It's possible that some condition may get set to one values, then later in the code you have a different mask and it gets replaced by a different value.

[Wesso]

my bad I was looking at the wrong matrix. Thanks a lot