Skip analysis when you have NaN or 0

I think I am having problems with my analysis result due to the presence of NaN and 0. How to skip analysis when I have NaN or 0?

p_fix = [];
for n = 4:size(t,1)
         X = [Ia(n-1,1) Ia(n-2,1) ; Ia(n-2,1) Ia(n-3,1)];
         future = [Ia(n,1) ; Ia(n-1,1)];
         C = X\future;
         Ia_future(n,1) = C(1,1)*Ia(n,1)+C(2,1)*Ia(n-1,1);
         PE(n,1)=Ia(n,1)+Ia_future(n,1);
         p(n,1)=(1+0.2)*max(PE(n-1,1));
        
     if  isempty(p_fix) && PE(n-1,1)>p(n,1)
         p_fix = p(n,1);
     if (mod(N,5)==0) && (PE(n-1,1)>p_fix) 
         fprintf('Accuses IC\n')
     else
         fprintf('Does not accuse IC\n')
         p_fix = [];
            
     end
    end

Basically I wanted this part to be ignored when (PE=NaN or 0) and (p = NaN or 0) happened. And perform the PE and p calculations again.

     if  isempty(p_fix) && PE(n-1,1)>p(n,1)
         p_fix = p(n,1);
     if (mod(N,5)==0) && (PE(n-1,1)>p_fix) 
         fprintf('Accuses IC\n')
     else
         fprintf('Does not accuse IC\n')
         p_fix = [];

Answer 1

[Image Analyst]

Do you want to break out of the loop and go somewhere else,

PE(n,1)=Ia(n,1)+Ia_future(n,1);
p(n,1)=(1+0.2)*max(PE(n-1,1));
if (isnan(PE(n, 1)) || PE(n, 1) == 0) && (isnan(p(n, 1)) || p(n, 1) == 0) 
    break; % Quit the loop totally
end

or do you want to continue with the next iteration of the loop?

PE(n,1)=Ia(n,1)+Ia_future(n,1);
p(n,1)=(1+0.2)*max(PE(n-1,1));
if (isnan(PE(n, 1)) || PE(n, 1) == 0) && (isnan(p(n, 1)) || p(n, 1) == 0) 
    continue;  % Skip to bottom of loop and continue the loop with the next iteration.
end

[Image Analyst]

Should the second and thirs ifs be nested under the first one? Probably not. Do this for me.

1. Type control-a (to select all the source code text in your editor.
2. Type control-i (to properly indent the if blocks).

Do they all look like they are nested as you want them to be? Probably not.

[Luccas S.]

Oh sorry, I ended up deleting the comment because I had found my mistake. That's right, it was the position of the end.

Answer 2

[William Rose]

@luccas santos,

a=1;  %try replacing "1" with "0" or "NaN"
if ~(isnan(a) || a==0)
    disp('a is not NaN and is not 0.')
else
    disp(a)
end 

a is not NaN and is not 0.

Try.