Check if the condition happened in previous cycles

I'm trying to make a condition to analyze every 5 cycles, but I need the value of p_fix (threshold) not to exceed PE. As I did, it only analyzes when it reaches 5 cycles, but I needed to check if PE>p was respected until it reaches 5 cycles again and again.

if mod(n,5)==0 && PE(n-1,1)>p_fix
            fprintf('Accuses IC\n')

Answer 1

[Matt J]

Update a boolean flag to keep track of it.

flag=true; %initial state
for n=1:N

    flag=flag & PE(n-1,1)>p_fix;
    
    if mod(n,5)==0 && flag
            fprintf('Accuses IC\n')
end

[Luccas S.]

In my case I can't declare the initial state before the for since the values of PE and p_fix are calculated inside it. And the initial state of p_fix is [].

complete code

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

Unrecognized function or variable 't'.

[Luccas S.]

Then he accuses the following error:

Operands to the || and && operators must be convertible to logical scalar values.

Error in linear_prediction (line 35)

if mod(n,5)==0 && flag

[Matt J]

I've edited your previous comment and Run the code in it. As you can see there, it doesn't produce the error you claim.

[Torsten]

@luccas santos

You use

flag=(PE(0,1)>p_fix);

in your code. But PE(0,1) does not exist and p_fix is not initialized to a real number. So flag is not initialized to logical.

[Luccas S.]

OK, yes. Now it worked, thank you all