Looking for ideas to solve a trivial problem. One must calculate each Period individually

This is a sketch to receive feedback from other sources, as the task is not easy.

We have the following interval, in which the maximum values (showned here as X=3000,81 Y=730 ; X =3005,16 Y=740 ; etc. ) should be saved independently; and at the same time measure its mean value. The data showned has been plot as plot(RR_location , RR), in which both RR_location and RR have a length of 14583x1 double.

There are 2 goals, which both of them are complementary from each other.

The first one is to measure each maximum peak for as long as to reach the next following one (this will be measure as a whole period, which will save in the parameter arrayMax the new maximum peaks (keeping the length of RR_location).

The second goal is the mean breathing rate {as an example for the first period: meanHF = (730 + 715 + 725 + 725 + 720 + 735 + 740) / 7 = 727,14}. This will then be save in finalHF{i} = meanHF / 60 (units: Beats per Minute).

My actual guess, which is wrong (no need to test it, its just a first sketch with some random parameters and comments floating around jeje), as I simply need some fresh air. :)

%array for only maximum values (this would be the output of the function
arrayMax = [];
%constant value for the actual Max value (or at least a guess)
newMax = 0;
firstMax= 0;
secondMax=0;

%Mean value Breathing Rate
MeanHF = 0;

%second output final HF


%actual HF
actualHF = 0;

%step Counter to do the mean value afterwards
counter = 0;

for i=1:length(RR_location) %this is for the whole database
    %The maximas showned in the graph are as follow:
    %Sample RR_location(i= 4429) => X=3000,81 ; Y=730
    %Sample RR_location(i= 4435) => X=3005,16 ; Y=740
    %Sample RR_location(i= 4440) => X=3008,8 ; Y= 740
    %Sample RR_location(i= 4444) => X=3011,74 ; Y= 750
    %Sample RR_location(i= 4453) => X=3018,15 ; Y= 745
    
    %This would be associated to a motion artifact,  aka a false value to be remove
    if(RR(i+1) > 1000) || (RR(i+1) < 500) 
        newMax = 0;
    else
        firstMax = RR(i+1);
    end    
    %And here is where my struggles arise (aka PROBLEM)
    if (RR(i+1) > firstMax)
        newMax = RR(i);
        counter = 0;
        
    elseif (RR(i+1) < firstMax)
        newMax = 0; %if it wasnt a Max peak, the arrayMax = 0;
        counter = counter + 1; %we increase the step counter;
    end
    
    arrayMax{i} = newMax;
    
    
    firstMax = RR(i);
    
    %if we find a maxPeak, we enter a new for-loop
    for j=1:10
        actualHF = RR(i);
        %count the number of samples taken
        if(RR(i+j+1)>RR(i+j))||(RR(i+j+1)>firstMax)
            counter = counter + 1;
        else
           secondMax = RR(i+j);
           break;
        end
    end
    
    meanHF = ( ) / 60;
    
    secondMax = firstMax;
end

If it helps, I can upload the X and Y values for the whole interval showned above. Just ask for it.

Any suggestion are welcome.

Thank you, David

[Image Analyst]

Is it "trivial" or "not easy"? Explain the contradiction.

Did you try findpeaks?

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