First number in foor loop

Hello, this is a part of my code, I have a image analysis, where the program reading frames (h), however is starts to read due to the image changing. This means, that the h is not always start with 1, but it can starts with 5 or higher number. That is the problem with the code, where I have the part h==1. So how to change the code, to have

if h= first number of h

I dont know how to get the first number of the loop and remember it, because the loop change it every time. please help

Or how to write values to avoid giving the initial value

for h=1:100  
    tabulka1=...

if h==1

     htabulka = tabulka1;

     else
     htabulka = [htabulka; tabulka1];
end
end

Answer 1

[Vilém Frynta]

Hello,

you already know the whole vector h. If you want it's first value, you can keep saving the h value into a vector, and then call it's first value.

[🇨🇿] Můžeš ukládat h postupně do vektoru, a pak si můžeš kdykoliv zavolat první hodnotu.

An example:

% Vector to save your "h" values
v = [];

for h = 5:100
    
    v(h) = h;           % save the current "h" value into a vector
    v = v(v > 0);       % ignore zero values
    
    if h == v(1)        % compare h to first value
          ...
    end

end

v(1) % the first number of iteration.

ans = 5

Hope I helped.

Answer 2

[Steven Lord]

Don't hard code the first number. Store it in a variable that you can use to define the loop and then to retrieve the starting value afterwards.

firstNumber = 5;
for k = firstNumber + (0:9) % 10 iterations starting at firstNumber
    fprintf("Loop body with k = %d.\n", k)
end

Loop body with k = 5.
Loop body with k = 6.
Loop body with k = 7.
Loop body with k = 8.
Loop body with k = 9.
Loop body with k = 10.
Loop body with k = 11.
Loop body with k = 12.
Loop body with k = 13.
Loop body with k = 14.

fprintf("First number in the loop was %d.\n", firstNumber)

First number in the loop was 5.

If you've written a function to do your processing, you could even specify firstNumber as one of the inputs to the function.