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How to append for each iteration?

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Maria Hart
Maria Hart on 18 Jan 2021
Commented: Maria Hart on 18 Jan 2021
Dear all,
the situation is the following: I have a projectlength of 3 years with an assumed number of hours of 10. For each hour the yield is calculated with a certain function (which I simplify in this example with the number 2). As the projectlength and hournumber should stay variable, I would like to first calculate the yield in each hour of the year (helper(k)) and then append the solutions of each projectyear below each other. So the yield should look like this:
I got the following so far, however, it does not work.
hournum = 10;
projectlength = 3;
yield = [];
for p=1:projectlength
for k = 1:hournum
helper(k)=2;
end
yield(p) = [yield(p); {helper(k)}];
end
Error Message: Index exceeds the number of array elements (0).

  2 Comments

Jan
Jan on 18 Jan 2021
Whenever you write "does not work", append the error message or a details description of the difference between your expectations and the obtained results.
Maria Hart
Maria Hart on 18 Jan 2021
Ok, thank you. I have added the error message.

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Accepted Answer

Jan
Jan on 18 Jan 2021
Edited: Jan on 18 Jan 2021
hournum = 10;
projectlength = 3;
yield = cell(1, projectlength); % Pre-allocate as cell
helper = zeros(1, hournum); % Pre-allocate as vector
for p = 1:projectlength
for k = 1:hournum
helper(k) = 2;
end
yield{p} = helper;
end

  3 Comments

Maria Hart
Maria Hart on 18 Jan 2021
I have tried your solution, but I get a 1x3 cell with thee times the 2 instead a 30x1 vector.
Maria Hart
Maria Hart on 18 Jan 2021
Thank you for your answer. If you would like to have instead a simple vector with all numbers beneath each other and no cell (I would like to plot all data), what would I do?
Maria Hart
Maria Hart on 18 Jan 2021
Found a way with cell2mat!

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