How to get multiple numbers in a row for an Array

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RTG 2020-3-2

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编辑： Jayla Wesley 2022-2-11

Right now I am calculating the Equation of Time for a class. The EOT depends on the day of the year however for my code I need an answer to repeat itself 24 times before moving on to the next answer. For example if my first EOT function reads out 1, I want this 1 to continue 24 times and then move on to the next EOT function and so on. Eventually I want to have a matrix of 1 x 8760. Is there a way to do this? Here is what I have so far.

for n=1:1:365

for j =1:24

N(n) = (n-1)*(360/365);

EOT = 229.2*(.000075+0.001868*cos(N)- 0.032077*sin(N) - 0.014615*cos(2*N)- 0.04089*sin(2*N));

end

Any help is appreciated as I am stuck and my EOT Matrix is still at 1x365. Thank you.

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TADA 2020-3-2

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you assign all of EOT each iteration, for the entire current value of N, which increases each iteration of the outer loop

If you want to get a 1x8760 you need to use indexing and set the values according to an index you will have to calculate using n and j

something like that:

EOT((n-1)*365 + j) = ...

you also don't use j (nor any random value) ever, so all 24 repeats will be identical

don't forget to preallocate your vectors for better performance:

EOT = zeros(1, 365*24);

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Answer 1

Reshma Nerella 2020-3-6

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Hi,

Since you are not finding any value varying with j in the inner for loop, you can remove it.

EOT = zeros(1, 365*24);
for n=1:365
            N(n) = (n-1)*(360/365);
            val  =  229.2*(.000075+0.001868*cos(N)- 0.032077*sin(N) - 0.014615*cos(2*N)- 0.04089*sin(2*N));                      
            EOT(1, (i-1)*24+1 : i*24) = val;
end