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Hi I am trying to find an output which is the sum of every 24 rows of a 8760x1 vector below as hourly_deficit. Therefore I am trying to sum rows 1:24, 25:49, 50:73 etc etc

I am trying to do this on the line below

for idx_number_panels = 1:length(number_panels) % range of PV panel units examined

for idx_number_turbines = 1:length(number_turbines) % range of wind turbine units examined

for idx_number_batteries = 1:length(number_batteries) % range of battery units examined

for h=2:8759 %# hours

hourly_deficit(idx_number_panels,idx_number_turbines,idx_number_batteries, h) = hourly_annual_demand(h) - (hourly_annual_PV(h)*number_panels(idx_number_panels)) - (hourly_annual_WT(h)*number_turbines(idx_number_turbines));

Matt Fig
on 24 Sep 2012

Edited: Matt Fig
on 24 Sep 2012

If A is 8760-by-1, and you want to find the sum of every 24 elements, such that you will end up with 365 sums, then do:

sum(reshape(A,24,365))

As an example you can see easier, get the sum of every two elements of a 10-by-1:

A = (1:10)'

B = reshape(A,2,5)

sum(B)

Daniel Shub
on 24 Sep 2012

Edited: Daniel Shub
on 24 Sep 2012

I am not sure what all the code you posted has to do with anything ...

If I have a 8760x1 array

x = randn(8760, 1);

I can reshape it to be 24x365 with

y = reshape(x, 24, 365);

and then sum each of the 365 columns

z = sum(y);

EDIT

You could also filter the data

z = filter(ones(24, 1), 1, x);

z = z(24:24:end);

Honglei Chen
on 24 Sep 2012

Edited: Honglei Chen
on 24 Sep 2012

Not sure what your end format is, but the following code adds every 24 rows and retain all the results in one column

reshape(sum(reshape(x,24,[])),[],1)

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