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sum every 24 rows in a vector

Andrew Alkiviades さんによって質問されました 2012 年 9 月 24 日
最新アクティビティ Mohamed Atef さんによって コメントされました 2019 年 1 月 19 日
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));

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3 件の回答

Matt Fig
Answer by Matt Fig
on 24 Sep 2012
Edited by Matt Fig
on 24 Sep 2012
 Accepted Answer

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)

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Thanks alot man

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Daniel Shub
Answer by Daniel Shub
on 24 Sep 2012
Edited by 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);

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Honglei Chen
Answer by Honglei Chen
on 24 Sep 2012
Edited by 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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