sum function and add using loop

I have a column vector with 31536001 number of rows.

e.g. [34 ;30; 14; 12; 54; 46; 47; 48; 49; ......10000]';

How can I write a command using a FOR loop, that adds the first 3600 elements together the "purpose is to change from seconds to hour", then the second element from 3601 s to 7200 second and so on.my data in seconds i need it in hour for one year i have 31536001 s and i need my factor to be 8760 hours

d=[1;2;3;4;5;6;................]

i need

[x1,x2,........]

which has

x1=(1+2+....3600)

thanks

[Jan]

Please use meaningful tags, because they are used to classify the questions. "matlab" is not helpful, beause all questions in a Matlab forum concern this topic.

What do you want to do with the last chunk, which does not contain 3600 elements?

Answer 1

[Jan]

num = floor(length(d) / 3600);
result = zeros(1, num);
for k = 1:num
  s = (k - 1) * 3600;
  result(k) = sum(d(1+s:3600+s));
end

Or faster, but without FOR:

len = floor(length(d) / 3600) * 3600;
e   = reshape(d(1:len), 3600, []);
result = sum(e);

[salahaldin]

thanks it works

[NURULAIN OTHMAN]

i want to ask. i have some data from 1 to 24000. then, i grouped them 1 to 1000,1001 to 2000, 2001 to 3000 and so on. then, i want to sum up data 1 with data 1001 with data 2001.. how can i do? Can help me?

[Jan]

@NURULAIN OTHMAN: Please open a new thread for a new question. Then I'd post this answer:

x = rand(1, 24000);
result = sum(reshape(x, 1000, numel(x) / 1000), 1);

Answer 2

[Honglei Chen]

Here is an example that you have 10 elements and you add every two of them.

x = rand(10,1);
sum(reshape(x,2,[])).'

[salahaldin]

thank

Answer 3

[Miroslav Balda]

Let your data be in vector data(1:31536001). Try this code:

k = 0; 
K = 0; 
x = zeros(1,8760); 
while k<8760, 
  k = k+1; 
  x(K) = sum(data(1+K:3600+K)); 
  K = K+3600; 
end

[salahaldin]

thank

Answer 4

[Andrei Bobrov]

data - your vector with size < 31536001 x 1 >;

n = numel(data);
[~,t] = histc(1:n,(0:3600:n) + eps(n));
out = accumarray(t(:),data(:));

or

out = sum(reshape([data;zeros(mod(-n,3600),1)],3600,[])).';

[salahaldin]

thank