sum function and add using loop

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salahaldin
salahaldin 2013-2-25
评论： Jan 2016-11-30
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
  1 个评论
Jan
Jan 2013-2-26
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?

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回答（4 个）

Jan
Jan 2013-2-26
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);
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NURULAIN OTHMAN
NURULAIN OTHMAN 2016-11-29
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
Jan 2016-11-30
@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);

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Honglei Chen
Honglei Chen 2013-2-25
Here is an example that you have 10 elements and you add every two of them.
x = rand(10,1);
sum(reshape(x,2,[])).'
Miroslav Balda
Miroslav Balda 2013-2-26
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
Andrei Bobrov
Andrei Bobrov 2013-2-26
编辑：Andrei Bobrov 2013-2-26
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,[])).';

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