How to find a mean value after each 360 degree?

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Hello everyone,
I urgently need your help. I have a data 50000x5 and my 1 column is degree such as 1.5,1.45,2,1.35 and so on.
I need to calculate a mean value after each 360 degree for whole data but do not know how to do.
If anyone can help I would be very glad.
Thanks in advance.
  4 Comments
darova
darova on 21 May 2020
Here is how i understand oyour question: You want to divide each cycle as following
My question: where is the points of breakup? 180? 360?
Ilkin Abdullayev
Ilkin Abdullayev on 21 May 2020
so, there is no known point for this purposes. Let me to explain.
In my data as you can see one column is the degree and for example if I do it in excel manualy, I sum up first 243 row (example) for angle and obtain 360 degree. Then next 360 degree is between 244 500 for example. The thing is the number of the rows that we need sum up for obtaining 360 degree is not the same always. So what I need, to find a mean value for each 360 degree. I hope now its clear.

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Accepted Answer

darova
darova on 21 May 2020
Try this
A = importdata('meanvalue.csv',',');
%%
B = cumsum(diff(A,2,1));
B(end+1,:) = -inf;
k = 1;
s = 1;
for i = 1:size(B,1)
if B(i,1) < -pi % end of cycle
s = s + 1; % group counter
A(k:i,3) = mean(A(k:i,2)); % calculate mean
A(k:i,4) = s; % assign group number
k = i; % index of next group
end
end
t = A(:,1);
r = cumsum(A(:,2));
[x,y] = pol2cart(t,r);
plot(x,y)
hold on
scatter(x,y,5,A(:,4),'fill')
hold off
  7 Comments
Ilkin Abdullayev
Ilkin Abdullayev on 22 May 2020
Yes, I agree with you. Just take a look picture that I have upload. You can notice from the size of data and also I will leave the result that you can see it does not give the mean just change the values. I mean normally we must have one number as a mean for whole rows but there is no this number.
I obtain the same graph that you obtain and do not change anything even use for the same data but no mean value which must be 1 number.
Ilkin Abdullayev
Ilkin Abdullayev on 22 May 2020
Sorry, I just noticed one thing which is mean value. Thank a lot . I really appreaciate your support

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More Answers (1)

Ilkin Abdullayev
Ilkin Abdullayev on 20 May 2020
I am just leving my data you may have look

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