I have to create a figure every 80 measurements of a row vector with length 31999 (Bn in the code). I tried to write this code but i receive only one figure with all the measurements (31999):
k= length(Bn);
for i= 1:80:(k-1)
Bn1(i) = L(i);
plot(Bn1);
end
any suggestion?

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Star Strider
Star Strider 2018년 11월 5일

1 개 추천

Try this:
repts = fix(numel(Bn)/80);
for k1 = 1:repts
figure(k1)
idxrng = (1:80) + 80*(k1-1);
plot(Bn(idxrng), L(idxrng))
grid
end

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Elisa Mammoliti
Elisa Mammoliti 2018년 11월 5일
thank you Star Rider, it works and it gives what i need. Also really thank you to mahdan ravi for the support!!!
Star Strider
Star Strider 2018년 11월 5일
My pleasure.
If my Answer helped you solve your problem, please Accept it!
Something like this could work, creating a cell array of the findchangepts outputs:
repts = fix(numel(Bn)/80);
for k1 = 1:repts
figure(k1)
idxrng = (1:80) + 80*(k1-1);
ipt{k1} = findchangepts(L(idxrng));
plot(Bn(idxrng), L(idxrng))
grid
end
Note: The ‘ipt’ indices would be with respect to each section of ‘L(idxrng)’ not the entire ‘L’ vector. To create that, you would have to add the ‘80*(k1-1)’ offset to each set. If you want to do that, create a second set of vectors with those offsets, then concatenate them into one vector of serial ‘ipt’ values.

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