Find Function in Matlab

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Karoline Qasem
Karoline Qasem 2016 年 10 月 23 日
コメント済み: Karoline Qasem 2016 年 10 月 24 日
hello,
I have say:
X=[0 1 0 0 2 0 0 0 0 0 4];
Y=[5 2 5 5 1 3 2 5 5 5 5];
somehow, Y is depending on X and whenever X is greater than 0, Y drops down. I am interested in seeing how many time steps does it take Y to go to its average value after each peak of X. In other words, I need to search for when is Y(i-1) = Y(i+a)+-10%
where i-1 is the day before X peak.
Example:
the first peak of X is 1
the day before it is where Y was 5
it looks like it took Y one time step to go back to this 5 after the peak of X
I hope I am clear in the question and really need help.
Thanks

採用された回答

Guillaume
Guillaume 2016 年 10 月 23 日
This should work:
xpeaks = find(X);
assert(xpeaks(1) ~= 1, 'Peak on first value, can''t go back before peak');
delay = arrayfun(@(peakloc) find(abs(Y(peakloc:end) - Y(peakloc-1)) <= Y(peakloc-1)/10, 1) - 1, xpeaks)
  3 件のコメント
Guillaume
Guillaume 2016 年 10 月 24 日
I would suspect that it is because Y never gets to within 10% of value just before the 12th peak, therefore find returns empty.
As per the error message set 'UniformOutput' to false in arrayfun so that it creates a cell array output (in order to be able to contain empty elements:
delay = arrayfun(@(peakloc) find(abs(Y(peakloc:end) - Y(peakloc-1)) <= Y(peakloc-1)/10, 1) - 1, ...
xpeaks, 'UniformOutput', false);
If you want, you can convert those empty elements into NaNs to get a matrix output:
delay(cellfun(@isempty, delay)) = {NaN}; %replace empty by NaN
delay = cell2mat(delay);
Karoline Qasem
Karoline Qasem 2016 年 10 月 24 日
Thank you very much

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その他の回答 (1 件)

Image Analyst
Image Analyst 2016 年 10 月 23 日
Try this:
X = [0 1 0 0 2 0 0 0 0 0 4]; % Not used.
Y = [5 2 5 5 1 3 2 5 5 5 5] % Our data.
% Find non-5 locations.
binaryVector = Y ~= 5
% Give an ID number to each non-5 region.
labeledVector = bwlabel(binaryVector);
% Measure the lengths of those regions.
% Requires the Image Processing Toolbox.
props = regionprops(labeledVector, 'Area');
% Compute the number of steps to return to 5
% once it has dropped down to a lower value:
numSteps = [props.Area]
% Get the mean of those numbers
meanStepCount = mean(numSteps)
You'll see:
Y =
5 2 5 5 1 3 2 5 5 5 5
binaryVector =
1×11 logical array
0 1 0 0 1 1 1 0 0 0 0
numSteps =
1 3
meanStepCount =
2
  5 件のコメント
Image Analyst
Image Analyst 2016 年 10 月 23 日
You can compute the difference from the prior value with diff
diffy = diff(y);
I'm not sure what value it must achieve until it's restored again. Is it the value right before it dipped?
Karoline Qasem
Karoline Qasem 2016 年 10 月 23 日
yes, i am looking for the value close to the one prior to the peak. I am sorry for not being clear, just my data is very noisy and hard to deal with since i am beginner to matlab.

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