selecting and extracting contiguous points as one event

2 vues (au cours des 30 derniers jours)
Stephen
Stephen le 7 Jan 2019
Commenté : Stephen le 8 Jan 2019
I'm sure there is probably a drop-dead simple way to do this, but my brain is fried at the moment and I can't think of one. I'm trying to identify and count numbers of events that exceed a certain value in a time series. For example, a plot of the attached file yields:Picture1.png
I'd like to be able to count the number of events that exceed the red line (y=0.13). Sometimes there is 1 point > 0.13 (e.g. the first event above at x=40) and that is easy to count as a single signal. But sometimes there are 2 or more contiguous points > 0.13 (e.g. the second event above, at x=56, 57), yet should be counted as just one "event", not as 2 events.
How can this be done? I've attached a file with the above signal.
  1 commentaire
nanren888
nanren888 le 7 Jan 2019
Perhaps you could count the number of upward level-crossings?

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Réponse acceptée

Andrei Bobrov
Andrei Bobrov le 7 Jan 2019
let a - your data - vector.
out = sum(diff([0;a(:) > .13]) == 1);

Plus de réponses (2)

nanren888
nanren888 le 7 Jan 2019
tt = linspace(0,12,1001);
s= sin(tt);
th = 0.4;
nnz(diff(s-th));
dd = [0,diff((s-th)>0)>0.5];
plot(tt,s,tt,dd);
grid('on');
nnz(dd)
ans =
2
fig.png
  1 commentaire
Stephen
Stephen le 7 Jan 2019
Thanks, nanren888! I'm having trouble applying your solution to a matrix of numbers (such as the excel spreadsheet that was attached with my Q) instead of a mathematical function such as the sin wave you show. sorry, I'm not as advanced as you are.

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Image Analyst
Image Analyst le 8 Jan 2019
If you have the Image Processing Toolbox, the simplest way is to just use the function that was made to do exactly that: bwlabel
[~, numRegions] = bwlabel(signal > 0.13);
Can't get any easier than that.
  1 commentaire
Stephen
Stephen le 8 Jan 2019
Thanks, Image Analyst! I will investigate whether I can get the Image Processing Toolbox. In any case, the answer by Andrei Bobrov also worked well.
I appreciate this help.

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