Puzzler: Count unique nonzero periods in a timeseries without a for loop

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given: a signal as a time series
return: discrete number of times is is holds a nonzero position
For example:
given: [0,1,0.5,0.7,0,1]
return: 2
given: [0,1,0,0,0,.3,1,0,0,1]
return: 3
given: [0,1]
return: 1
is there a way to do this task in a vector manner as opposed to a for loop that counts nonzero periods as it encounters them and flicks a counter.

Réponse acceptée

Azzi Abdelmalek
Azzi Abdelmalek le 19 Juil 2016
Modifié(e) : Azzi Abdelmalek le 19 Juil 2016
a=[0,1,0,0,0,.3,1,0,0,1]
out=numel(strfind([0 logical(a)],[0 1]))

Plus de réponses (1)

Image Analyst
Image Analyst le 19 Juil 2016
Yes
[~, numRegions] = bwlabel(yourVector ~= 0);
  1 commentaire
Will Kinsman
Will Kinsman le 19 Juil 2016
sorry dont have imaging processing toolbox. but damn. nice solution.

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