how to find cycle in data
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Max Bernstein
am 30 Sep. 2011
Kommentiert: Walter Roberson
am 15 Jul. 2017
Hello, I have a set of data points similar to one below named "data". The first column is time, second column is voltage. One voltage cycles start at V>=70 and ends at V<=70, and can varies from 70-150. There are two cycles in this example. How can I find the indices where the cycle starts and ends? I tried to use the the find command but it returns all indices of the values w/in that range, not where it starts or ends.
Thanks!
cycle=find(data(:,2)>=70 & data(:,2) <= 150)
0 0
0.001 0
0.002 10
0.003 60
0.004 70
0.005 75
0.006 150
0.007 72
0.008 150
0.009 100
0.01 130
0.011 100
0.012 90
0.013 70
0.014 50
0.015 6
0.016 0
0.017 0
0.018 60
0.019 10
0.02 77
0.021 92
0.022 100
0.023 130
0.024 120
0.025 85
0.026 65
0.027 20
0.028 0
0.029 0
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Akzeptierte Antwort
Walter Roberson
am 30 Sep. 2011
logind = [false, data(:,2)>=70 & data(:,2) <= 150, false];
cyclestart = strfind([false true], logind);
cycleend = strfind([true false], logind) - 1;
cyclestart(K) will be the index of the beginning of cycle #K, and cycleend(K) will be the index of the end of cycle #K. The code protects against boundary conditions (cycle starts at beginning of data or ends at the end of data)
5 Kommentare
Walter Roberson
am 15 Jul. 2017
data = [ 0 0
0.001 0
0.002 10
0.003 60
0.004 70
0.005 75
0.006 150
0.007 72
0.008 150
0.009 100
0.01 130
0.011 100
0.012 90
0.013 70
0.014 50
0.015 6
0.016 0
0.017 0
0.018 60
0.019 10
0.02 77
0.021 92
0.022 100
0.023 130
0.024 120
0.025 85
0.026 65
0.027 20
0.028 0
0.029 0];
logind = [false; data(:,2)>=70 & data(:,2) <= 150; false] .';
cyclestart = strfind(logind, [false true]);
cycleend = strfind(logind, [true false]) - 1;
This code has a small difference compared to the original specification: it does not consider the cycle ended until the data is below 70, whereas the original specification said that it ended at <= 70. With the original specification, the 70 in row 5 would have to be considered to both start and end a cycle, and then the next entry would have to be considered to be a different cycle, unless a further restriction was added that cycles are to be at least length 2.
Weitere Antworten (1)
Andrei Bobrov
am 30 Sep. 2011
more variant
d1 = data(:,2)>=70 & data(:,2) <= 150;
b = find([true;diff(d1)~=0;true]);
b2 = [b(1:end-1),b(2:end)-1];
out = b2(d1(b(1:end-1)),:)
3 Kommentare
Walter Roberson
am 30 Sep. 2011
d1 would be the only large array explicitly generated: Andrei's b and b2 and out arrays would have only as many entries as there are cycles.
However, data(:,2)>=70 & data(:,2) <= 150 is going to construct a number of intermediate arrays that are going to be large.
If I recall correctly, someone recently added a File Exchange contribution written in MEX in order to do fast find of the elements within a given range. Unfortunately, I do not recall whom now, and when I poke around the File Exchange, I do not come up with anything. Perhaps someone else will remember.
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