how to find cycle in data

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

Answer 1

[Walter Roberson]

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)

[Max Bernstein]

Thanks for your help. I tried it but I get the error

??? Error using ==> strfind

Input strings must have one row.

[Walter Roberson]

Ah, columns... I'm surprised the horzcat() didn't complain.

logind = [false; data(:,2)>=70 & data(:,2) <= 150; false] .';

[Max Bernstein]

Thank you, this works with small modification. It needs to use the "findstr" function in order to work. Thanks a lot for your help!

[Christine Nee]

What did you add that made it work? Im returning no values

[Walter Roberson]

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.

Answer 2

[Andrei Bobrov]

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)),:)

[Max Bernstein]

Thank you for your help. This method works but if possible, I would prefer to not having to make more arrays. The data files I have will be around 1-2GB and there's not enough memory on the computer to process all the arrays.

[Walter Roberson]

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.

[Max Bernstein]

This method works too. thanks!