moving average of unevenly spaced time data
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Research
le 19 Fév 2018
Commenté : Julie van der Hoop
le 1 Avr 2019
Hello,
I have the following data, and need to average the signal data as per defined window size. Any help?
signal= [0.1 0.2 0.15 0.3]; % volt reading
time_signal = [3.5 5.5 6.5 10.5];% seconds after event, the time when signal was recorded
time_regular = [ *1 2 3 4* 5 6 7 8 *9 10 11 12 13*]; time in seconds
Window_size is 4, 4 and 5. I want to average the signal data that falls within first 4 seconds, then next 4 seconds and finally in next 5 seconds. I highlighted the window sizes in time_regular array.
The answer is 0.1, 0.175 and 0.3. The calculation is done as follows. In first 4 seconds there is only one signal data point that falls within first 4 seconds, therefore first value is 0.1. In next 4 seconds, there are two signal data points (0.2 and 0.15), therefore the average of these two points is 0.175. In next 5 seconds, there is only one signal data point (0.3), therefore the average is 0.3. I can do this using for loop, but I have 100s of window sizes and big data.
Any matlab function can do this trick?
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Andrei Bobrov
le 19 Fév 2018
Modifié(e) : Andrei Bobrov
le 19 Fév 2018
t = randi([3 5],6,1); % your time intervals in seconds
n = cumsum(t);
s = rand(n(end),1);
s = s.*(s < .3); % your signal
ii = zeros(n(end),1);
ii(n - t + 1) = 1;
s(s == 0) = nan;
out = accumarray(cumsum(ii),s,[],@(x)mean(x,'omitnan'));
4 commentaires
Andrei Bobrov
le 19 Fév 2018
Other varint here 't' with don't integers
t = cumsum(.6*rand(40,1) + .5); % time
s = .3*rand(40,1) + .4;
s = s.*(s > .5); % your signal
k = randi([3 6],6,1);
tt = k/sum(k);
ii = [0;cumsum(t(end)*tt)]; % time-intervals boundaries
jj = discretize(t,ii);
s(s == 0) = nan;
out = accumarray(jj(:),s,[],@(x)mean(x,'omitnan'));
Julie van der Hoop
le 1 Avr 2019
Can I ask why the k is it setting unequal time bins? Isn't the goal to have equal time bins across the series (i.e., compute over 4 seconds)?
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