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Efficiently calculating sum-thresholds across vector

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I have a 1xN vector of samples, for which I need to calculate the indices of when a moving sum exceeds a threshold.
One could think of the samples as current that is charging a capacitor, and once the capacitor is charged to the threshold, a pulse is fired and the capacitor is reset. I need to find the times when the pulse is fired.
I have a working loop implementation below, however, it is relatively slow and I'd like to find a more efficient method. Is there a vectorized or built-in method for finding these pulse times?
samples;
threshold = 50;
pulseIdx = zeros(size(samples));
runningSum = 0;
for i = 1:length(samples)
runningSum = runningSum + samples(i);
if runningSum > threshold
pulseIdx(i) = 1;
runningSum = runningSum - threshold; % allow spillover to next cycle
end
end

Accepted Answer

Turlough Hughes
Turlough Hughes on 22 Sep 2020
Edited: Turlough Hughes on 22 Sep 2020
You could do the following but with the caveat that it assumes samples are all positive values
runningSum = cumsum(samples);
idx = find(diff(mod(runningSum,threshold))<0)+1;
Or in the case where they are not all positive, perhaps something like the following:
idx = find(diff(mod(runningSum,50))<-0.5*threshold)+1

More Answers (2)

Ameer Hamza
Ameer Hamza on 22 Sep 2020
Edited: Ameer Hamza on 22 Sep 2020
Try something like this
rng(0);
samples = rand(1, 1000);
threshold = 10;
samples_sum = cumsum(samples);
idx = [];
while any(samples_sum > threshold)
i = find(samples_sum > threshold, 1);
idx = [idx i];
samples_sum = samples_sum - samples_sum(i-1);
end
idx are indexes of samples where sum exceed threshold.

Bruno Luong
Bruno Luong on 22 Sep 2020
Edited: Bruno Luong on 22 Sep 2020
This works regardless the sign of sample
% Test data
samples = 5*rand(1,10000);
samples = samples+0.5*randn(size(samples)); % with noise
% Your method
threshold = 50;
pulseIdx = zeros(size(samples));
runningSum = 0;
for i = 1:length(samples)
runningSum = runningSum + samples(i);
if runningSum > threshold
pulseIdx(i) = 1;
runningSum = runningSum - threshold; % allow spillover to next cycle
end
end
% My method
cs = cummax(cumsum(samples));
n = floor(cs(end)/threshold);
bdr = (1:n)*threshold;
[~,~,loc]=histcounts(bdr,[0,cs]);
pulseIdx2 = zeros(size(samples));
pulseIdx2(loc(loc>0)) = 1;
% This should return 1, excepted when numerical errors spoil thing
isequal(pulseIdx,pulseIdx2)

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