Fast and simple trend

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Azura Hashim
Azura Hashim on 17 Dec 2017
Commented: Azura Hashim on 18 Dec 2017
I need a fast and simple way to calculate the trend of a variable at each point in time for data over the preceding 1 hour. All I need is whether the trend is increasing or decreasing and to what degree. At the moment I am using fitlm to return the slope for each row but have found this to be too slow. Below is a simple example but my application has a much bigger dataset and I need at least an order of magnitude speedup. Appreciate some help please, especially if there are ways to vectorize the calculation. Thank you.
for row=1:length(x)
startrow=find(time >= time(row)-1,1);
%calculate slope if there are more 2 or more data points
if row > startrow
mdl = fitlm(temptime,tempx);

Accepted Answer

the cyclist
the cyclist on 17 Dec 2017
Edited: the cyclist on 17 Dec 2017
You can do the fit directly with matrix operations. It should be roughly a gazillion times faster.
coeffs = [ones(size(temptime')) temptime']\tempx';
slopes(row) = coeffs(2);
There are presumably other efficiencies if you restructure your data ahead such that you do not need to do the transposes, or create the "ones" matrix inside the loop.
  1 Comment
Azura Hashim
Azura Hashim on 18 Dec 2017
Thank you, this worked well!

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More Answers (1)

Jan on 17 Dec 2017
What about the faster polyfit:
time = [0.2,0.8,0.9,1.1,1.2,1.7,1.8,2.1,2.2];
x = [0.2,0.4,0.5,0.7,1.1,0.7,0.6,1.7,2.1];
slopes = NaN(length(x), 1);
for row = 1:length(x)
startrow = find(time >= time(row)-1,1);
if row > startrow
P = polyfit(time(startrow:row), x(startrow:row), 1);
slopes(row) = P(1);
If this is still too slow, use a leaner version of polyfit:
function p = LeanPolyFit1(x, y)
V = [x(:), ones(numel(x), 1)];
% Solve least squares problem:
[Q, R] = qr(V, 0);
p = transpose(R \ (transpose(Q) * y(:)));

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