Fast and simple trend

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Azura Hashim
Azura Hashim 2017-12-17
评论： Azura Hashim 2017-12-18
Hi,
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.
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=repmat(NaN,length(x),1);
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
temptime=time(startrow:row);
tempx=x(startrow:row);
mdl = fitlm(temptime,tempx);
slopes(row)=mdl.Coefficients.Estimate(2);
end
end

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采纳的回答

the cyclist
the cyclist 2017-12-17
编辑：the cyclist 2017-12-17
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.

更多回答（1 个）

Jan
Jan 2017-12-17
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);
end
end
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(:)));
end

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