Fast and simple trend

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

Answer 1

[the cyclist]

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.

[Azura Hashim]

Thank you, this worked well!

Answer 2

[Jan]

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