Polynomial Fit from a high Deviation set of feature experiments
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I am doing a number of experiments to evaluate a feature in different points in time, the result of each experiment have a high error value or a big standard deviation. I know that by taking a point from each experiment I should be able to create a linear fit. So I need to create an algorithm which is able to pick the right feature evaluations in order to form a linear fit with minimum error (with the least mean square error from the line as possible).
For example I have the following experiments:
exp1 = [10,9,11,8];
exp2 = [5,7,5,9];
exp3 = [2,4,3,2];
exp4 = [-4,-2,-3,-1];
I want to pick one point (single feature evaluation) from each experiment, and from the points taken from every experiment, to make a linear fit with minimum deviation as possible.
So in the example case In order to create a line without any deviation I will choose the set of points [8,5,2,-1]. Each point is measured in different experiment and doing a linear fit I can make a perfect line of feature vs experiment number. If I'll choose point 11 from exp1 and -4 from exp4 the line would have a higher deviation (MSE) from the line chosen above.
Any help?
Thanks in advance.
2 Comments
Matt J
on 3 May 2022
So in the example case In order to create a line without any deviation I will choose the set of points [8,5,2,-1].
I hope you know that the solution is not unique. Another solution here is [11,7,3,-1],
plot([11,7,3,-1],'-o')
Accepted Answer
Matt J
on 3 May 2022
Edited: Matt J
on 3 May 2022
expData = {[10,9,11,8];
[5,7,5,9];
[2,4,3,2];
[-4,-2,-3,-1]};
[Y{1:4}]=ndgrid(expData{:});
N=numel(Y);
Y=reshape( cat(N+1,Y{:}) ,[],N)';
X=[1:N;ones(1,N)]';
SE=vecnorm(X*(X\Y)-Y,2,1);%L2 error
minSE=min( SE ); %min. L2 error
AllSolutions=unique(Y(:,abs(SE-minSE)<=1e-10*minSE)','rows')'
2 Comments
Matt J
on 3 May 2022
Edited: Matt J
on 3 May 2022
The equation for a line is y=m*x+b, which can be written in matrix multiplication form,
[x,1]*[m;b]=y
If you have many x(i) and y(i), this becomes,
[x1,1;
x2,1;
...
xN,1]*[m;b]=[y1;y2;y3;...;yN]
The Nx2 matrix on the left hand side is X, the Nx1 vector on the right hand side is Y, and the solution for [m;b] is X\Y.
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