Polynomial Fit from a high Deviation set of feature experiments

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.

[Matt J]

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')

[kuku hello]

Thanks, thats ok the solution doesn't have to be unique. If I have the option to see all of the options it would be great.

Answer 1

[Matt J]

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')'

AllSolutions = 4×2

8 11 5 7 2 3 -1 -1

[kuku hello]

Wow thanks it looks like it works but can I have further information about the role of X? I'll admit this code seems to be high level programming :P.

[Matt J]

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.