How to fit data to a function form

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Fynn Oppermann
Fynn Oppermann el 19 de En. de 2022
Comentada: Star Strider el 19 de En. de 2022
Apologies if my use of terminology is wrong.
I essentially have four equal sized data sets (eg. x y z w) and I want find an equation that describes one of those data sets as a funtion of the other three. To begin with I'm assuming this is a linear relationship i.e. x = a*y + b*z + c*w.
Is there a way to fit these data sets to find the values of the coefficients a b c?
Thanks for your help!

Respuesta aceptada

Star Strider
Star Strider el 19 de En. de 2022
Yes! The mldivide,\ funciton will do this.
x = randn(10,1);
y = randn(10,1);
z = randn(10,1);
w = randn(10,1);
DM = [y(:) z(:) w(:)]; % Design Matrix
abc = DM \ x(:)
abc = 3×1
-0.0250 -0.2847 -0.4305
Results = table(x(:),DM*abc,x(:)-DM*abc, 'VariableNames',{'Original x','Regressed x','Difference'})
Results = 10×3 table
Original x Regressed x Difference __________ ___________ __________ 0.13267 0.10844 0.024228 -2.5397 -1.2413 -1.2984 0.05946 0.71663 -0.65717 0.82979 1.1043 -0.2745 -0.11678 -0.37399 0.25721 0.42652 -0.55752 0.98404 0.26181 0.62586 -0.36404 1.0023 -0.039926 1.0422 0.57563 -0.16844 0.74407 -0.92174 -0.34643 -0.57531
meanDifference = mean(Results.Difference)
meanDifference = -0.0118
To get statistics on the fit, use regress, fitlm, or similar functions.
.
  2 comentarios
Fynn Oppermann
Fynn Oppermann el 19 de En. de 2022
Thank you!
Would this method still work if my function changes to be nonlinear?
Star Strider
Star Strider el 19 de En. de 2022
My pleasure!
It will only work for linear relations, however coding it for nonlilnear relations would be straightforward and may not involve anythng other than core MATLAB. For a nonlinear regression, one approach would be to use fminsearch if other Toolboxes are not available.
Example —
x = randn(10,1);
y = randn(10,1);
z = randn(10,1);
w = randn(10,1);
yzw = [y(:) z(:) w(:)]; % Contatenate Column Vectors
% % % FUNCTION: exp(a*y) * sin(2*pi*b*z) + c*w
objfcn = @(b,iv) exp(b(1)*iv(:,1)) .* sin(2*pi*b(2).*iv(:,2)) + b(3)*iv(:,3)
objfcn = function_handle with value:
@(b,iv)exp(b(1)*iv(:,1)).*sin(2*pi*b(2).*iv(:,2))+b(3)*iv(:,3)
B = fminsearch(@(b) norm(x(:) - objfcn(b,yzw)), rand(3,1))
B = 3×1
0.3240 1.2573 0.0702
fprintf(1, '\n\ta = %9.4f\n\tb = %9.4f\n\tc = %9.4f\n',B)
a = 0.3240 b = 1.2573 c = 0.0702
Results = table(x(:), objfcn(B,yzw), x(:)-objfcn(B,yzw), 'VariableNames',{'Original x','Regressed x','Difference'})
Results = 10×3 table
Original x Regressed x Difference __________ ___________ __________ 1.1085 0.83031 0.27821 0.90088 0.47772 0.42316 -0.67653 0.63336 -1.3099 0.0039623 -0.88416 0.88813 1.1665 0.61915 0.54735 0.80068 0.90119 -0.10051 0.12086 -0.39058 0.51144 -0.59784 0.17719 -0.77503 0.88238 0.58316 0.29923 0.033881 -0.74814 0.78202
Here, ‘iv’ is the independent variable matrix, composed of the independent variable vectors. This allows a single variable to be passed to any of the curve-fitting (parameter estimation) functions, as their syntax requires, while fitting every independent variable.
Since there are only three parameters, the fminsearch function can likely do a decent approximaton to the desired parameter estimates.
.

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