How can I do non-linear regression for three varietals?
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Hi All,
I have matrix with three variables (x,y,z) I would like to get best non linear regression for these variables using like this equation: Eq=a*x+b*y+c*z+d
How can I get the constants and correlation coefficient?
Thanks in advance,
Riyadh
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Star Strider
2017년 1월 23일
The equation you posted is linear. Assuming it is a stand-in for a nonlinear equation, the usual way of fitting a function of several variables is to create a matrix of the incependent variables and passing that as one argument to the objective and fitting functions.
Example:
% % % MAPPING: x = xyz(:,1), y = xyz(:,2), z = xyz(:,3), a = b(1), b= B(2), c = b(3), d = b(4)
xyz = [x(:) y(:) z(:)];
Eq = @(b,xyz) b(1).*xyz(:,1) + b(2).*xyz(:,2) + b(3)*zyz(:,3) + b(4);
Then just use them as arguments to whatever fitting function you want (such as nlinfit or lsqcurvefit).
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Star Strider
2017년 1월 25일
My pleasure.
You have described a linear model. I would do something like this:
Prms = [ones(size(SPM(:))), S(:), A(:)]\SPM(:);
a = Prms(1)
b = Prms(2)
c = Prms(3)
The core MATLAB linsolve function and the Statistics and Machine Learning Toolbox regress and glmfit functions (and several others) are also options.
That will work if your matrix is not sparse. If it is sparse, use the lsqr function.
See the documentation for the various functions to understand how to use them.
추가 답변 (1개)
the cyclist
2017년 1월 23일
편집: the cyclist
2017년 1월 23일
Maybe this will help?
% Here is an example of using nlinfit(). For simplicity, none of
% of the fitted parameters are actually nonlinear!
% Define the data to be fit
x = (0:0.25:10)'; % Explanatory variables
y = x.^2;
z = x.^3;
E = 5 + 3*x + 7*y + 11*z; % Response variable (if response were perfect)
E = E + 500*randn((size(x)));% Add some noise to response variable
% Define function that will be used to fit data
% (F is a vector of fitting parameters)
f = @(F,X) F(1) + F(2).*X(:,1) + F(3).*X(:,2) + F(4).*X(:,3);
F_fitted = nlinfit([x y z],E,f,[1 1 1 1]);
% Display fitted coefficients
disp(['F = ',num2str(F_fitted)])
% Plot the data and fit
figure
plot(z,E,'*',z,f(F_fitted,[x y z]),'g');
legend('data','fit','Location','NorthWest')
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the cyclist
2017년 1월 23일
I edited my example, so that it now uses three explanatory variables: x,y,z.
(It is not important that I happened to used x itself to define y and z.)
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