Non-linear curvefitting in MATLAB

2013 10 8
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更新时间：2013 10 9
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Hey guys! I'm some given some huge set of data. I am trying to fit a set of data into a model of functional form as described below:
z(x, y) = c0. * x^0 * y^2 + c1. * x^1 * y^1 + c2. * x^2 *y^1
where c0, c1, c2 are the coefficients to be found.
My attempt is to use the nlinfit function to solve it.
So far I have tried:
% i have just added a small portion of my data
a= [ 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001,0.011, 0.011, 0.011, 0.011, 0.011, 0.011, 0.011, 0.011, 0.011, 0.011];
x = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
y = x.* a;
z = [ -.304860225, .170315374, .343019354, .370114906, .373180536, .36719579, .363397853, .363417755, .366962504, .379710865, -.304860225, .170315374, .343019354, .370114906, .373180536, .36719579, .363397853, .363417755, .366962504, .379710865];
model= c0.* (x(:).^0).* (y(:).^2) + c1.* (x(:).^1).* (y(:).^1) + c2.* (x(:).^2).* (y(:).^0)
[c0 c1 c2] = [0.001 0.007 0.788]
C= nlinfit( [x,y], z, 'model', [0.001 0.007 0.788])
% Here x,y are independent variables and z is dependent variable.
How can one set these initial values for the coefficients? I'm not getting how to pass the arguments. I'm getting this error "??? Undefined function or variable 'c0' ". Please help!!!
Thanks in advance, Syeda

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 采纳的回答

Greg Heath
Greg Heath 2013-10-9

0 个投票

The solution is trivial because you have a linear system of equations for the 3 coefficients
A*c = b;
c = A\b
Hope this helps
Thank you for formally accepting my answer
Greg

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Syeda
Syeda 2013-10-9
Thankyou Greg! Now, if I put these three co-efficients back into my proposed model. and re-check for random x,y values again. It gives error:
x=1;
y=0.001;
z = 0 * x^0 * y^2 + 0.0000 * x^1 * y^1 + 0.0056 * x^0 * y^2
Any advice?
Matt J
Matt J 2013-10-9
编辑：Matt J 2013-10-9
It's really not ideal to fit polynomials bluntly using backslash. That's why MATLAB offers the more robust POLYFIT and why the File Exchange offers a variety of polynomial fitters.
Syeda
Syeda 2013-10-9
yes!! My results are not accurate by using this method.

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更多回答（1 个）

Matt J
Matt J 2013-10-8

1 个投票

Using a nonlinear solver for a linear fitting problem seems like the wrong way to go. A better option might be
http://www.mathworks.com/matlabcentral/fileexchange/34765-polyfitn

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Syeda
Syeda 2013-10-8
Thankyou for replying Matt.
I tried John D'Errico's polyfitn()as well. But, how can I fit my proposed model in this function?
model = [ ( x(:).^0 ).*( y(:).^2 ), ( x(:).^1 ).*( y(:).^1 ), ( x(:).^2 ).*( y(:).^0 ) ]
p = polyfitn([x,y], z, 'model');
It gives an error:
??? Error using ==> polyfitn at 137 indepvar and depvar are of inconsistent sizes.
Matt J
Matt J 2013-10-8
I've never used the tool myself. Perhaps you need to columnize z,
p = polyfitn([x,y], z(:), 'model');
Syeda
Syeda 2013-10-8
It still gives the same error.
Anyways, thankyou for replying!!
Matt J
Matt J 2013-10-8
Maybe this
p = polyfitn([x(:),y(:)], z(:), {'y', 'x*y', 'x^2*y'});
Syeda
Syeda 2013-10-8
编辑：Syeda 2013-10-8
This works for me
p = polyfitn([x(:),y(:)], z(:), 'model')
But, it is giving only one coefficient. I want 3 coefficients..
Syeda
Syeda 2013-10-8
编辑：Syeda 2013-10-8
I tried both equations, but it gives error!
p = polyfitn([x(:),y(:)], z(:), {'y', 'x*y', 'x^2*y'});
p = polyfitn([x(:),y(:)], z(:), {'x*y^2', 'x*y', 'x^2*y'});
??? Error using ==> polyfitn>parsemodel at 319 Variable is not a valid name: '2'
Error in ==> polyfitn at 151 [modelterms,varlist] = parsemodel(modelterms,p);
Matt J
Matt J 2013-10-8
I can't see why the syntax
p = polyfitn([x(:),y(:)], z(:), 'model')
would work. And it is obviously not working, since it gives you the wrong number of coefficients.
Syeda
Syeda 2013-10-8
yes! it is not working as well. I guess there is some syntax problem in my code.
Matt J
Matt J 2013-10-8
Dunno. Instead of 'x^2*y' maybe you should try either 'x*x*y' or 'y*x^2'.
Syeda
Syeda 2013-10-9
No, it still gives error.
??? Undefined function or method 'polyfitn' for input arguments of type 'cell'

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