Curve fitting f(x,y) result
6 views (last 30 days)
Show older comments
Hi,
I used Curve fitting tool to get the equation of the regression line of my area. So I have some X and Y which give some Z.
The curve fitting tool provide me a linear model like this :
Linear model Poly54:
f(x,y) = p00 + p10*x + p01*y + p20*x^2 + p11*x*y + p02*y^2 + p30*x^3 + p21*x^2*y+ p12*x*y^2 + p03*y^3 + p40*x^4 + p31*x^3*y + p22*x^2*y^2+ p13*x*y^3 + p04*y^4 + p50*x^5 + p41*x^4*y + p32*x^3*y^2+ p23*x^2*y^3 + p14*x*y^4
But when I try to obtain a Z value from this equation it's not good at all. For example with X = 60 and Y = 60 , Z = 300 but with the formula I have a result like xxxE+14
Actually, I'm looking for an equation which can provide me a Z value whatever X and Y such as f(x,y) = z The area looks like this : http://www.hostingpics.net/viewer.php?id=329914example.jpg
Thank you if you can help
0 Comments
Answers (7)
Richard Willey
on 20 Apr 2012
I'm attaching code that shows a couple different ways to solve your problem.
I prefer the second option. The R^2 is slightly better and the model is simplier. (Note that option 2 as coded uses Statistics Toolbox rather than Curve Fitting Toolbox)
%%Input some data
x = [0 10 20 30 35 40 45 50 55 60];
y = [60 120 180 240 300];
z = [299.8 299.1 296.9 293.4 291.4 289.9 288.5 283.3 276.4 272.1;
299.8 299.1 296.9 293.4 291.4 289.9 288.5 283.3 276.4 272.1;
299.8 299.1 296.9 293.4 291.4 289.9 288.5 283.3 276.4 272.1;
299.8 299.1 296.9 293.4 291.4 289.9 288.5 283.3 276.4 272.1;
299.8 299.1 296.9 293.4 291.4 289.9 288.5 283.3 276.4 272.1];
% Observe that the data is constant in one dimension.
% The variable Y has no predictive power
% This means we can treat this as a curve fitting problem rather
% than a surface fitting problem
x = x';
z = mean(z, 1)';
% Visualize our data
scatter(x,z)
%%Generate a fit
% Option one
[foo, GoF] = fit(x,z, 'poly2')
hold on
plot(foo)
%%Option two: Fit a piecewise linear model
% Looking at the data, it appears as if we might generate a better fit by
% assuming a piecewise linear model with a "kink" at the 7th data point
% Start by centering the data such that the 7th data point is equal to zero
New_x = x - 45;
% Create a dummy variable that flags whether we are before or after the 7th
% data point
dummy = New_x > 0
% Create an interaction term
interaction = New_x .* dummy
% Generate our regression model
X = [ones(size(x)), New_x, interaction];
b = regress(z, X)
plot(x, X*b)
2 Comments
Ibrahim Maassarani
on 28 Feb 2019
Is b the function?
I got:
b =
1.0e+03 *
7.6023
0.0043
-0.0144
How should I read this?!
Walter Roberson
on 28 Feb 2019
You should give the command
format long g
and then you should display b again.
Richard Willey
on 20 Apr 2012
Hi there
The "Poly54" fit type is specifying a 5th order polynomial in one direction and a 4th order polynomial in the other.
From the looks of things, you have roughly 40 data points. Your model has 20 terms (many of which are high order). I suspect that you're overfitting like crazy, which means that your model can perform very poorly if used to generate a prediction for out of sample data.
Based on the chart that you provided, you can probably cobble together a simple custom equation with a linear model in one direction and a cubic in the other. Something like
Y = b0 + b1*X1 + b2*X2^3
0 Comments
Richard Willey
on 20 Apr 2012
Hi ZazOufUMl
A "Poly31" model is different than the one that I suggested. (Poly31 will contain a number of cross terms).
Any chance that you can post your raw X, Y, and Z data?
1 Comment
Walter Roberson
on 20 Apr 2012
http://www.mathworks.com/matlabcentral/answers/7924-where-can-i-upload-images-and-files-for-use-on-matlab-answers
See Also
Categories
Find more on Linear and Nonlinear Regression in Help Center and File Exchange
Products
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!