How to get R^2 of a 5th order polyfit?

I need to find how poorly the 5th order fit is for:

            [p,S] = polyfit(x,y0,5);
            yp = polyval(p,x);

An R^2 would be perfect, but I can not understand the answers I'm finding. I don't think I want correlations.

R^2.

Thanks!

Answer 1

[John D'Errico]

polyfit does not return an R^2. A good idea, because IMHO, R^2 is of little value in determining if the fit is good. Does the fit look good? If not, then who cares what R^2 tells you? And if the fit looks like crap, then again, do you need R^2? Reliance on a single number is a bad idea. But people demand it.

You can download my polyfitn from the file exchange , which does compute an R^2.

https://www.mathworks.com/matlabcentral/fileexchange/34765-polyfitn

x = linspace(0,1,50);
y = exp(x) + randn(size(x))/100;

P5 = polyfitn(x,y,5)
P5 = 
  struct with fields:

        ModelTerms: [6×1 double]
      Coefficients: [-0.34052 0.37377 0.44485 0.11449 1.1189 0.99134]
      ParameterVar: [1.223 7.7232 6.3053 0.95902 0.023737 5.4246e-05]
      ParameterStd: [1.1059 2.7791 2.511 0.97929 0.15407 0.0073652]
               DoF: 44
                 p: [0.7596 0.89362 0.8602 0.90746 4.7259e-09 3.4139e-59]
                R2: 0.99963
        AdjustedR2: 0.99959
              RMSE: 0.0096032
          VarNames: {'X1'}

P5.R2
ans =
      0.99963

plot(x,y,'ro',x,polyvaln(P5,x),'b-')

If you have the curve fitting toolbox, this would have worked as well:

[mdl,stuff] = fit(x',y','poly5')
mdl = 
     Linear model Poly5:
     mdl(x) = p1*x^5 + p2*x^4 + p3*x^3 + p4*x^2 + p5*x + p6
     Coefficients (with 95% confidence bounds):
       p1 =     -0.3405  (-2.569, 1.888)
       p2 =      0.3738  (-5.227, 5.975)
       p3 =      0.4448  (-4.616, 5.506)
       p4 =      0.1145  (-1.859, 2.088)
       p5 =       1.119  (0.8084, 1.429)
       p6 =      0.9913  (0.9765, 1.006)
stuff = 
  struct with fields:

             sse: 0.0046111
         rsquare: 0.99963
             dfe: 44
      adjrsquare: 0.99959
            rmse: 0.010237

Finally, with a little more effort, you could have used the stats toolbox. You would need to do a bit more to use regress and regstats however. But you would get more complete information about the model then too.

[David Pesetsky]

r_sqr=power(corr2(yp,y0),2) gives the same value as the adj R^2 from fit. So that one is good enough.

Humans aren't around to lay eyes on hundreds of fits. We want to kick out only the "bad-ish" ones :)