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.
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.
