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Hello,

Is there a way to show the R^2 value when fitting a curve to data with polyfit? I know if I have two outputs, I get a structure

P =

R: [2x2 double]

df: 3

normr: 0.1782

EDU>> P.R

ans =

-4.3248 -2.1613

0 0.5734

but I don't know how to interpret this. I just want to know what the R^2 value is from a least squares fit. What are the numbers in the array of P.R? Thanks!

Orion
on 20 Oct 2014

Hi,

Did you read the help of polyfit ?

[p,S] = polyfit(x,y,n) returns the polynomial coefficients p and a structure S for use with polyval to obtain error estimates or predictions. Structure S contains fields R, df, and normr, for the triangular factor from a QR decomposition of the Vandermonde matrix of x, the degrees of freedom, and the norm of the residuals, respectively. If the data y are random, an estimate of the covariance matrix of p is (Rinv*Rinv')*normr^2/df, where Rinv is the inverse of R. If the errors in the data y are independent normal with constant variance, polyval produces error bounds that contain at least 50% of the predictions.

in your case, P must be the 2nd output argument.

Star Strider
on 20 Oct 2014

hoppingbuffalo
on 7 Nov 2018

Good answer except it's the corrcoef function. I just want to add if your data is two column vectors then the off-diagonal elements of the 2x2 matrix corrcoef returns is what we conventionally think of as the correlation coefficient. That off-diagonal element squared is R^2.

Both polyfit and corrcoef are order N algorithms so both run very fast. Legendre and Gauss performed fitting by hand circa 1800. You can run polyfit and corrcoef one right after the other. If you don't want to use corrcoef you have to do a little extra work to get the output structure of polyfit to a correlation coefficient. The corrcoef documentation shows how to connect the covariance matrix to the correlation coefficients.

I'm a bit surprised that Mathworks doesn't have polyfit output the correlation coefficient matrix. Most people want the correlation coefficient and not the QR decomposition of the Vandermonde matrix of x.

Sarah
on 29 Nov 2018

Hello ,

can someone confirm, is R here (if squared) the regression coefficient of the fit polynomial?

John D'Errico
on 29 Nov 2018

NO. NO. NO. R can be used to obtain an approximate (estimated) covariance matrix of the parameters, althought it is NOT the covariace matrix itself. And R is definitely NOT the regression coefficients.

x = rand(10,1);y = rand(10,1);

[P,S] = polyfit(x,y,1)

P =

0.119244718933779 0.392023572442104

S =

struct with fields:

R: [2×2 double]

df: 8

normr: 0.808027325351178

The regression coefficients are contained in the first returned argument, here P.

S.R is a 2x2 upper triangular matrix, that contains information about the uncertainty in the model parameters as estimated.

Squaring the matrix S.R will not give you the frequently bandied about parameter R^2 either! This will work, in case you want that number:

1 - (S.normr/norm(y - mean(y)))^2

ans =

0.022268658204696

which is the correct value of R^2 for this problem.

Sarah
on 4 Dec 2018

John D'Errico
on 4 Dec 2018

Yes. That expression will work to produce R^2. As long as your model has a constant term in it, and all models that polyfit would produce have a constant term, so that point is a given.

As a check, I'll compare that expression to what I get from my own code, polyfitn.

x = rand(10,1);y = rand(10,1);

[P,S] = polyfit(x,y,1)

P =

0.269646796107357 0.322887986613605

S =

struct with fields:

R: [2×2 double]

df: 8

normr: 0.920139196035294

1 - (S.normr/norm(y - mean(y)))^2

ans =

0.0877303425800532

mdl = polyfitn(x,y,1)

mdl =

struct with fields:

ModelTerms: [2×1 double]

Coefficients: [0.269646796107357 0.322887986613605]

ParameterVar: [0.0945091695169763 0.0448722642822286]

ParameterStd: [0.307423436837493 0.211830744421646]

DoF: 8

p: [0.405990496340959 0.165946560438871]

R2: 0.0877303425800532

AdjustedR2: -0.0263033645974402

RMSE: 0.29097356238677

VarNames: {'X1'}

But you can also find the same expression for R^2 if you look online, Wikipedia for example, with have the same expression.

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