Polyfit isn't returning linear gradient and intercept

testx = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

testx = 1×10

1 2 3 4 5 6 7 8 9 10

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testy =[4, 6, 8, 10, 12, 14, 16, 18, 20, 22]

testy = 1×10

4 6 8 10 12 14 16 18 20 22

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[PolyCoeff, S, mu] = polyfit(testx, testy, 1)

PolyCoeff = 1×2

6.0553 13.0000

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S = struct with fields:
           R: [2×2 double]
          df: 8
       normr: 7.9936e-15
    rsquared: 1

mu = 2×1

5.5000 3.0277

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So my question is - why isn't PolyCoeff returning (1) as 2, and (2) as 2? Where are 6 and 13 coming from? am I missing something really silly and obvious?

Answer 1

[dpb]

testx = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

testx = 1×10

1 2 3 4 5 6 7 8 9 10

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testy =[4, 6, 8, 10, 12, 14, 16, 18, 20, 22]

testy = 1×10

4 6 8 10 12 14 16 18 20 22

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[PolyCoeff] = polyfit(testx, testy, 1)

PolyCoeff = 1×2

2.0000 2.0000

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What you're missing is that according to the documentation for polyfit, when you ask for the optional third output of the mean,

[p,S,mu] = polyfit(x,y,n) also returns mu, which is a two-element vector with centering and scaling values. mu(1) is mean(x), and mu(2) is std(x). Using these values, polyfit centers x at zero and scales it to have unit standard deviation,

[Jolene]

Thank you for your answer, I realised shortly after I posted (Sod's law) that the third output argument was causing the issue... however, I don't see why calling this third argument changes the values of the polynomial coefficients, why would it do that?

I see that the documentation states that x is rescaled and normalised to zero, I guess I didn't realise that meant it would also retroactively change the coefficient output. I'll go back through my scripts and remove the mu argument.

Thanks again, appreciate you taking the time for my silly question! I will have to read and understand documentation more closely in future! I took that it was such a basic function for granted!

[Torsten]

testx = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
testy = [4, 6, 8, 10, 12, 14, 16, 18, 20, 22];
m = mean(testx)

m = 5.5000

s = std(testx)

s = 3.0277

[PolyCoeff, S, mu] = polyfit(testx, testy, 1)

PolyCoeff = 1×2

6.0553 13.0000

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S = struct with fields:
           R: [2×2 double]
          df: 8
       normr: 7.9936e-15
    rsquared: 1

mu = 2×1

5.5000 3.0277

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PolyCoeff(1) * (testx - mu(1)) / mu(2) + PolyCoeff(2)

ans = 1×10

4.0000 6.0000 8.0000 10.0000 12.0000 14.0000 16.0000 18.0000 20.0000 22.0000

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testy

testy = 1×10

4 6 8 10 12 14 16 18 20 22

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●

%Recover coefficients of PolyCoeff(1) * testx  + PolyCoeff(2)
PolyCoeff(1)/mu(2)

ans = 2.0000

PolyCoeff(1)*(-mu(1))/mu(2) + PolyCoeff(2)

ans = 2.0000

[Jolene]

This is really clear, thanks so much for taking the time.

[dpb]

"...I didn't realise that meant it would also retroactively change the coefficient output. "

Oh, sorry. I see @Torsten already responded, but I'll add aonther comment.

testx = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
testy = [4, 6, 8, 10, 12, 14, 16, 18, 20, 22];
testz = zscore(testx)                           % standardize the independent variable

testz = 1×10

-1.4863 -1.1560 -0.8257 -0.4954 -0.1651 0.1651 0.4954 0.8257 1.1560 1.4863

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[mean(testz) std(testz)]

ans = 1×2

0 1

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and, indeed, as advertised mean, std deviation.

Now,

polyfit(testz,testy,1)                          % fit vs z instead of vs x

ans = 1×2

6.0553 13.0000

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and you see you get the coefficients returned by polyfit in the three-output case. It is the same as if the coefficients were scaled after fitting y vs x but polyfit actually uses the z-score values as the independent variable to improve numerical precision for higher-order polynomials, This is shown in the Alogorithms section of the documentation.

"I will have to read and understand documentation more closely in future! I took that it was such a basic function for granted!"

<LOL> Yes, Mathworks works very diligently to ensure that computations are as accurate as possible so what seems trivial on the surface may have very in-dpeth implementation consequences.

In the case of polyfit being one of the historical functions that has been included since the very early days, its implementation if beginning today would be unlikely to have this behavior at the user level of changing the algorithm based on the expected return variable(s) but would be implemented with named parameter pairs to control the behavior. But, it remains as is for backward compatibility.