Am I using corrcoef correctly?

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Oliver Lundblad
Oliver Lundblad am 18 Sep. 2021
Kommentiert: Star Strider am 18 Sep. 2021
I want to find R^2 for my polyfit approximation and I read on this form that corrcoef can be used to find R^2. I am however not sure if I have the correct inputs or if I interpret the output correctly? Can someone confirm if I have correctly used corrcoef in this code and that R^2 equals 0.6327 for my polyfit approximation?
H=log([0.01 0.02 0.03 0.04 0.05]);
T=log([0.645 0.647 0.646 0.646 0.647]);
p=polyfit(H,T,1);%derivative=0.0013
f=polyval(p,H);
plot (H,T,'o',H,f,'-');
%legend('data','linear fit')
axis([-4.8 -2.8 -0.45 -0.42])
xlabel("ln(h/m)")
ylabel ("ln(T/s)")
grid on
[R1,P1] = corrcoef(H,T) %R^2=0.6327

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Star Strider
Star Strider am 18 Sep. 2021
Am I using corrcoef correctly?
Yes, except for squaring the result, and that will produce .
H=log([0.01 0.02 0.03 0.04 0.05]);
T=log([0.645 0.647 0.646 0.646 0.647]);
p=polyfit(H,T,1);%derivative=0.0013
f=polyval(p,H);
plot (H,T,'o',H,f,'-');
%legend('data','linear fit')
axis([-4.8 -2.8 -0.45 -0.42])
xlabel("ln(h/m)")
ylabel ("ln(T/s)")
grid on
[R1,P1] = corrcoef(H,T) %R^2=0.6327
R1 = 2×2
1.0000 0.6327 0.6327 1.0000
P1 = 2×2
1.0000 0.2520 0.2520 1.0000
Rsq = R1.^2
ans = 2×2
1.0000 0.4003 0.4003 1.0000
And fitlm agrees —
mdl = fitlm(H,T)
mdl =
Linear regression model: y ~ 1 + x1 Estimated Coefficients: Estimate SE tStat pValue _________ __________ _______ __________ (Intercept) -0.43194 0.0033633 -128.43 1.0409e-06 x1 0.0012893 0.00091105 1.4152 0.25196 Number of observations: 5, Error degrees of freedom: 3 Root Mean Squared Error: 0.00116 R-squared: 0.4, Adjusted R-Squared: 0.2 F-statistic vs. constant model: 2, p-value = 0.252
.
  4 Kommentare
Oliver Lundblad
Oliver Lundblad am 18 Sep. 2021
It works now. thanks for telling me which product to install.
Star Strider
Star Strider am 18 Sep. 2021
As always, my pleasure!
.

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