excel linear regression trouble vs excel linear regression
5 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
I am trying to find the equation for the linear regression line of x and y.
I was able to get the linear regression from excel, but I am trying to find it with matlab.
Excel says the linear regression equation is y = -0.003x + 1.7919.
x = 5.92 22.75 73.26 227.56 308.74 589.54 613.66
y = 2.550865 1.869146 1.1623 0.567358 0.459001 0.248734 0.225807
beta = regress(y',x')
% I had to rotate the arrays otherwise the function would not give an answer
the function spits out beta = 7.9666e-04.
0 commentaires
Réponses (1)
John D'Errico
le 13 Avr 2019
Modifié(e) : John D'Errico
le 13 Avr 2019
Simple enough.
polyfit(x,y,1)
ans =
-0.002965 1.7919
Or:
regress(y',[x',ones(length(x),1)])
ans =
-0.002965
1.7919
Or:
[x',ones(length(x),1)]\y'
ans =
-0.002965
1.7919
I could go on for at least a half dozen others. Don't dare me. ;-) With the curve fitting toolbox...
fit(x',y','poly1')
ans =
Linear model Poly1:
ans(x) = p1*x + p2
Coefficients (with 95% confidence bounds):
p1 = -0.002965 (-0.005117, -0.0008127)
p2 = 1.792 (1.03, 2.554)
>> lsqr([x',ones(length(x),1)],y')
lsqr converged at iteration 2 to a solution with relative residual 0.34.
ans =
-0.002965
1.7919
>> lsqlin([x',ones(length(x),1)],y')
ans =
-0.002965
1.7919
>> lsqminnorm([x',ones(length(x),1)],y')
ans =
-0.002965
1.7919
>> pinv([x',ones(length(x),1)])*y'
ans =
-0.002965
1.7919
>> lscov([x',ones(length(x),1)],y')
ans =
-0.002965
1.7919
Wanna bet I can't come up with a half dozen more? Save your money. :)
0 commentaires
Voir également
Catégories
En savoir plus sur Linear and Nonlinear Regression dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!