fitting 2d data set fit function is not working

2 vues (au cours des 30 derniers jours)
Asliddin Komilov
Asliddin Komilov le 1 Mar 2020
Hi,
this is my data set and I wanted to try to get a fit for it.
well, did not work with
fit([x,y],z,'poly23');
first it said: Dimensions of matrices being concatenated are not consistent.
then I made x and y the same length and
it said: Y must be a column vector.
And now I am stack. help please.
thanks

Réponse acceptée

Thiago Henrique Gomes Lobato
Modifié(e) : Thiago Henrique Gomes Lobato le 3 Mar 2020
The problem is that your z data is defined in a grid while your x and y define only the vectors of this grid. If you first actually create the grid you will be able to create the model
[xmesh,ymesh] = meshgrid(x,y);
a = fit([xmesh(:),ymesh(:)],z(:),'poly23');
figure,surf(xmesh,ymesh,z),shading interp
hold on
plot3(xmesh(:),ymesh(:), a([xmesh(:),ymesh(:)]),'*' )
  2 commentaires
Asliddin Komilov
Asliddin Komilov le 2 Mar 2020
Modifié(e) : Asliddin Komilov le 3 Mar 2020
thanks, meshgrid is new to me, I will do some reading.
what is the formula for the surface, why this one is not working:
axTc = a(1) + a(2) *x + a(3) *y + a(4) *x^2 + a(5) *x*y + a(6) *y^2 + a(7) *x^2*y + a(8) *x*y^2 + a(9) *y^3;
Thiago Henrique Gomes Lobato
a is your linear model, but you can't acess the coefficients as a(x), but rather: a.p00, a.p01,etc... You can check the equation by looking at the output of the model
a
Linear model Poly23:
a(x,y) = p00 + p10*x + p01*y + p20*x^2 + p11*x*y + p02*y^2 + p21*x^2*y +
p12*x*y^2 + p03*y^3
Coefficients (with 95% confidence bounds):
p00 = 0.4365 (0.405, 0.4679)
p10 = 0.02601 (0.01672, 0.0353)
p01 = -0.0005043 (-0.0008066, -0.000202)
p20 = -0.09332 (-0.09619, -0.09046)
p11 = 0.0002262 (0.0001687, 0.0002837)
p02 = 1.986e-08 (-9.467e-07, 9.865e-07)
p21 = -7.72e-07 (-9.884e-06, 8.34e-06)
p12 = -2.696e-08 (-1.176e-07, 6.368e-08)
p03 = 8.97e-12 (-1.019e-09, 1.037e-09)
x and y must have the same length and you will need to use dot operators ( .*, .^) if you want to write the formula, although it is much easier to just give the x and y to your model as I did in the response.

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Plus de réponses (1)

Asliddin Komilov
Asliddin Komilov le 4 Mar 2020
thanks, it is perfect but the fit gives this warning:
Warning: Equation is badly conditioned. Remove repeated data points or try centering and scaling.
does it influence the accuracy?
  2 commentaires
Thiago Henrique Gomes Lobato
Sometimes yes, but in your case no. In your case you got an equation that basically perfect fits your curve.
Asliddin Komilov
Asliddin Komilov le 5 Mar 2020
thank you

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