fitting 2d data set fit function is not working

2 views (last 30 days)
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

Accepted Answer

Thiago Henrique Gomes Lobato
Edited: Thiago Henrique Gomes Lobato on 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 Comments
Asliddin Komilov
Asliddin Komilov on 2 Mar 2020
Edited: Asliddin Komilov on 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.

Sign in to comment.

More Answers (1)

Asliddin Komilov
Asliddin Komilov on 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 Comments
Thiago Henrique Gomes Lobato
Sometimes yes, but in your case no. In your case you got an equation that basically perfect fits your curve.

Sign in to comment.

Categories

Find more on Get Started with Curve Fitting Toolbox in Help Center and File Exchange

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!