Curve fitter, how to start?

Question

Sergio 2024-2-26

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2086998-curve-fitter-how-to-start

评论： Star Strider 2024-2-26

data.xlsx

Hi, I just imported curve fitter to the my MATLAB. I have these coordinates from the xls file , and I wish to import them directly in curve fitter. But that doesn't open it. Can this be done easily?

Any help appreciated

Thanks

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Star Strider 2024-2-26

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2086998-curve-fitter-how-to-start#answer_1416688

在 MATLAB Online 中打开

data.xlsx

Try this —

T1 = readtable('data.xlsx');

Var1 = str2double(cellfun(@(x)regexp(x, '\d*', 'match'), T1.Var1));

T1 = removevars(T1,'Var1');

T1 = addvars(T1, Var1, 'Before','Var2')

T1 = 5×2 table

Var1 Var2 ____ ____ 39 167 40 192 41 240 42 343 43 497

fitf1 = fit(T1.Var1, T1.Var2, 'poly1')

fitf1 =

Linear model Poly1: fitf1(x) = p1*x + p2 Coefficients (with 95% confidence bounds): p1 = 81.1 (32.16, 130) p2 = -3037 (-5045, -1030)

figure

plot(fitf1, T1.Var1, T1.Var2)

grid

fitf2 = fit(T1.Var1, T1.Var2, 'poly2')

fitf2 =

Linear model Poly2: fitf2(x) = p1*x^2 + p2*x + p3 Coefficients (with 95% confidence bounds): p1 = 22.36 (14.35, 30.36) p2 = -1752 (-2409, -1096) p3 = 3.45e+04 (2.105e+04, 4.795e+04)

figure

plot(fitf2, T1.Var1, T1.Var2)

grid

.

2 个评论
显示无隐藏无

Sergio 2024-2-26

Thanks, the exponential form was the desired! Thanks!

Star Strider 2024-2-26

As always, my pleasure!

Not all the estimated parameters in the exponential or power functions were significantly different from zero (the confidence limits have opposite signs) so I only posted the models with significant parameters. (I tried several different models with 3 or fewer parameters, since there are only 4 data points, only models with 3 or fewer paramters will give reliable results in this instance.)

请先登录，再进行评论。