Extraction of values from a cfit object

Question

Antonio 2013-4-3

3
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/69634-extraction-of-values-from-a-cfit-object

评论： David Ebert 2023-2-8

Dear, I used the fit function to interpolate two set of experimental data (X,Y) with a function fun previously defined with fittype. Fun contains 3 constant A, k1 and k2. After i estimates three constants with by using fit as follow: [fitresult, gof]=fit(X,Y,fun) I obitaining an object containing the fit the followin results: fitresult =

     General model:
     fitresult(x) = 562.16*exp(-k2*x)+(k1*A/(k2-k1))*(exp(-k1*x)-exp(-k2*x))
     Coefficients (with 95% confidence bounds):
       A =        1202  (481.5, 1921)
       k1 =           5  (-119, 129)
       k2 =      0.2374  (-0.1202, 0.595)

This is a cfit object. My question is: How can extract the values of A, k1 and k2 from the above object. I need to use the estimated value for different steps of my script.

Thaks

Antonio

4 个评论
显示 2更早的评论隐藏 2更早的评论

John D'Errico 2020-7-27

@pinar: read my answer.

David Ebert 2023-2-8

@Yougu Yes! Perfect! Thank you! :)

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

John D'Errico 2020-7-27

11
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/69634-extraction-of-values-from-a-cfit-object#answer_471292

编辑：John D'Errico 2020-7-27

在 MATLAB Online 中打开

When you do not know how to pull some information from a class, the best thing is to try using methods to learn what you can do.

mdl = fit(rand(10,1),rand(10,1),'poly2')
mdl = 
     Linear model Poly2:
     mdl(x) = p1*x^2 + p2*x + p3
     Coefficients (with 95% confidence bounds):
       p1 =     -0.8558  (-4.362, 2.651)
       p2 =      0.6407  (-3.374, 4.655)
       p3 =      0.4551  (-0.5222, 1.432)

We have here a very simple fit result. As you see, this works, but it may not be clear how to have known it would work. Sometimes just trying things is sufficient.

mdl.p1
ans =
        -0.855805940768612
mdl.p2
ans =
         0.640653729424836
mdl.p3
ans =
         0.455095631339414

And as much as I want to say this should be the logical thing to try, or that it is shown in the documentation, a reading of the docs for fit does not show me a spot where that is explicitly stated.

But if we want a clear list of the functions we can use to work with mdl, or extract information from it, we use methods.

methods(mdl)
Methods for class cfit:
argnames       coeffnames     dependnames    fitoptions     integrate      numcoeffs      probnames      type           
category       coeffvalues    differentiate  formula        islinear       plot           probvalues     
cfit           confint        feval          indepnames     numargs        predint        setoptions

Hmm, so coeffnames is a method we can use. do you see, I was able to extract the names of the coefficients using one of the methods?

coeffnames(mdl)
ans =
  3×1 cell array
    {'p1'}
    {'p2'}
    {'p3'}

One of the other methods listed was coeffvalues. The name alone should give me a hint that coeffvalues might help in our quest.

coeffvalues(mdl)
ans =
        -0.855805940768612         0.640653729424836         0.455095631339414

So we can simply and programmatically use the coeffvalues method to do what we wanted, and there WAS a simple way to have learned that.

A followup question asked how to extact the confidence intervals. Again, LOOK AT THE METHODS. One of them is named confint. Does that help? I think it might.

confint(mdl)
ans =
         -4.36215292171444         -3.37378195209185        -0.522170718930676
          2.65054104017722          4.65508941094152           1.4323619816095

The point is, when you have a class that you have never seen before, you can learrn much by using methods on the class object.

If you wish to get direct help, then we could have done this:

help cfit/coeffvalues
 coeffvalues Coefficient values.
    coeffvalues(FUN) returns the values of the coefficients of the
    CFIT object FUN as a row vector.

The information is there on how to extract what you want.