cfit object doesn't work with 'plot' or 'differentiate'
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Hello,
I stored a cfit obj to hard disc. It looks like:
General model:
ans(xd) = off+amp./(1+exp(-a.*(xd-b)))
Coefficients (with 95% confidence bounds):
off = -4.047 (-4.054, -4.041)
amp = 8.179 (8.169, 8.189)
a = 4.053 (4.034, 4.073)
b = 2.271 (2.27, 2.273)
In the process of creating the obj I was able to plot it and to calculate the derivatives.
Now I want to come back to these data and tried:
>>plot(cfObj)
but receive:
Error using cfit/plot (line 111)
Error evaluating CFIT function:
Error while trying to evaluate CFIT model: obj:
Error while trying to evaluate FITTYPE function obj:
Argument must contain a string or function_handle.
For:
>>differntiate(cfObj, x)
I receive:
Error using cfit/feval (line 31)
Error while trying to evaluate CFIT model: fitobj:
Error while trying to evaluate FITTYPE function obj:
Argument must contain a string or function_handle.
Error in cfit/differentiate (line 33)
f1 = feval(fitobj,x+ms);
Can anyone please help out with that? Sure I can recreate the fit from the coefficients, but I think it is not meant to be like that.
Thanks,
Kai
2 commentaires
Onno Broekmans
le 1 Juil 2013
Your fit results are definitely not useless. You can use coeffvalues(cfObj) to get the coefficient values from the fit manually, and then plot them using:
fitCoeffs = num2cell(coeffvalues(cfObj));
fitFun = @(off,amp,a,b,xd) off+amp./(1+exp(-a.*(xd-b)));
plot(x, feval(fitFun, fitCoeffs{:}, x, '-r');
(Warning: untested code, so there might be minor syntax errors in there. I'm assuming your x data is in the variable 'x').
Note that this is what plot(cfObj,x) does internally, so the results are exactly the same :)
Kai
le 1 Juil 2013
Réponse acceptée
Plus de réponses (1)
Andrei Bobrov
le 26 Juin 2013
Modifié(e) : Andrei Bobrov
le 26 Juin 2013
x = (0:.01:4)'; %
y = -4 + 8./(1+exp(4.*(x - 2))) + .5*randn(numel(x),1); % your data
ftfun = fit(x,y,fittype(@(a,b,c,d,x)a + b./(1+exp(c.*(x - d)))),...
'Startpoint',[1 1 1 1])
dft = differentiate(ftfun,x);
plot(x,[y,ftfun(x),dft])
1 commentaire
Kai
le 26 Juin 2013
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