How to fit a defined function?

13 vues (au cours des 30 derniers jours)
Sourabh Jain
Sourabh Jain le 10 Nov 2022
Commenté : Star Strider le 10 Nov 2022

I want to fit a custom function to my experimental data. For simplicity, I have some arbitary x & y values and a very simple linear function. I write the following code:

clear all;

x = [1 2 3 4 5]'; % x data
y = [.8 4 10 18 23]'; % y data

Y = lsqcurvefit(fun,1,x,y) % fitting function 'fun' defined below to find parameter 'a'

function y = fun(a,x)
y = a*x; % just a simple function for example, in actual problem, it is a long complicated function with various parameters
end

I get the following error: 'Not enough input arguements.'

I know this particular simple function can be defined as anonymous function and be fitted but I don't it that way.

Réponse acceptée

Star Strider
Star Strider le 10 Nov 2022
The ‘fun’ function must be presented to lsqcurvefit as a function handle using the ‘@’ operator —
x = [1 2 3 4 5]'; % x data
y = [.8 4 10 18 23]'; % y data
Y = lsqcurvefit(@fun,1,x,y) % fitting function 'fun' defined below to find parameter 'a'
Local minimum found. Optimization completed because the size of the gradient is less than the value of the optimality tolerance.
Y = 4.1055
function y = fun(a,x)
y = a*x; % just a simple function for example, in actual problem, it is a long complicated function with various parameters
end
See What Is a Function Handle? for details.
.
  2 commentaires
Sourabh Jain
Sourabh Jain le 10 Nov 2022
Thank you Star Strider for your answer. It solved my problem.
Star Strider
Star Strider le 10 Nov 2022
As always, my pleasure!

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Plus de réponses (2)

VBBV
VBBV le 10 Nov 2022
Modifié(e) : VBBV le 10 Nov 2022
To define a and then call in function
  2 commentaires
VBBV
VBBV le 10 Nov 2022
Then use @
Y = lsqcurvefit(@fun,1,x,y)
Sourabh Jain
Sourabh Jain le 10 Nov 2022
Thank you for your respone, it helped me.

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Torsten
Torsten le 10 Nov 2022
Modifié(e) : Torsten le 10 Nov 2022
A simpler way for this problem, but I guess your "real" model is more complicated:
x = [1 2 3 4 5]'; % x data
y = [.8 4 10 18 23]'; % y data
a = x\y
a = 4.1055
  1 commentaire
Sourabh Jain
Sourabh Jain le 10 Nov 2022
Thank you Torsten for this intelligent answer, I really like your perspective.
Yes, my actual problem is not so simple, it is a transfer matrix model.

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