Re: [Matplotlib-users] absolute error of curve fit parameters

On Tue, Aug 16, 2011 at 2:15 AM, 4ndre <and...@gm...> wrote:

>
> Hi,
>
> I'm a little bit desperated. I googled for days to find an appropriate way
> to calculate the absolute errors of parameters of non-linear fits. I found
> different sites which suggest different, sometimes more or less similar
> ways. But the results don't make sense. Here is a simple example:
>
> from pylab import *
> from scipy.optimize import curve_fit
> def f(x,a,b):
>        return a*x+b
> def fit(n=1000):
>        x,y=array([0,1,1]),array([1,2,4])
>        xf=[(min(x)+i*(max(x)-min(x))/float(n)) for i in range(0,int(n)+1)]
>        a,acov=curve_fit(f,x,y)
>        yf=[f(*tuple([i]+list(a))) for i in xf]
>        meansq=sum([pow(y[i]-f(x[i],a[0],a[1]),2)/(len(x)) for i in
> range(0,len(x))])
>        redchi2=sum([pow(y[i]-f(x[i],a[0],a[1]),2)/(len(x)-len(a)) for i in
> range(0,len(x))])
>        print 'standard dev:',sqrt(diag(acov))
>        print 'abs. meansq err.:',[acov[i][i]*sqrt(meansq) for i in
> range(0,len(a))]
>        print 'abs. reduced meansq err.:',[acov[i][i]*sqrt(redchi2) for i in
> range(0,len(a))]
> fit()
>
> I included 3 ways I found. The used arrays have 3 points, where for x=1 I
> give 2 different y values. I would say,
> a=2+/-1, b=1+/-0,
> but no way yields to a similar result! How can I trust this function on
> more
> complicated, really non-linear functions? Or is simply this example wrong?
> Or are the methods not right?
> I'm very thankful for any help!
>
> Best regards,
> André
>

Andre,

Typically, I do something like the following:

from scipy.optimize import polyfit, polyval
import numpy as np.

# x and y are arrays of floats

a, b = polyfit(x, y, 1)  # first-order polynomial
y_fit = polyval([a, b], x)
mean_abs_err = np.mean(np.abs(y_fit - y))


I hope that helps!
Ben Root