On Tue, Aug 16, 2011 at 2:15 AM, 4ndre <andre.dank...@gmail.com> 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
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