Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
530 views
in Technique[技术] by (71.8m points)

python - How do I put a constraint on SciPy curve fit?

I'm trying to fit the distribution of some experimental values with a custom probability density function. Obviously, the integral of the resulting function should always be equal to 1, but the results of simple scipy.optimize.curve_fit(function, dataBincenters, dataCounts) never satisfy this condition. What is the best way to solve this problem?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

You can define your own residuals function, including a penalization parameter, like detailed in the code below, where it is known beforehand that the integral along the interval must be 2.. If you test without the penalization you will see that what your are getting is the conventional curve_fit:

enter image description here

import matplotlib.pyplot as plt
import scipy
from scipy.optimize import curve_fit, minimize, leastsq
from scipy.integrate import quad
from scipy import pi, sin
x = scipy.linspace(0, pi, 100)
y = scipy.sin(x) + (0. + scipy.rand(len(x))*0.4)
def func1(x, a0, a1, a2, a3):
    return a0 + a1*x + a2*x**2 + a3*x**3

# here you include the penalization factor
def residuals(p,x,y):
    integral = quad( func1, 0, pi, args=(p[0],p[1],p[2],p[3]))[0]
    penalization = abs(2.-integral)*10000
    return y - func1(x, p[0],p[1],p[2],p[3]) - penalization

popt1, pcov1 = curve_fit( func1, x, y )
popt2, pcov2 = leastsq(func=residuals, x0=(1.,1.,1.,1.), args=(x,y))
y_fit1 = func1(x, *popt1)
y_fit2 = func1(x, *popt2)
plt.scatter(x,y, marker='.')
plt.plot(x,y_fit1, color='g', label='curve_fit')
plt.plot(x,y_fit2, color='y', label='constrained')
plt.legend(); plt.xlim(-0.1,3.5); plt.ylim(0,1.4)
print 'Exact   integral:',quad(sin ,0,pi)[0]
print 'Approx integral1:',quad(func1,0,pi,args=(popt1[0],popt1[1],
                                                popt1[2],popt1[3]))[0]
print 'Approx integral2:',quad(func1,0,pi,args=(popt2[0],popt2[1],
                                                popt2[2],popt2[3]))[0]
plt.show()

#Exact   integral: 2.0
#Approx integral1: 2.60068579748
#Approx integral2: 2.00001911981

Other related questions:


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...