如何对 SciPy 曲线拟合施加约束? [英] How do I put a constraint on SciPy curve fit?

问题描述

我正在尝试使用自定义概率密度函数拟合一些实验值的分布.显然，结果函数的积分应该总是等于1，但是简单的scipy.optimize.curve_fit(function, dataBincenters, dataCounts) 的结果永远不满足这个条件.解决此问题的最佳方法是什么?

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?

推荐答案

您可以定义自己的残差函数，包括一个惩罚参数，如下面的代码中详述，其中事先知道沿区间的积分必须为2..如果你在没有惩罚的情况下进行测试，你会看到你得到的是传统的 curve_fit:

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:

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

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