如何对SciPy曲线拟合施加约束？

Question

我试图用自定义概率密度函数拟合一些实验值的分布.显然,所得到的函数的积分应该总是等于1,但简单scipy.optimize.curve_fit(功能,dataBincenters,dataCounts)的结果从未满足该条件.解决这个问题的最佳方法是什么？

Answer 1

您可以定义自己的残差函数,包括惩罚参数,如下面的代码中详细说明的那样,事先知道沿着区间的积分必须是2..如果你在没有惩罚的情况下进行测试,你会发现你所获得的是传统的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

其他相关问题:

• SciPy LeastSq Fit Fit of Fit Estimator

Answer 2

这是一个几乎完全相同的片段,仅使用curve_fit.

import matplotlib.pyplot as plt
import numpy as np
import scipy.optimize as opt
import scipy.integrate as integr


x = np.linspace(0, np.pi, 100)
y = np.sin(x) + (0. + np.random.rand(len(x))*0.4)

def Func(x, a0, a1, a2, a3):
    return a0 + a1*x + a2*x**2 + a3*x**3

# modified function definition with Penalization
def FuncPen(x, a0, a1, a2, a3):
    integral = integr.quad( Func, 0, np.pi, args=(a0,a1,a2,a3))[0]
    penalization = abs(2.-integral)*10000
    return a0 + a1*x + a2*x**2 + a3*x**3 + penalization


popt1, pcov1 = opt.curve_fit( Func, x, y )
popt2, pcov2 = opt.curve_fit( FuncPen, x, y )

y_fit1 = Func(x, *popt1)
y_fit2 = Func(x, *popt2)

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

#Exact   integral: 2.0
#Approx integral1: 2.66485028754
#Approx integral2: 2.00002116217