python - scipy curve_fit 奇怪的结果

Question

我正在尝试使用 scipy 的 curve_fit 拟合分布。我尝试拟合一个单分量指数函数，结果几乎是一条直线(见图)。我还尝试了两分量指数拟合，似乎效果很好。两个分量仅意味着方程的一部分使用不同的输入参数重复。无论如何，这是单组件拟合函数:

def Exponential(Z,w0,z0,Z0):
    z = Z - Z0
    termB = (newsigma**2 + z*z0) / (numpy.sqrt(2.0)*newsigma*z0)
    termA = (newsigma**2 - z*z0) / (numpy.sqrt(2.0)*newsigma*z0)
    return w0/2.0 * numpy.exp(-(z**2 / (2.0*newsigma**2))) * (numpy.exp(termA**2)*erfc(termA) + numpy.exp(termB**2)*erfc(termB))

安装完成

fitexp = curve_fit(Exponential,newx,y2)

然后我尝试了一些东西，只是为了尝试一下。我取了两个组件拟合的两个参数，但没有在计算中使用它们。

def ExponentialNew(Z,w0,z0,w1,z1,Z0):
    z = Z - Z0
    termB = (newsigma**2 + z*z0) / (numpy.sqrt(2.0)*newsigma*z0)
    termA = (newsigma**2 - z*z0) / (numpy.sqrt(2.0)*newsigma*z0)
    return w0/2.0 * numpy.exp(-(z**2 / (2.0*newsigma**2))) * (numpy.exp(termA**2)*erfc(termA) + numpy.exp(termB**2)*erfc(termB))

突然间这就起作用了。

现在，我的问题是。为什么？正如您所看到的，拟合度的计算绝对没有差异。它只是获得两个未使用的额外变量。这不应该得到相同的结果吗？

@Andras_Deak一个实际的例子:

from scipy.special import erfc
import numpy
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit

#setup data
x = [-58.,-54.,-50.,-46.,-42.,-38.,-34.,-30.,-26.,-22.,-18.,-14.,-10.,-6.,-2.,2.,6.,10.,14.,18.,22.,26.,30.,34.,38.,42.,46.,50.,54.,58.]
y = [23.06763817, 16.89802085, 17.83258379, 16.63446237, 13.81878965, 12.97965839, 14.30451789, 16.98288216, 22.26811491, 28.56756908, 33.06990344, 38.59842098, 54.19860393, 86.37381604, 137.47253315, 199.49724512, 238.66047662, 219.89405445, 160.68820199, 103.88901303, 65.92405727, 43.84596266, 31.5395342, 25.9610156, 22.71683709, 18.06740651, 13.85362374, 11.12867065, 10.36502799, 11.31855619]
y_err = [17.9823065, 4.13684885, 1.66490726, 2.4109372, 2.93359141, 1.9701747, 3.19214881,  3.65593012, 2.89089074, 3.58922121, 4.25505348, 4.72728874, 6.77736567, 11.3888196, 21.87771722, 39.0087495, 56.6910311, 51.7592369, 26.39750958, 10.62678862, 7.85893395, 8.11741621, 7.91731416, 7.07739132, 5.41818744, 6.11286843, 8.27070757, 7.85323065, 4.26885499, 0.9047867]

#function to fit
def Exponential2(Z, w0, z0, w1, z1, Z0):
    z = Z - Z0
    s = 3.98098937586
    a = z**2 / (2.0*s**2)
    b = (s**2 + z*z0) / (numpy.sqrt(2.0)*s*z0)
    c = (s**2 - z*z0) / (numpy.sqrt(2.0)*s*z0)
    d = (s**2 + z*z1) / (numpy.sqrt(2.0)*s*z1)
    e = (s**2 - z*z1) / (numpy.sqrt(2.0)*s*z1)
    return w0/2.0 * numpy.exp(-a) * (numpy.exp(c**2)*erfc(c) + numpy.exp(b**2)*erfc(b)) + w1/2.0 * numpy.exp(-a) * (numpy.exp(e**2)*erfc(e) + numpy.exp(d**2)*erfc(d))


#derive and set initial guess
ymaxpos = x[numpy.where(y==numpy.max(y))[0]]
p0_2 = [numpy.max(y),5,numpy.max(y)/2.0,20,ymaxpos]

#fit
fitexp2 = curve_fit(Exponential2,x,y,p0=p0_2,sigma=y_err)

#get results
w0err = numpy.sqrt(numpy.diag(fitexp2[1]))[0]
z0err = numpy.sqrt(numpy.diag(fitexp2[1]))[1]
w1err = numpy.sqrt(numpy.diag(fitexp2[1]))[2]
z1err = numpy.sqrt(numpy.diag(fitexp2[1]))[3]
w0 = fitexp2[0][0]
z0 = fitexp2[0][1]
w1 = fitexp2[0][2]
z1 = fitexp2[0][3]
Z0 = fitexp2[0][4]
#new x array for smoother curve
smoothx = numpy.arange(-58,59,0.1)
y2 = Exponential2(smoothx,w0,z0,w1,z1,Z0)

print 'Exponential 2: w0: '+str(w0.round(3))+' +/- '+str(w0err.round(3))+' \t z0: '+str(z0.round(3))+' +/- '+str(z0err.round(3))+' \t w1: '+str(w1.round(3))+' +/- '+str(w1err.round(3))+' \t\t z1: '+str(z1.round(3))+' +/- '+str(z1err.round(3))

#plot
fig = plt.figure()
ax = fig.add_subplot(111)
ax.errorbar(x,y,y_err,fmt='o',markersize=2,label='data')
ax.plot(smoothx,y2,label='fit',color='red')
ax.grid()
ax.legend()
plt.show()

正如您所看到的，该图确实看起来不错，但返回值 z1 完全不切实际。

Exponential 2: w0: 312.608 +/- 36.764    z0: 8.263 +/- 1.158     w1: 12.689 +/- 9.138        z1: 1862257.883 +/- 45201809883.8

Answer 1

根据我的经验curve_fit有时可以发挥作用并坚持使用参数的初始值。我怀疑在你的情况下添加一些假参数会改变相关参数初始化方式的启发式(尽管这与文档中的声明相矛盾，即在没有给出初始值的情况下，它们都默认为 1)。

如果您为拟合参数指定合理的边界和初始值(我的意思是 p0 和 bounds 关键字)，这对获得可靠的拟合有很大帮助。默认起始值​​应全部为 1 这一事实表明，对于大多数用例，默认值不会削减它。