scipy curve_fit奇怪的结果

Question

我正在尝试使用scipy的curve_fit进行分配。我试图拟合一个单分量指数函数，这导致了几乎直线（见图）。我也试过一个双组分指数拟合，似乎很好用。两个分量只意味着等式的一部分用不同的输入参数重复。无论如何，这是一个组件拟合函数：

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))

并用

完成拟合

fitexp = curve_fit(Exponential,newx,y2)

然后我试了一下，试试看。我采用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))

突然之间有效。

现在，我的意思是。为什么？如您所见，拟合的计算完全没有区别。它只会获得两个未使用的额外变量。这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，情节确实看起来不错，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有时会起作用并坚持参数的初始值。我怀疑在你的情况下添加一些假参数改变了相关参数如何被初始化的启发式（尽管这与文档的声明相矛盾，即没有给出初始值，它们都默认为1）。

如果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），它可以帮助您获得可靠的拟合。默认起始值​​应该都是1的事实表明，对于大多数用例，默认值不会削减它。