卡方用於多項式擬合的最佳順序

Question

我有以下代碼，其中DGauss是一個生成期望值的函數。 另一方面，這兩個數組允許我生成一個分布，將其作為觀測值。 該代碼基於觀察到的值，提取描述其趨勢的多項式（針對第七度）。

import matplotlib.pyplot as plt
import numpy as np
from scipy.optimize import curve_fit

def DGauss(x,I1,I2,sigma1,sigma2):
    return I1*np.exp(-x*x/(2*sigma1*sigma1)) + I2*np.exp(-x*x/(2*sigma2*sigma2))

Pos = np.array([3.28, 3.13, 3.08, 3.03, 2.98, 2.93, 2.88, 2.83, 2.78, 2.73, 2.68,
       2.63, 2.58, 2.53, 2.48, 2.43, 2.38, 2.33, 2.28, 2.23, 2.18, 2.13,
       2.08, 2.03, 1.98, 1.93, 1.88, 1.83, 1.78, 1.73, 1.68, 1.63, 1.58,
       1.53, 1.48, 1.43, 1.38, 1.33, 1.28, 1.23, 1.18, 1.13, 1.08, 1.03,
       0.98, 0.93, 0.88, 0.83, 0.78, 0.73, 0.68, 0.63, 0.58, 0.53, 0.48,
       0.43, 0.38, 0.33, 0.28, 0.23, 0.18, 0.13, 0.08, 0.03])

Val = np.array([0.00986279, 0.01529543, 0.0242624 , 0.0287456 , 0.03238484,
       0.03285927, 0.03945234, 0.04615091, 0.05701618, 0.0637672 ,
       0.07194268, 0.07763934, 0.08565687, 0.09615262, 0.1043281 ,
       0.11350606, 0.1199406 , 0.1260062 , 0.14093328, 0.15079665,
       0.16651464, 0.18065023, 0.1938894 , 0.2047541 , 0.21794024,
       0.22806706, 0.23793043, 0.25164404, 0.2635118 , 0.28075974,
       0.29568682, 0.30871501, 0.3311846 , 0.34648062, 0.36984661,
       0.38540666, 0.40618835, 0.4283945 , 0.45002014, 0.48303911,
       0.50746062, 0.53167057, 0.5548792 , 0.57835128, 0.60256181,
       0.62566436, 0.65704847, 0.68289386, 0.71332794, 0.73258027,
       0.769608  , 0.78769989, 0.81407275, 0.83358852, 0.85210239,
       0.87109068, 0.89456217, 0.91618782, 0.93760247, 0.95680234,
       0.96919757, 0.9783219 , 0.98486193, 0.9931429 ])

f = np.linspace(-9,9,2*len(Pos))    

plt.errorbar(Pos, Val, xerr=0.02, yerr=2.7e-3, fmt='o')
popt, pcov = curve_fit(DGauss, Pos, Val)
plt.plot(xfull, DGauss(f, *popt), '--', label='Double Gauss')

x = Pos
y = Val
#z, w = np.polyfit(x, y, 7, full=False, cov=True)
p = np.poly1d(z)
u = np.array(p)
xp = np.linspace(1, 6, 100)
_ = plt.plot(xp, p(xp), '-', color='darkviolet')
x = symbols('x')
list = u[::-1]
poly = sum(S("{:7.3f}".format(v))*x**i for i, v in enumerate(list))
eq_latex = sympy.printing.latex(poly)
print(eq_latex)

#LOOP SUGGESTED BY @Fourier
dof = [1,2,3,4,5,6,7,8,9,10]
for i in dof:
    z = np.polyfit(x, y, i, full=False, cov=True)
    chi = np.sum((np.polyval(z, x) - y) ** 2)
    chinorm = chi/i
    plt.plot(chinorm)

我現在想做的是通過改變多項式的階數來擬合，從而找出哪一個是我需要良好擬合且不超過自由參數數量的最小階數。 特別是，我想使它適合不同的階數並繪制卡方圖，必須針對自由度的數量對其進行標准化。 有人可以幫我嗎？

謝謝！

Answer 1

根據發布的代碼，這應該可以滿足您的目的：

chiSquares = []

dofs = 10
for i in np.arange(1,dofs+1):
    z = np.polyfit(x, y, i, full=False, cov=False)
    chi = np.sum((np.polyval(z, x) - y) ** 2) / np.std(y) #ideally you should divide this using an error for Val array
    chinorm = chi/i
    chiSquares.append(chinorm)

plt.plot(np.arange(1,dofs+1),chiSquares)

如果從圖中看不出來，則可以進一步使用F檢驗來檢查實際需要多少自由度：

n = len(y)
for d, (rss1,rss2) in enumerate(zip(chiSquares,chiSquares[1:])):
    p1 = d + 1
    p2 = d + 2        
    F = (rss1-rss2/(p2-p1)) / (rss2/(n-p2))
    p = 1.0 - scipy.stats.f.cdf(F,p1,p2)
    print 'F-stats: {:.3f},  p-value: {:.5f}'.format(F,p)