Question

我有以下代码，其中DGauss是一个生成期望值的函数。另一方面，这两个数组允许我生成一个分布，我将其作为观察值。该代码基于观察到的值，提取描述其趋势的多项式（在七度的时刻）。

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)

我现在要做的是通过改变多项式的阶数来拟合，以确定哪个是我需要具有良好拟合的最小阶数而不是超过自由参数的数量。特别是，我想用不同的顺序进行拟合并绘制卡方形，其必须相对于自由度的数量进行归一化。有人可以帮助我吗？

谢谢！

Answer 1

根据发布的代码，这应该适用于您的目的：

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)

如果情节不明显，您可以进一步使用F测试来检查真正需要多少自由度：

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)

使用多项式

1 个答案: