独特的趋势曲线拟合

如何解决独特的趋势曲线拟合

我有这样的数据：

x = np.array([   0.,3.,3.3,10.,18.,43.,80.,120.,165.,210.,260.,310.,360.,410.,460.,510.,560.,610.,660.,710.,760.,809.5,859.,908.5,958.,1007.5,1057.,1106.5,1156.,1205.5,1255.,1304.5,1354.,1403.5,1453.,1502.5,1552.,1601.5,1651.,1700.5,1750.,1799.5,1849.,1898.5,1948.,1997.5,2047.,2096.5,2146.,2195.5,2245.,2294.5,2344.,2393.5,2443.,2492.5,2542.,2591.5,2640.,2690.,2740.,2789.67,2839.33,2891.5 ])
y = array([ 1.45,1.65,5.8,6.8,8.0355,8.0379,8.04,8.0505,8.175,8.3007,8.4822,8.665,8.8476,9.0302,9.528,9.6962,9.864,10.032,10.2,10.9222,11.0553,11.1355,11.2228,11.3068,11.3897,11.4704,11.5493,11.6265,11.702,11.7768,11.8491,11.9208,11.9891,12.0571,12.1247,12.1912,12.2558,12.3181,12.3813,12.4427,12.503,12.5638,12.6226,12.6807,12.7384,12.7956,12.8524,12.9093,12.9663,13.0226,13.0786,13.1337,13.1895,13.2465,13.3017,13.3584,13.4156,13.4741,13.5311,13.5899,13.6498,13.6533,13.657,13.6601])

看起来像这样：

我需要针对这种趋势进行曲线拟合。我正在使用移动平均线进行平滑处理，如下所示：

其中洋红色是 MA，我使用多项式（5th Ordo），如下所示：

其中蓝色是多项式的结果。我尝试了更高的ordo，但结果越来越糟糕。我怎样才能得到第一个指向 (0,0) 并看起来像这样（如黑色曲线）的结果？

这是我的代码：

import numpy as np
from scipy import interpolate
def movingaverage(interval,window_size):
     window= np.ones(int(window_size))/float(window_size)
     print(window)
     return np.convolve(interval,window,'same')
y_av = movingaverage(y,2)
X = np.arange(0,np.max(x),30).ravel()
yinter = interpolate.interp1d(x,y_av)(X)
z = np.poly1d(np.polyfit(x,y_av,5))
Y = z(X)
plt.figure(1)
plt.plot(xm,ym,'*-r')
plt.plot(xm,'.-m')
plt.plot(X,Y,'*-b')

解决方法

为此，您应该使用基于某些假设（不仅是多项式函数）的分析函数（带参数）。您可以使用 curve_fit 形式 scipy.optimize 查找最适合您输入数据的分析函数的未知参数。

例如：

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

# your analytical function (theoretical function) with parameters: a,b (or more)
def your_analytical_func(x,a,b):
    return a * np.log(x + b) # this is just for example 
# or using anonymous (lambda) function
# your_analytical_func = lambda x,b: a * np.log(x + b)


# Fit for the parameters a,b (or more) of the function your_analytical_func:
popt,pcov = curve_fit(your_analytical_func,x,y)

plt.plot(x,y,'r.',label='incoming data')
plt.plot(x,your_analytical_func(x,*popt),'-',color="black",label='fit: your_analytical_func(x,a=%5.3f,b=%5.3f)' % tuple(popt))
plt.legend()