拟合S型曲线(Python)

问题描述

我正在尝试用S型曲线和三次多项式来拟合我的数据(成本与收入之比)，然后找出拐点/收益递减的点。

这是我到目前为止拥有的代码，适应性不是很好。任何建议都将是非常有用的，谢谢！

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

def sigmoid(x, a, b):
    y = 1 / (1 + np.exp(-b*(x-a)))
    return y

xdata = [ 404.91,  731.89,  804.23,    0.  ,  954.72,  954.72,  954.72, 744.54,  744.54,  498.5 ,  355.03,
         359.61,    0.  ,    0.  , 0.  ,  753.77, 1116.02,  557.07,  589.06,  761.86,  722.97, 162.69, 
         354.47,  474.  ,  306.83,  538.57,  134.26,  134.26, 134.26,  134.26,  134.26,  652.29, 1296.26,
         547.78,  845.22, 872.62,  881.59,  556.23,  500.2 ,  569.97,  679.46,  679.46, 623.08,  628.33,
         754.88, 2014.12, 1870.43, 1444.69,  826.05, 1071.03,  816.74]
ydata = [ 6986.97, 36591.27, 23702.95,  6380.01, 26873.68, 19398.27,24693.5 , 18435.52, 19066.1 ,  8534.14,  8534.14,  8534.14,
          2032.07,   567.26,  7544.64, 21051.07, 21051.07, 18592.84,18592.84, 18592.84, 19566.14,  4787.51,  7269.55,
         11596.66, 9083.43, 13260.51,  6280.95,  4112.17,  6004.46,  7613.15, 6436.83, 10726.22, 20430.67,  8265.88,
         15344.32, 30246.91,29928.96, 12215.02,  7776.27,  9714.94, 16642.3 , 29493.06,15496.04, 15496.04, 15496.04,
         33397.61, 33397.61, 33397.61, 22525.93, 22525.93, 48941.98]

#fit 3rd order polynomial
p = np.poly1d(np.polyfit(x, y, 3))
second_deriv = p.deriv(2)
inflection = -second_deriv[0]/second_deriv[1]
print("polynomial inflection point:", inflection)

#fitting a sigmoid curve
popt, pcov = curve_fit(sigmoid, xdata, ydata,  method='dogbox', p0=[1000, 0.6])
estimated_k, estimated_x0 = popt
print("sigmoid inflection point:", estimated_x0)

x = np.linspace(0, int(max(xdata)), len(ydata))
y = sigmoid(x, *popt)*max(ydata)

t = np.linspace(0, int(np.max(xdata)), int(np.max(xdata)))
plt.plot(xdata, ydata, 'o', label='data')
plt.plot(p(t), 'b-', label="polynomial")
plt.plot(x,y, label='sigmoid')
plt.legend(loc='best')
plt.show()

---

解决方案

数据高度分散。根据所选的拟合标准(最小均方误差、最小均方相对误差或最小平均绝对误差等)，可获得一致的结果。另外，对于常用的迭代方法软件，猜测方便的参数初值可能是不确定的。

所以，我站在这里不是为了回答你的问题，而是将你的结果与另一种非常不同和不同寻常的拟合方法的结果进行比较。该方法不需要迭代，也不需要参数的初始猜测。有关理论，请参阅https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales中的第37-38页