为什么会这样scipy.optimize.curveu fit曲线拟合不符合数据？

数值算法往往更好地工作时，没有饲料非常小（或
（大）数字。
在本例中，图表显示数据的x和y值非常小。如果
如果缩放它们，拟合效果会更好：

xData = np.load('xData.npy')*10**5yData = np.load('yData.npy')*10**5

---

from __future__ import divisionimport osos.chdir(os.path.expanduser('~/tmp'))import numpy as npimport scipy.optimize as optimizeimport matplotlib.pyplot as pltdef func(x,a,b,c):   return a*np.exp(-b*x)-cxData = np.load('xData.npy')*10**5yData = np.load('yData.npy')*10**5print(xData.min(), xData.max())print(yData.min(), yData.max())trialX = np.linspace(xData[0], xData[-1], 1000)# Fit a polynomial fitted = np.polyfit(xData, yData, 10)[::-1]y = np.zeros(len(trialX))for i in range(len(fitted)):   y += fitted[i]*trialX**i# Fit an exponentialpopt, pcov = optimize.curve_fit(func, xData, yData)print(popt)yEXP = func(trialX, *popt)plt.figure()plt.plot(xData, yData, label='Data', marker='o')plt.plot(trialX, yEXP, 'r-',ls='--', label="Exp Fit")plt.plot(trialX, y, label = '10 Deg Poly')plt.legend()plt.show()

Note that after rescaling

xData

and

yData

, the parameters returned by

curve_fit

must also be rescaled. In this case,

a

,

b

and

c

each must be
divided by 10**5 to obtain fitted parameters for the original data.

---

One objection you might have to the above is that the scaling has to be chosen
rather “carefully”. (Read: Not every reasonable choice of scale works!)

You can improve the robustness of

curve_fit

by providing a reasonable
initial guess for the parameters. Usually you have some a priori knowledge
about the data which can motivate ballpark / back-of-the envelope type guesses
for reasonable parameter values.

For example, calling

curve_fit

with

guess = (-1, 0.1, 0)popt, pcov = optimize.curve_fit(func, xData, yData, guess)

helps improve the range of scales on which

curve_fit

succeeds in this case.