Python 为什么 scipy.optimize.curve_fit 不适合数据？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/15624070/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Why does scipy.optimize.curve_fit not fit to the data?
提问by user1696811
I've been trying to fit an exponential to some data for a while using scipy.optimize.curve_fit but i'm having real difficulty. I really can't see any reason why this wouldn't work but it just produces a strait line, no idea why!
我一直在尝试使用 scipy.optimize.curve_fit 对某些数据进行指数拟合，但我遇到了真正的困难。我真的看不出有任何理由为什么这不起作用，但它只会产生一条海峡，不知道为什么！
Any help would be much appreciated
任何帮助将非常感激
from __future__ import division
import numpy
from scipy.optimize import curve_fit
import matplotlib.pyplot as pyplot
def func(x,a,b,c):
return a*numpy.exp(-b*x)-c
yData = numpy.load('yData.npy')
xData = numpy.load('xData.npy')
trialX = numpy.linspace(xData[0],xData[-1],1000)
# Fit a polynomial
fitted = numpy.polyfit(xData, yData, 10)[::-1]
y = numpy.zeros(len(trailX))
for i in range(len(fitted)):
y += fitted[i]*trialX**i
# Fit an exponential
popt, pcov = curve_fit(func, xData, yData)
yEXP = func(trialX, *popt)
pyplot.figure()
pyplot.plot(xData, yData, label='Data', marker='o')
pyplot.plot(trialX, yEXP, 'r-',ls='--', label="Exp Fit")
pyplot.plot(trialX, y, label = '10 Deg Poly')
pyplot.legend()
pyplot.show()
xData = [1e-06, 2e-06, 3e-06, 4e-06,
5e-06, 6e-06, 7e-06, 8e-06,
9e-06, 1e-05, 2e-05, 3e-05,
4e-05, 5e-05, 6e-05, 7e-05,
8e-05, 9e-05, 0.0001, 0.0002,
0.0003, 0.0004, 0.0005, 0.0006,
0.0007, 0.0008, 0.0009, 0.001,
0.002, 0.003, 0.004, 0.005,
0.006, 0.007, 0.008, 0.009, 0.01]
yData =
[6.37420666067e-09, 1.13082012115e-08,
1.52835756975e-08, 2.19214493931e-08, 2.71258852882e-08, 3.38556130078e-08, 3.55765277358e-08,
4.13818145846e-08, 4.72543475372e-08, 4.85834751151e-08, 9.53876562077e-08, 1.45110636413e-07,
1.83066627931e-07, 2.10138415308e-07, 2.43503982686e-07, 2.72107045549e-07, 3.02911771395e-07,
3.26499455951e-07, 3.48319349445e-07, 5.13187669283e-07, 5.98480176303e-07, 6.57028222701e-07,
6.98347073045e-07, 7.28699930335e-07, 7.50686502279e-07, 7.7015576866e-07, 7.87147246927e-07,
7.99607141001e-07, 8.61398763228e-07, 8.84272900407e-07, 8.96463883243e-07, 9.04105135329e-07,
9.08443443149e-07, 9.12391264185e-07, 9.150842683e-07, 9.16878548643e-07, 9.18389990067e-07]
采纳答案by unutbu
Numerical algorithms tend to work better when not fed extremely small (or large) numbers.
当不输入极小（或大）的数字时，数值算法往往会更好地工作。
In this case, the graph shows your data has extremely small x and y values. If you scale them, the fit is remarkable better:
在这种情况下，图表显示您的数据具有极小的 x 和 y 值。如果缩放它们，拟合效果会更好：
xData = np.load('xData.npy')*10**5
yData = np.load('yData.npy')*10**5
from __future__ import division
import os
os.chdir(os.path.expanduser('~/tmp'))
import numpy as np
import scipy.optimize as optimize
import matplotlib.pyplot as plt
def func(x,a,b,c):
return a*np.exp(-b*x)-c
xData = np.load('xData.npy')*10**5
yData = np.load('yData.npy')*10**5
print(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 exponential
popt, 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 xDataand yData, the parameters returned by curve_fitmust also be rescaled. In this case, a, band ceach must be divided by 10**5 to obtain fitted parameters for the original data.
注意，在对xDataand进行缩放后yData，返回的参数curve_fit也必须进行缩放。在这种情况下，a，b和c每个必须除以 10**5 才能获得原始数据的拟合参数。
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!)
您可能对上述提出的一个反对意见是必须相当“谨慎”地选择缩放比例。（阅读：并非所有合理的比例选择都有效！）
You can improve the robustness of curve_fitby providing a reasonable initial guess for the parameters. Usually you have some a prioriknowledge about the data which can motivate ballpark / back-of-the envelope type guesses for reasonable parameter values.
您可以curve_fit通过为参数提供合理的初始猜测来提高 的稳健性。通常，您对数据有一些先验知识，这些知识可以激发对合理参数值的大致/信封类型猜测。
For example, calling curve_fitwith
例如，调用curve_fit与
guess = (-1, 0.1, 0)
popt, pcov = optimize.curve_fit(func, xData, yData, guess)
helps improve the range of scales on which curve_fitsucceeds in this case.
有助于改善curve_fit在这种情况下成功的规模范围。
回答by Johann
A (slight) improvement to this solution, not accounting for a priori knowledge of the data might be the following: Take the inverse-mean of the data set and use that as the "scale factor" to be passed to the underlying leastsq() called by curve_fit(). This allows the fitter to work and returns the parameters on the original scale of the data.
对此解决方案的（轻微）改进，不考虑数据的先验知识可能如下：取数据集的逆均值并将其用作“比例因子”以传递给底层的 leastsq()由 curve_fit() 调用。这允许拟合器工作并以数据的原始比例返回参数。
The relevant line is:
相关行是：
popt, pcov = curve_fit(func, xData, yData)
which becomes:
变成：
popt, pcov = curve_fit(func, xData, yData,
diag=(1./xData.mean(),1./yData.mean()) )
Here is the full example which produces this image:
这是生成此图像的完整示例：
from __future__ import division
import numpy
from scipy.optimize import curve_fit
import matplotlib.pyplot as pyplot
def func(x,a,b,c):
return a*numpy.exp(-b*x)-c
xData = numpy.array([1e-06, 2e-06, 3e-06, 4e-06, 5e-06, 6e-06,
7e-06, 8e-06, 9e-06, 1e-05, 2e-05, 3e-05, 4e-05, 5e-05, 6e-05,
7e-05, 8e-05, 9e-05, 0.0001, 0.0002, 0.0003, 0.0004, 0.0005,
0.0006, 0.0007, 0.0008, 0.0009, 0.001, 0.002, 0.003, 0.004, 0.005
, 0.006, 0.007, 0.008, 0.009, 0.01])
yData = numpy.array([6.37420666067e-09, 1.13082012115e-08,
1.52835756975e-08, 2.19214493931e-08, 2.71258852882e-08,
3.38556130078e-08, 3.55765277358e-08, 4.13818145846e-08,
4.72543475372e-08, 4.85834751151e-08, 9.53876562077e-08,
1.45110636413e-07, 1.83066627931e-07, 2.10138415308e-07,
2.43503982686e-07, 2.72107045549e-07, 3.02911771395e-07,
3.26499455951e-07, 3.48319349445e-07, 5.13187669283e-07,
5.98480176303e-07, 6.57028222701e-07, 6.98347073045e-07,
7.28699930335e-07, 7.50686502279e-07, 7.7015576866e-07,
7.87147246927e-07, 7.99607141001e-07, 8.61398763228e-07,
8.84272900407e-07, 8.96463883243e-07, 9.04105135329e-07,
9.08443443149e-07, 9.12391264185e-07, 9.150842683e-07,
9.16878548643e-07, 9.18389990067e-07])
trialX = numpy.linspace(xData[0],xData[-1],1000)
# Fit a polynomial
fitted = numpy.polyfit(xData, yData, 10)[::-1]
y = numpy.zeros(len(trialX))
for i in range(len(fitted)):
y += fitted[i]*trialX**i
# Fit an exponential
popt, pcov = curve_fit(func, xData, yData,
diag=(1./xData.mean(),1./yData.mean()) )
yEXP = func(trialX, *popt)
pyplot.figure()
pyplot.plot(xData, yData, label='Data', marker='o')
pyplot.plot(trialX, yEXP, 'r-',ls='--', label="Exp Fit")
pyplot.plot(trialX, y, label = '10 Deg Poly')
pyplot.legend()
pyplot.show()
回答by benlala
the model a*exp(-b*x)+cfit well the data, but I suggest a little modification:
use this instead
该模型a*exp(-b*x)+c非常适合数据，但我建议稍作修改：
改用这个
a*x*exp(-b*x)+c
a*x*exp(-b*x)+c
good luck
祝你好运
