通过 scipy curve_fit 的两个数据点拟合指数函数

Question

我想用常数 pw 拟合指数函数 y=x ** pw 以拟合两个数据点。 scipy curve_fit 函数应该优化 adj1 和 adj2。我尝试使用下面的代码，但无法正常工作。曲线不通过数据点。我该如何解决？

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

def func(x, adj1,adj2):
    return np.round(((x+adj1) ** pw) * adj2, 2)

x = [0.5,0.85] # two given datapoints to which the exponential function with power pw should fit
y = [0.02,4]

pw=15
popt, pcov = curve_fit(func, x, y)

xf=np.linspace(0,1,50)

plt.figure()
plt.plot(x, y, 'ko', label="Original Data")
plt.plot(xf, func(xf, *popt), 'r-', label="Fitted Curve")
plt.show()

Answer 1

这里是解决方案。我认为对于曲线拟合，lmfit 是 scipy.

的一个很好的替代方案

from lmfit import minimize, Parameters, Parameter, report_fit
import numpy as np

# create data to be fitted
xf = [0.5,0.85] # two given datapoints to which the exponential function with power pw should fit
yf = [0.02,4]

# define objective function: returns the array to be minimized
def fcn2min(params, x, data):
    pw = params['pw'].value
    adj1 = params['adj1'].value
    adj2 = params['adj2'].value

    model = adj1 * np.power(x + adj2, pw)
    return model - data

pw=2

# create a set of Parameters
params = Parameters()
params.add('pw',   value= pw, vary=False)
params.add('adj1', value= 1)
params.add('adj2', value= 1)


# do fit, here with leastsq model
result = minimize(fcn2min, params, args=(xf, yf))

# calculate final result
final = yf + result.residual

# write error report
report_fit(result.params)
adj1=result.params['adj1']
adj2=result.params['adj2']

# try to plot results
x = np.linspace(0, 1, 100)
y = adj1 * np.power(x + adj2, pw)

import pylab
pylab.plot(xf, yf, 'ko')
pylab.plot(x, y, 'r')
pylab.show()

Answer 2

如果您只想从两个数据点找到 objective 函数中的两个参数，这不一定是最小二乘拟合的问题：只需求解联立方程 y1 = b(x1 +a)^p and y2 = b(x2+a)^p for the parameters a and b:

import numpy as np
import matplotlib.pyplot as plt

def func(x, adj1,adj2):
    return ((x+adj1) ** pw) * adj2

x = [0.5,0.85] # two given datapoints to which the exponential function with power pw should fit
y = [0.02,4]

pw = 15
A = np.exp(np.log(y[0]/y[1])/pw)
a = (x[0] - x[1]*A)/(A-1)
b = y[0]/(x[0]+a)**pw

xf=np.linspace(0,1,50)
plt.figure()
plt.plot(x, y, 'ko', label="Original Data")
plt.plot(xf, func(xf, a, b), 'r-', label="Fitted Curve")
plt.show()

当然，结果函数将准确地通过这两个点。

Answer 3

只是因为轮法正在扼杀 curve_fit 搜索 space 的能力。 p0 的小扰动总是会给出相同的结果，因此它会立即停止搜索并且总是 return 无论你给它什么作为起点（默认 p0 = [1.,1.]）。解决方案是简单地从您的函数中删除 np.round。

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

def func(x, adj1,adj2):
    return ((x+adj1) ** pw) * adj2

x = [0.5,0.85] # two given datapoints to which the exponential function with power pw should fit
y = [0.02,4]

pw=15
popt, pcov = curve_fit(func, x, y)

xf=np.linspace(0,1,50)

plt.figure()
plt.plot(x, y, 'ko', label="Original Data")
plt.plot(xf, func(xf, *popt), 'r-', label="Fitted Curve")
plt.show()

通过 scipy curve_fit 的两个数据点拟合指数函数

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