寻找更好的scipycurve_fit! - 指数函数?

Finding a better scipy curve_fit! -- exponential function?

我正在尝试使用 scipy curve_fit 来拟合数据。我 相信 负指数可能是最好的,因为这对我的一些其他(类似生成的)数据很有效——但我取得了次优的结果。

我对数据集进行了标准化以避免提供初始值,并按如下方式应用指数函数:

import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
data = np.array([[0.32,0.38],[0.61,0.32],[0.28,0.50],[0.60,0.32],[0.26,0.45],[0.19,0.57],[0.61,0.32],[0.59,0.29],[0.39,0.42],[0.61,0.32],[0.20,0.46],[0.24,0.45],[0.59,0.29],[0.39,0.42],[0.56,0.39],[0.32,0.43],[0.38,0.44],[0.54,0.34],[0.61,0.32],[0.20,0.46],[0.28,0.51],[0.54,0.34],[0.60,0.32],[0.30,0.42],[0.28,0.43],[0.14,0.57],[0.24,0.54],[0.39,0.42],[0.20,0.56],[0.56,0.39],[0.24,0.54],[0.33,0.37],[0.33,0.51],[0.20,0.46],[0.32,0.39],[0.20,0.56],[0.19,0.57],[0.32,0.39],[0.30,0.42],[0.33,0.50],[0.54,0.34],[0.28,0.50],[0.32,0.39],[0.28,0.43],[0.27,0.42],[0.56,0.39],[0.19,0.57],[0.19,0.57],[0.60,0.32],[0.44,0.41],[0.27,0.42],[0.19,0.57],[0.24,0.38],[0.24,0.54],[0.61,0.32],[0.39,0.40],[0.30,0.41],[0.19,0.57],[0.14,0.57],[0.32,0.43],[0.14,0.57],[0.59,0.29],[0.44,0.41],[0.30,0.41],[0.32,0.38],[0.61,0.32],[0.20,0.46],[0.20,0.56],[0.30,0.41],[0.33,0.36],[0.14,0.57],[0.19,0.57],[0.46,0.38],[0.36,0.44],[0.61,0.32],[0.31,0.48],[0.60,0.32],[0.39,0.40],[0.14,0.57],[0.44,0.41],[0.24,0.49],[0.41,0.40],[0.19,0.57],[0.19,0.57],[0.31,0.49],[0.31,0.43],[0.35,0.35],[0.20,0.46],[0.54,0.34],[0.20,0.56],[0.39,0.44],[0.33,0.36],[0.20,0.56],[0.30,0.41],[0.56,0.39],[0.31,0.48],[0.28,0.51],[0.14,0.57],[0.61,0.32],[0.30,0.50],[0.20,0.56],[0.19,0.57],[0.59,0.31],[0.20,0.56],[0.27,0.42],[0.29,0.48],[0.56,0.39],[0.32,0.39],[0.20,0.56],[0.59,0.29],[0.24,0.49],[0.56,0.39],[0.60,0.32],[0.35,0.35],[0.28,0.50],[0.46,0.38],[0.14,0.57],[0.54,0.34],[0.32,0.38],[0.26,0.45],[0.26,0.45],[0.39,0.42],[0.19,0.57],[0.28,0.51],[0.27,0.42],[0.33,0.50],[0.54,0.34],[0.39,0.40],[0.19,0.57],[0.33,0.36],[0.22,0.44],[0.33,0.51],[0.61,0.32],[0.28,0.51],[0.25,0.50],[0.39,0.40],[0.34,0.35],[0.59,0.31],[0.31,0.49],[0.20,0.46],[0.39,0.46],[0.20,0.50],[0.32,0.39],[0.30,0.41],[0.23,0.44],[0.29,0.53],[0.28,0.50],[0.31,0.48],[0.61,0.32],[0.54,0.34],[0.28,0.53],[0.56,0.39],[0.19,0.57],[0.14,0.57],[0.59,0.29],[0.29,0.48],[0.44,0.41],[0.27,0.51],[0.50,0.29],[0.14,0.57],[0.60,0.32],[0.32,0.39],[0.19,0.57],[0.24,0.38],[0.56,0.39],[0.14,0.57],[0.54,0.34],[0.61,0.38],[0.27,0.53],[0.20,0.46],[0.61,0.32],[0.27,0.42],[0.27,0.42],[0.20,0.56],[0.30,0.41],[0.31,0.51],[0.32,0.39],[0.31,0.51],[0.29,0.48],[0.20,0.46],[0.33,0.51],[0.31,0.43],[0.30,0.41],[0.27,0.44],[0.31,0.51],[0.29,0.48],[0.35,0.35],[0.46,0.38],[0.28,0.51],[0.61,0.38],[0.31,0.49],[0.33,0.51],[0.59,0.29],[0.14,0.57],[0.31,0.51],[0.39,0.40],[0.32,0.39],[0.20,0.56],[0.55,0.31],[0.56,0.39],[0.24,0.49],[0.56,0.39],[0.27,0.50],[0.60,0.32],[0.54,0.34],[0.19,0.57],[0.28,0.51],[0.54,0.34],[0.56,0.39],[0.19,0.57],[0.59,0.31],[0.37,0.45],[0.19,0.57],[0.44,0.41],[0.32,0.43],[0.35,0.48],[0.24,0.49],[0.26,0.45],[0.14,0.57],[0.59,0.30],[0.26,0.45],[0.26,0.45],[0.14,0.57],[0.20,0.50],[0.31,0.45],[0.27,0.51],[0.30,0.41],[0.19,0.57],[0.30,0.41],[0.27,0.50],[0.34,0.35],[0.30,0.42],[0.27,0.42],[0.27,0.42],[0.34,0.35],[0.35,0.35],[0.14,0.57],[0.45,0.36],[0.26,0.45],[0.56,0.39],[0.34,0.35],[0.19,0.57],[0.30,0.41],[0.19,0.57],[0.26,0.45],[0.26,0.45],[0.59,0.29],[0.19,0.57],[0.26,0.45],[0.32,0.39],[0.30,0.50],[0.28,0.50],[0.32,0.39],[0.59,0.29],[0.32,0.51],[0.56,0.39],[0.59,0.29],[0.61,0.38],[0.33,0.51],[0.22,0.44],[0.33,0.36],[0.27,0.42],[0.20,0.56],[0.28,0.51],[0.31,0.48],[0.20,0.56],[0.61,0.32],[0.24,0.54],[0.59,0.29],[0.32,0.43],[0.61,0.32],[0.19,0.57],[0.61,0.38],[0.55,0.31],[0.19,0.57],[0.31,0.46],[0.32,0.52],[0.30,0.41],[0.28,0.51],[0.28,0.50],[0.60,0.32],[0.61,0.32],[0.27,0.50],[0.59,0.29],[0.41,0.47],[0.39,0.42],[0.20,0.46],[0.19,0.57],[0.14,0.57],[0.23,0.47],[0.54,0.34],[0.28,0.51],[0.19,0.57],[0.33,0.37],[0.46,0.38],[0.27,0.42],[0.20,0.56],[0.39,0.42],[0.30,0.47],[0.26,0.45],[0.61,0.32],[0.61,0.38],[0.35,0.35],[0.14,0.57],[0.35,0.35],[0.28,0.51],[0.61,0.32],[0.24,0.54],[0.54,0.34],[0.28,0.43],[0.24,0.54],[0.30,0.41],[0.56,0.39],[0.23,0.52],[0.14,0.57],[0.26,0.45],[0.30,0.42],[0.32,0.43],[0.19,0.57],[0.45,0.36],[0.27,0.42],[0.29,0.48],[0.28,0.43],[0.27,0.51],[0.39,0.44],[0.32,0.49],[0.24,0.49],[0.56,0.39],[0.20,0.56],[0.30,0.42],[0.24,0.38],[0.46,0.38],[0.28,0.50],[0.26,0.45],[0.27,0.50],[0.23,0.47],[0.39,0.42],[0.28,0.51],[0.24,0.49],[0.27,0.42],[0.26,0.45],[0.60,0.32],[0.32,0.43],[0.39,0.42],[0.28,0.50],[0.28,0.52],[0.61,0.32],[0.32,0.39],[0.24,0.50],[0.39,0.40],[0.33,0.36],[0.24,0.38],[0.54,0.33],[0.19,0.57],[0.61,0.32],[0.33,0.36],[0.19,0.57],[0.30,0.41],[0.19,0.57],[0.34,0.35],[0.24,0.42],[0.27,0.42],[0.54,0.34],[0.54,0.34],[0.24,0.49],[0.27,0.42],[0.56,0.39],[0.19,0.57],[0.20,0.50],[0.14,0.57],[0.30,0.41],[0.30,0.41],[0.33,0.36],[0.26,0.45],[0.26,0.45],[0.23,0.47],[0.32,0.39],[0.27,0.53],[0.30,0.41],[0.20,0.46],[0.34,0.35],[0.34,0.35],[0.14,0.57],[0.46,0.38],[0.27,0.42],[0.36,0.44],[0.17,0.51],[0.60,0.32],[0.27,0.42],[0.20,0.56],[0.24,0.49],[0.41,0.40],[0.61,0.38],[0.19,0.57],[0.28,0.50],[0.23,0.52],[0.61,0.32],[0.39,0.46],[0.33,0.51],[0.19,0.57],[0.39,0.44],[0.56,0.39],[0.35,0.35],[0.28,0.43],[0.54,0.34],[0.36,0.44],[0.14,0.57],[0.61,0.38],[0.46,0.38],[0.61,0.32],[0.19,0.57],[0.54,0.34],[0.27,0.53],[0.33,0.51],[0.31,0.51],[0.59,0.29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x = data[:,0]
y = data[:,1]

def func(x,a,b,c):
    return a * np.exp(-b*x) + c

popt, pcov = curve_fit(func, x, y)
a, b, c = popt

x_line = np.arange(min(x), max(x), 0.01)
x_line =np.reshape(x_line,(-1,1))

y_line = func(x_line, a, b, c)
y_line = np.reshape(y_line,(-1,1))

plt.scatter(x,y)
plt.plot(x_line,y_line)

plt.show()

如您所见,拟合偏离了高 x 值。 example plot 我知道那里有很多类似的问题,而且我已经阅读了很多 - 但我的数学技能并不出色,所以我正在努力为我的特定问题想出更好的解决方案。

我不了解指数函数 - 谁能提出更好的建议? 我需要为数百个数据集半自动执行此操作,因此理想情况下我想要尽可能灵活的东西。

非常感谢任何帮助!

p.s。对于 post 如此大的样本数据集,我感到抱歉 - 但我认为这种问题需要实际数据,我不想 post 链接到看起来可疑的文件.. =)

curve_fit 无法让您很好地控制合身性,您可能想看看更通用但使用起来更复杂的 least_squareshttps://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.least_squares.html

在这里你可以控制很多东西。 curve_fit 确实给了你一个 sigma 参数,它允许你在你的点数中加权 'uncertainty'。这里的技巧是将较低的不确定性分配给 x=1 周围拟合较差的点。通过给它较低的不确定性,装配工将更加努力地适应它们。

经过一些试验,如果将 ...curve_fit... 行替换为

uncertainty = np.exp(-5*x*x)
popt, pcov = curve_fit(func, x, y, sigma = uncertainty)

我得到了以下合身

您可以尝试使用上面的 uncertainty 向量来改善这一点

这不是最佳解决方案,但它应该适用于数据中的任何类型的密度分布。这个想法是通过计算沿 x 轴的局部平均值以具有均匀分布的点来对数据进行给定次数的重新采样。

#!/usr/bin/python3.6
import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit

def my_floor(a, precision=0):
    return np.round(a - 0.5 * 10**(-precision), precision)

data = 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x = data[:,0]
y = data[:,1]

#---------------------------ADD THIS---------------------------

# Define how to resample the data 
n_bins = 20 # choose how many samples to use 
bin_size = (max(x) - min(x))/n_bins

# Prepare empty arrays for resampled x and y 
x_res, y_res = [],[] 

# Resample the data with consistent density 
for i in range(n_bins-1): 
    lower = x >= min(x)+i*bin_size
    higher = x < min(x)+(i+1)*bin_size
    x_res.append(np.mean(x[np.where(lower & higher)]))
    y_res.append(np.mean(y[np.where(lower & higher)]))

#------------------------------------------------------

def func(x,a,b,c):
    return a * np.exp(-b*x) + c

popt, pcov = curve_fit(func, x_res, y_res)
a, b, c = popt

x_line = np.arange(min(x), max(x), 0.01)
x_line = np.reshape(x_line,(-1,1))

y_line = func(x_line, a, b, c)
y_line = np.reshape(y_line,(-1,1))

plt.scatter(x,y,alpha=0.5)
plt.scatter(x_res, y_res)
plt.plot(x_line,y_line, c='red')

plt.show()

给出输出: