如何用Python提升误差函数拟合质量？目标R²>0.99

我正在对lenJ1数据集进行误差函数拟合，但拟合曲线效果不佳，希望优化拟合质量使R²值大于0.99。

现有实现代码 #

import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
from sklearn.metrics import r2_score

# Define the error function
def error_function(x, a, b, c, d):
    return a + b * np.exp(-(x - c)**2 / (2.0 * d**2))

# Data
lenJ1 = [337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 336, 336, 334, 334, 333, 331, 331, 331, 329, 329, 328, 324, 323, 315, 308, 294, 283, 273, 264, 244, 234, 222, 217, 205, 188, 181, 174, 162, 151, 133, 126, 117, 112, 105, 96, 95, 87, 80, 73, 62, 59, 58, 52, 40, 35, 33, 31, 30, 29, 29, 26, 23, 21, 19, 18, 16, 15, 15, 15, 15, 15, 15, 14, 13, 12, 12, 12, 12, 12, 11, 10, 8, 8, 7, 7, 7, 6, 6, 5, 5, 5, 5, 5, 5, 5, 4, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 0]
lenJ1 = [1 - (i / 1012) for i in lenJ1]

t_start = 0.0
t_end = 1e2
timesteps = int(1e3 * t_end) + 1
t1 = np.linspace(t_start, t_end, timesteps)
print(timesteps)

# Provide adjusted initial guesses for parameters based on the data characteristics
initial_guess = [1, 1, 1, 1]

# Fit the error function to the data with adjusted initial guesses
popt, pcov = curve_fit(error_function, t1[:len(lenJ1)]/1.0330677596203013, lenJ1, p0=initial_guess, bounds=([0, 0, 0, 0], [100, 100, 100, 100]))

# Predicted values using the fitted curve
predicted_values = error_function(t1[:len(lenJ1)]/1.0330677596203013, *popt)

# Calculate R^2 value
r_squared = r2_score(lenJ1, predicted_values)

# Plot the data and the fitted curve
plt.plot(t1[:len(lenJ1)], lenJ1, 'b-', label='Data')
plt.plot(t1[:len(lenJ1)], error_function(t1[:len(lenJ1)]/1.0330677596203013, *popt), 'r--', label='Fitted Curve')

plt.xlabel('t1')
plt.ylabel('Values')
plt.title('Error Function Fit')
plt.legend()
plt.show()

# Print the parameters of the fitted error function
print("Parameters (a, b, c, d):", popt)
print("R^2 value:", r_squared)

当前拟合结果 #

优化方案及修改后代码 #

核心优化点 #

• 精准初始参数猜测：根据数据趋势调整初始值，匹配误差函数的物理意义：
  • a：数据最终渐近值（接近0）
  • b：初始值与渐近值的差值（≈0.67，由1-337/1012计算得出）
  • c：数据开始下降的中心位置（≈0.03，对应前27个稳定点的结束位置）
  • d：控制下降速度的宽度参数（≈0.05）
• 缩小参数边界：根据数据范围限制参数搜索区间，避免拟合陷入局部最优
• 增加迭代次数：设置maxfev=10000，确保拟合有足够次数收敛

修改后代码 #

import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
from sklearn.metrics import r2_score

# Define the error function
def error_function(x, a, b, c, d):
    return a + b * np.exp(-(x - c)**2 / (2.0 * d**2))

# Data
lenJ1 = [337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 337, 336, 336, 334, 334, 333, 331, 331, 331, 329, 329, 328, 324, 323, 315, 308, 294, 283, 273, 264, 244, 234, 222, 217, 205, 188, 181, 174, 162, 151, 133, 126, 117, 112, 105, 96, 95, 87, 80, 73, 62, 59, 58, 52, 40, 35, 33, 31, 30, 29, 29, 26, 23, 21, 19, 18, 16, 15, 15, 15, 15, 15, 15, 14, 13, 12, 12, 12, 12, 12, 11, 10, 8, 8, 7, 7, 7, 6, 6, 5, 5, 5, 5, 5, 5, 5, 4, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 0]
lenJ1 = [1 - (i / 1012) for i in lenJ1]

t_start = 0.0
t_end = 1e2
timesteps = int(1e3 * t_end) + 1
t1 = np.linspace(t_start, t_end, timesteps)
print(timesteps)

# 基于数据特征的精准初始猜测
initial_guess = [0.01, 0.67, 0.03, 0.05]

# 缩小参数边界，匹配数据范围
lower_bounds = [0, 0, 0, 0.01]
upper_bounds = [0.1, 0.7, 0.1, 0.2]

# 拟合时增加迭代次数，确保收敛
popt, pcov = curve_fit(
    error_function, 
    t1[:len(lenJ1)]/1.0330677596203013, 
    lenJ1, 
    p0=initial_guess, 
    bounds=(lower_bounds, upper_bounds),
    maxfev=10000
)

# 计算预测值与R²
predicted_values = error_function(t1[:len(lenJ1)]/1.0330677596203013, *popt)
r_squared = r2_score(lenJ1, predicted_values)

# 绘图
plt.plot(t1[:len(lenJ1)], lenJ1, 'b-', label='Data')
plt.plot(t1[:len(lenJ1)], predicted_values, 'r--', label='Fitted Curve')
plt.xlabel('t1')
plt.ylabel('Values')
plt.title('Optimized Error Function Fit')
plt.legend()
plt.show()

# 输出结果
print("Parameters (a, b, c, d):", popt)
print("R^2 value:", r_squared)

内容的提问来源于stack exchange，提问作者user23495528