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如何计算python中非线性曲线拟合的决定系数(R2)和均方根误差(RMSE)。以下代码一直执行到曲线拟合。那么如何计算R2和RMSE呢?

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

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

x = np.linspace(0,4,50)
y = func(x, 2.5, 1.3, 0.5)
yn = y + 0.2*np.random.normal(size=len(x))

popt, pcov = curve_fit(func, x, yn)

plt.figure()
plt.plot(x, yn, 'ko', label="Original Noised Data")
plt.plot(x, func(x, *popt), 'r-', label="Fitted Curve")
plt.legend()
plt.show()
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2 回答 2

5

你可以这样做:

print "Mean Squared Error: ", np.mean((y-func(x, *popt))**2)

ss_res = np.dot((yn - func(x, *popt)),(yn - func(x, *popt)))
ymean = np.mean(yn)
ss_tot = np.dot((yn-ymean),(yn-ymean))
print "Mean R :",  1-ss_res/ss_tot

这是直接采用定义,例如在维基百科中: http ://en.wikipedia.org/wiki/Coefficient_of_determination#Definitions

于 2013-11-21T08:30:32.147 回答
0

Martin Böschen,不是yyn在这里:

np.mean((y-func(x, *popt))**2)

并阅读有关均方根误差 (RMSE) 的信息:http ://en.wikipedia.org/wiki/Regression_analysis

residuals = yn - func(x,*popt)
print "RMSE",(scipy.sum(residuals**2)/(residuals.size-2))**0.5

现在它计算为 Excel 2003 分析工具库。

于 2014-11-24T01:15:06.403 回答