从 polyfit 中找出不确定性

Question

我使用 2 阶的简单 polyfit 来拟合示例数据中的一行：

np.polyfit(x, y, 2)

其中 returns 个系数。

现在我想找到拟合线的不确定性，并尝试使用 cov 参数，其中 returns 3x3 协方差矩阵：

np.polyfit(x, y, 2, cov=True)

但我不确定如何计算不确定性，根据我的 Google 搜索应该通过对协方差矩阵的对角线求平方来计算。

Answer 1

此问题已由 P.H "Estimating Errors in Least-Squares Fitting" 解决。 Richter, 1995, TDA 进度报告 42-122.

从报告来看，这一段对你来说可能已经足够了

The first instance considered above, namely, determining the error of one or more fitting parameters, has a straightforward answer given in terms of the diagonal elements of the covariance matrix of the fit, and is well known.

您感兴趣的对角线元素例如：

x = linspace(0,1,1000)
# comment and uncomment the last term to see how the fit appears in the figure,
# and how the covariances of the single polynomial coefficients vary in turn.
y = cos(x)*x**2+x+sin(x-1.) #+(x*1.3)**6
p,cov = polyfit(x,y,2,cov=True)
plot(x,y,'b')
plot(x,polyval(p,x),'r')
print sqrt(diag(cov))

更一般地说，参考文献解决了多项式系数中的这个误差如何也是因变量 y 作为自变量 x 的函数的误差。来自报告：

It is the purpose of this article to discuss the above errors and, in particular, to present results that will permit one to determine the standard error of the fit as a function of the independent variable, as well as to establish confidence limits for these errors.

Answer 2

为了您的方便，我根据 gg349 的回答为 Python 3 制作了一个完整的示例。

import numpy as np
import matplotlib.pyplot as plt 

x = np.linspace(0,1,1000)
# comment and uncomment the last term to see how the fit appears in the figure,
# and how the covariances of the single polynomial coefficients vary in turn.
y = np.cos(x) * x**2 + x + np.sin(x - 1.) \
#     + (x * 1.3)**6

p, cov = np.polyfit(x, y, 2, cov=True)

plt.plot(x, y)
plt.plot(x, np.polyval(p,x))
plt.show()

print(np.sqrt(np.diag(cov)))

从 polyfit 中找出不确定性

Find uncertainty from polyfit

python

numpy