numpy polyfit中使用的权重值是多少，拟合的误差是什么 [英] What are the weight values to use in numpy polyfit and what is the error of the fit

问题描述

我正在尝试对numpy中的某些数据进行线性拟合.

I'm trying to do a linear fit to some data in numpy.

Ex(其中w是该值的样本数量，即对于点(x=0, y=0)，我只有1次测量，而该测量值是2.2，但对于点(1,1)，我具有3.5的2次测量.

Ex (where w is the number of samples I have for that value, i.e. for the point (x=0, y=0) I only have 1 measurement and the value of that measurement is 2.2, but for the point (1,1) I have 2 measurements with a value of 3.5.

x = np.array([0, 1, 2, 3])
y = np.array([2.2, 3.5, 4.6, 5.2])
w = np.array([1, 2, 2, 1])

z = np.polyfit(x, y, 1, w = w)

因此，现在的问题是: 在这些情况下，在Polyfit中使用w=w是正确的还是应该使用我应该使用的w = sqrt(w)?

So, now the question is: is it correct to use w=w in polyfit for these cases or should I use w = sqrt(w) of what should I use?

此外，如何从polyfit中获取拟合误差?

Also, how can I get the fit error from polyfit?

推荐答案

如果测量结果呈正态分布，则每个值的不确定性将与1/sqrt(n)成比例，其中n是测量次数.您想用不确定性的倒数来衡量您的健康状况，所以第二个猜测是最好的:w=np.sqrt(n)

If you have normally distributed measurements, then your uncertainty in each value would be proportional to 1/sqrt(n) where n is the number of measurements. You want to weigh your fit by the inverse of your uncertainty, so your second guess is best: w=np.sqrt(n)

要获取参数的协方差，请同时输入cov=True.

To get the covariance on your parameters, also give cov=True.

x = np.array([0, 1, 2, 3])
y = np.array([2.2, 3.5, 4.6, 5.2])
n = np.array([1, 2, 2, 1])

p, c = np.polyfit(x, y, 1, w=np.sqrt(n), cov=True)

cov矩阵的对角线是每个参数的单独方差，当然，非对角线是协方差.因此，最有可能需要拟合误差"的是这些对角线的平方根:

The diagonals of your cov matrix are the individual variances on each parameter, and of course the off-diagonals are the covariances. So most likely what you want for "fit error" is the square root of these diagonals:

e = np.sqrt(np.diag(c))

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