估计python中分布参数的置信区间 [英] Estimate confidence intervals for parameters of distribution in python

问题描述

是否有一个内置函数可以为python软件包中的参数估计提供置信区间，或者这是我需要手动实现的功能?我正在寻找类似于matlabs gevfit http://www.mathworks.com/help/的东西stats/gevfit.html .

Is there a built in function that will provide the confidence intervals for parameter estimates in a python package or is this something I will need to implement by hand? I am looking for something similar to matlabs gevfit http://www.mathworks.com/help/stats/gevfit.html.

推荐答案

引导程序可用于估计任何函数的置信区间( np.mean ， st.genextreme.fit 等)，还有一个Python库: scikits.bootstrap .

The bootstrap can be used to estimate confidence intervals of any function (np.mean, st.genextreme.fit, etc.) of a sample, and there is a Python library: scikits.bootstrap.

此处提供了问题作者相关的问题中的数据:

Here for the data from the question author's related question:

import numpy as np, scipy.stats as st, scikits.bootstrap as boot
data = np.array([ 22.20379411,  22.99151292,  24.27032696,  24.82180626,
  25.23163221,  25.39987272,  25.54514567,  28.56710007,
  29.7575898 ,  30.15641696,  30.79168255,  30.88147532,
  31.0236419 ,  31.17380647,  31.61932755,  32.23452568,
  32.76262978,  33.39430032,  33.81080069,  33.90625861,
  33.99142006,  35.45748368,  37.0342621 ,  37.14768791,
  38.14350221,  42.72699534,  44.16449992,  48.77736737,
  49.80441736,  50.57488779])

st.genextreme.fit(data)   # just to check the parameters
boot.ci(data, st.genextreme.fit)

结果是

(-0.014387281261850815, 29.762126238637851, 5.8983127779873605)
array([[ -0.40002507,  26.93511496,   4.6677834 ],
       [  0.19743722,  32.41834882,   9.05026202]])

引导程序在我的计算机上需要大约三分钟的时间；默认情况下， boot.ci 使用10,000个引导程序迭代( n_samples )，请参见

The bootstrap takes about three minutes on my machine; by default, boot.ci uses 10,000 bootstrap iterations (n_samples), see code or help(boot.ci), and st.genextreme.fit is not superfast.

boot.ci 的置信区间与MATLAB的

The confidence intervals from boot.ci do not match the ones from MATLAB's gevfit exactly. E.g., MATLAB gives a symmetric one [-0.3032, 0.3320] for the first parameter (0.0144).

这篇关于估计python中分布参数的置信区间的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！