如何计算python中列表的方差?

How can I calculate the variance of a list in python?(如何计算python中列表的方差?)

问题描述

如果我有这样的列表:

results=[-14.82381293, -0.29423447, -13.56067979, -1.6288903, -0.31632439,
          0.53459687, -1.34069996, -1.61042692, -4.03220519, -0.24332097]

我想在 Python 中计算此列表的方差，即均值的平方差的平均值.

I want to calculate the variance of this list in Python which is the average of the squared differences from the mean.

我该怎么办?访问列表中的元素来进行计算让我无法获得平方差.

How can I go about this? Accessing the elements in the list to do the computations is confusing me for getting the square differences.

推荐答案

您可以使用 numpy 的内置函数 var:

You can use numpy's built-in function var:

import numpy as np

results = [-14.82381293, -0.29423447, -13.56067979, -1.6288903, -0.31632439,
          0.53459687, -1.34069996, -1.61042692, -4.03220519, -0.24332097]

print(np.var(results))

这给你 28.822364260579157

如果 - 无论出于何种原因 - 您不能使用 numpy 和/或您不想为其使用内置函数，您也可以使用例如手动"计算它列表理解:

If - for whatever reason - you cannot use numpy and/or you don't want to use a built-in function for it, you can also calculate it "by hand" using e.g. a list comprehension:

# calculate mean
m = sum(results) / len(results)

# calculate variance using a list comprehension
var_res = sum((xi - m) ** 2 for xi in results) / len(results)

这会给你相同的结果.

如果您对标准差感兴趣，可以使用numpy.std:

If you are interested in the standard deviation, you can use numpy.std:

print(np.std(results))
5.36864640860051

@Serge Ballesta 很好地解释了方差 n 和 n 之间的区别-1.在 numpy 中，您可以使用选项 ddof 轻松设置此参数；它的默认值是 0，所以对于 n-1 的情况你可以简单地做:

@Serge Ballesta explained very well the difference between variance n and n-1. In numpy you can easily set this parameter using the option ddof; its default is 0, so for the n-1 case you can simply do:

np.var(results, ddof=1)

@Serge Ballesta 的回答中给出了手工"解决方案.

The "by hand" solution is given in @Serge Ballesta's answer.

两种方法都产生 32.024849178421285.

你也可以为std设置参数:

np.std(results, ddof=1)
5.659050201086865

这篇关于如何计算python中列表的方差?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编程学习网！