python中的方差分析使用带有statsmodels或scipy的pandas数据框? [英] ANOVA in python using pandas dataframe with statsmodels or scipy?

问题描述

我想使用 Pandas 数据框来细分一个变量的方差.

I want to use the Pandas dataframe to breakdown the variance in one variable.

例如，如果我有一个名为度数"的列，并且我为不同的日期、城市和夜晚与白天建立了索引，我想找出这个系列中变异的哪一部分来自交叉- 部分城市变化，有多少来自时间序列变化，有多少来自夜间与白天.

For example, if I have a column called 'Degrees', and I have this indexed for various dates, cities, and night vs. day, I want to find out what fraction of the variation in this series is coming from cross-sectional city variation, how much is coming from time series variation, and how much is coming from night vs. day.

在 Stata 中，我会使用固定效果并查看 R^2.希望我的问题有意义.

In Stata I would use Fixed effects and look at the R^2. Hopefully my question makes sense.

基本上，我想要做的是通过其他三个列找到学位"的方差分析细分.

Basically, what I want to do, is find the ANOVA breakdown of "Degrees" by three other columns.

推荐答案

我设置了一个直接对比来测试他们，发现他们的假设可以稍微不同 ，从统计学家那里得到了一个提示，这里是一个关于熊猫的方差分析的例子数据框匹配 R 的结果:

I set up a direct comparison to test them, found that their assumptions can differ slightly , got a hint from a statistician, and here is an example of ANOVA on a pandas dataframe matching R's results:

import pandas as pd
import statsmodels.api as sm
from statsmodels.formula.api import ols


# R code on R sample dataset

#> anova(with(ChickWeight, lm(weight ~ Time + Diet)))
#Analysis of Variance Table
#
#Response: weight
#           Df  Sum Sq Mean Sq  F value    Pr(>F)
#Time        1 2042344 2042344 1576.460 < 2.2e-16 ***
#Diet        3  129876   43292   33.417 < 2.2e-16 ***
#Residuals 573  742336    1296
#write.csv(file='ChickWeight.csv', x=ChickWeight, row.names=F)

cw = pd.read_csv('ChickWeight.csv')

cw_lm=ols('weight ~ Time + C(Diet)', data=cw).fit() #Specify C for Categorical
print(sm.stats.anova_lm(cw_lm, typ=2))
#                  sum_sq   df            F         PR(>F)
#C(Diet)    129876.056995    3    33.416570   6.473189e-20
#Time      2016357.148493    1  1556.400956  1.803038e-165
#Residual   742336.119560  573          NaN            NaN

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