零截距模型的lm()中的R平方 [英] R-squared in lm() for zero-intercept model

问题描述

我在R中运行 lm()，这是摘要的结果:

 多个R平方:0.8918，调整后R平方:0.8917F统计量:9和10283 DF上的9416，p值:<2.2e-16 

看来这是一个很好的模型，但是如果我手动计算R ^ 2，我会得到:

  model = lm(S〜0 + C + HA + L1 + L2，data = train)pred =预测(模型，火车)rss<-sum((model $ fitted.values-train $ S)^ 2)tss<-sum((train $ S-平均值(train $ S))^ 2)1-RSS/TSS## [1] 0.247238rSquared(train $ S，(train $ S-model $ fitted.values))## [，1]## [1，] 0.247238 

怎么了?

  str(train [，c('S'，'Campionato'，'HA'，'L1'，'L2')])类别"tbl_df"，"tbl"和"data.frame":10292 obs.5个变量中:$ S:num 19 18 9 12 12 8 21 24 9 8 ...$ C:有6个等级的因子"D"，"E"，"F"，"I"，..:4 4 4 4 4 4 4 4 4 4 ...$ HA:有2个等级的系数"A"，"H":1 2 1 1 2 1 2 2 1 2 ...$ L1:数量0.99 1.41 1.46 1.43 1.12 1.08 1.4 1.45 0.85 1.44 ...$ L2:编号1.31 0.63 1.16 1.15 1.29 1.31 0.7 0.65 1.35 0.59 ... 

解决方案

您正在运行一个没有截距的模型(公式右侧的〜0).对于这些类型的模型，R ^ 2的计算存在问题，并且会产生误导性的值.这篇文章对此进行了很好的解释: https://stats.stackexchange.com/a/26205/99681

I run an lm() in R and this is the results of the summary:

Multiple R-squared:  0.8918,    Adjusted R-squared:  0.8917 
F-statistic:  9416 on 9 and 10283 DF,  p-value: < 2.2e-16

and it seems that it is a good model, but if I calculate the R^2 manually I obtain this:

model=lm(S~0+C+HA+L1+L2,data=train)
pred=predict(model,train)
rss <- sum((model$fitted.values - train$S) ^ 2)
tss <- sum((train$S - mean(train$S)) ^ 2)
1 - rss/tss
##[1] 0.247238
rSquared(train$S,(train$S-model$fitted.values))
##          [,1]
## [1,] 0.247238

What's wrong?

str(train[,c('S','Campionato','HA','L1','L2')])
Classes ‘tbl_df’, ‘tbl’ and 'data.frame':   10292 obs. of  5 variables:
 $ S         : num  19 18 9 12 12 8 21 24 9 8 ...
 $ C         : Factor w/ 6 levels "D","E","F","I",..: 4 4 4 4 4 4 4 4 4 4 ...
 $ HA        : Factor w/ 2 levels "A","H": 1 2 1 1 2 1 2 2 1 2 ...
 $ L1        : num  0.99 1.41 1.46 1.43 1.12 1.08 1.4 1.45 0.85 1.44 ...
 $ L2        : num  1.31 0.63 1.16 1.15 1.29 1.31 0.7 0.65 1.35 0.59 ...

解决方案

You are running a model without the intercept (the ~0 on the right hand side of your formula). For these kinds of models the calculation of R^2 is problematic and will produce misleading values. This post explains it very well: https://stats.stackexchange.com/a/26205/99681

这篇关于零截距模型的lm()中的R平方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！