处理 R 线性回归中的嵌套变量
Dealing with nested variables in R linear regression
我有一个包含一些嵌套变量的数据集。例如，我有以下变量：
一辆车的speed，是否有另一辆车跟随它other_car，如果有另一辆车，两辆车之间的距离distance。虚拟数据集：
speed <- c(30,50,60,30,33,54,65,33,33,54,65,34,45,32)
other_car <- c(0,1,0,0,0,1,1,1,1,0,1,0,1,0)
distance <- c(NA,20,NA,NA,NA,21,5,15,17,NA,34,NA,13,NA)
dft <- data.frame(speed,other_car,distance)
我想以嵌套变量的形式将变量 other_car 和 distance 包含在模型中，即如果有汽车，还要考虑距离。按照此处提到的方法： https://stats.stackexchange.com/questions/372257/how-do-you-deal-with-nested-variables-in-a-regression-model ，我尝试了以下操作：
dft <- data.frame(speed,other_car,distance)
dft$other_car<-factor(dft$other_car)
lm_speed <- lm(speed ~ dft$other_car + dft$other_car:dft$distance)
summary(lm_speed)
这给出了以下错误：
Error in contrasts<-(*tmp*, value = contr.funs[1 + isOF[nn]]) : contrasts can be applied only to factors with 2 or more levels
有什么想法吗？
这是因为当other_car==0时，距离都等于NA，see:
dft$distance[dft$other_car==0]
[1] NA NA NA NA NA NA NA
您可以指定一个常数距离来替换 NA 代替 other_car==0，这样模型就可以使用因子 other_car==0 并发现该距离对该子集没有影响：
dft$distance[dft$other_car==0]<-0
dft$other_car<- factor(dft$other_car)
lm_speed <- lm(speed ~ other_car + other_car:distance, data = dft)
summary(lm_speed)
Call:
lm(formula = speed ~ other_car + other_car:distance, data = dft)
Residuals:
Min 1Q Median 3Q Max
-16.015 -8.500 -3.876 8.894 21.000
Coefficients: (1 not defined because of singularities)
Estimate Std. Error t value Pr(>|t|)
(Intercept) 39.0000 5.0405 7.737 8.96e-06 ***
other_car1 4.6480 13.0670 0.356 0.729
other_car0:distance NA NA NA NA
other_car1:distance 0.3157 0.6133 0.515 0.617
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 13.34 on 11 degrees of freedom
Multiple R-squared: 0.1758, Adjusted R-squared: 0.026
F-statistic: 1.174 on 2 and 11 DF, p-value: 0.3452
另一种解决方法是将 factor 转换为 numeric，但是 isn't the same model:
speed <- c(30,50,60,30,33,54,65,33,33,54,65,34,45,32)
other_car <- c(0,1,0,0,0,1,1,1,1,0,1,0,1,0)
distance <- c(NA,20,NA,NA,NA,21,5,15,17,NA,34,NA,13,NA)
dft <- data.frame(speed,other_car,distance)
dft$other_car<- as.numeric(factor(dft$other_car))
lm_speed <- lm(speed ~ other_car + other_car:distance, data = dft)
summary(lm_speed)
Call:
lm(formula = speed ~ other_car + other_car:distance, data = dft)
Residuals:
2 6 7 8 9 11 13
0.03776 3.72205 19.77341 -15.38369 -16.01511 10.61782 -2.75227
Coefficients: (1 not defined because of singularities)
Estimate Std. Error t value Pr(>|t|)
(Intercept) 43.6480 12.9010 3.383 0.0196 *
other_car NA NA NA NA
other_car:distance 0.1579 0.3281 0.481 0.6508
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 14.27 on 5 degrees of freedom
(7 observations deleted due to missingness)
Multiple R-squared: 0.04424, Adjusted R-squared: -0.1469
F-statistic: 0.2314 on 1 and 5 DF, p-value: 0.6508
这表明速度随着与另一辆车的距离增加而增加（或者相反，当另一辆车离得太近时，司机往往会减速）。
我有一个包含一些嵌套变量的数据集。例如，我有以下变量：
一辆车的speed，是否有另一辆车跟随它other_car，如果有另一辆车，两辆车之间的距离distance。虚拟数据集：
speed <- c(30,50,60,30,33,54,65,33,33,54,65,34,45,32)
other_car <- c(0,1,0,0,0,1,1,1,1,0,1,0,1,0)
distance <- c(NA,20,NA,NA,NA,21,5,15,17,NA,34,NA,13,NA)
dft <- data.frame(speed,other_car,distance)
我想以嵌套变量的形式将变量 other_car 和 distance 包含在模型中，即如果有汽车，还要考虑距离。按照此处提到的方法： https://stats.stackexchange.com/questions/372257/how-do-you-deal-with-nested-variables-in-a-regression-model ，我尝试了以下操作：
dft <- data.frame(speed,other_car,distance)
dft$other_car<-factor(dft$other_car)
lm_speed <- lm(speed ~ dft$other_car + dft$other_car:dft$distance)
summary(lm_speed)
这给出了以下错误：
Error in
contrasts<-(*tmp*, value = contr.funs[1 + isOF[nn]]) : contrasts can be applied only to factors with 2 or more levels
有什么想法吗？
这是因为当other_car==0时，距离都等于NA，see:
dft$distance[dft$other_car==0]
[1] NA NA NA NA NA NA NA
您可以指定一个常数距离来替换 NA 代替 other_car==0，这样模型就可以使用因子 other_car==0 并发现该距离对该子集没有影响：
dft$distance[dft$other_car==0]<-0
dft$other_car<- factor(dft$other_car)
lm_speed <- lm(speed ~ other_car + other_car:distance, data = dft)
summary(lm_speed)
Call:
lm(formula = speed ~ other_car + other_car:distance, data = dft)
Residuals:
Min 1Q Median 3Q Max
-16.015 -8.500 -3.876 8.894 21.000
Coefficients: (1 not defined because of singularities)
Estimate Std. Error t value Pr(>|t|)
(Intercept) 39.0000 5.0405 7.737 8.96e-06 ***
other_car1 4.6480 13.0670 0.356 0.729
other_car0:distance NA NA NA NA
other_car1:distance 0.3157 0.6133 0.515 0.617
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 13.34 on 11 degrees of freedom
Multiple R-squared: 0.1758, Adjusted R-squared: 0.026
F-statistic: 1.174 on 2 and 11 DF, p-value: 0.3452
另一种解决方法是将 factor 转换为 numeric，但是 isn't the same model:
speed <- c(30,50,60,30,33,54,65,33,33,54,65,34,45,32)
other_car <- c(0,1,0,0,0,1,1,1,1,0,1,0,1,0)
distance <- c(NA,20,NA,NA,NA,21,5,15,17,NA,34,NA,13,NA)
dft <- data.frame(speed,other_car,distance)
dft$other_car<- as.numeric(factor(dft$other_car))
lm_speed <- lm(speed ~ other_car + other_car:distance, data = dft)
summary(lm_speed)
Call:
lm(formula = speed ~ other_car + other_car:distance, data = dft)
Residuals:
2 6 7 8 9 11 13
0.03776 3.72205 19.77341 -15.38369 -16.01511 10.61782 -2.75227
Coefficients: (1 not defined because of singularities)
Estimate Std. Error t value Pr(>|t|)
(Intercept) 43.6480 12.9010 3.383 0.0196 *
other_car NA NA NA NA
other_car:distance 0.1579 0.3281 0.481 0.6508
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 14.27 on 5 degrees of freedom
(7 observations deleted due to missingness)
Multiple R-squared: 0.04424, Adjusted R-squared: -0.1469
F-statistic: 0.2314 on 1 and 5 DF, p-value: 0.6508
这表明速度随着与另一辆车的距离增加而增加（或者相反，当另一辆车离得太近时，司机往往会减速）。