R中子样本的平均值的预测

问题描述 投票:0回答:1

这是我的代码:

install.packages("wooldridge")
library(wooldridge)
data(wage1)
wage1reg<-lm(lwage ~ educ + exper + tenure, data = wage1) 

我正在尝试根据其平均值预测这三个变量的拟合值。什么是正确的代码?这些东西不起作用:

wagemeans<-data.frame(c(colMeans(data.frame(wage1$educ, wage1$exper, wage1$tenure))))
predict(wage1reg, newdata=wagemeans)

此外,如果我尝试其他操作,例如

wage1.subset1.means<-data.frame(colMeans(subset(wage1, select=c(educ, exper, tenure))))
predict(wage1reg, newdata=wage1.subset1.means)

也不起作用。

查找三个变量均值的预测值的最佳方法是什么?

r predict
1个回答
0
投票

工资平均值数据框的格式不适合预报()。改用它

library(wooldridge)
data(wage1)

wage1reg<-lm(lwage ~ educ + exper + tenure, data = wage1) 

wagemeans<-data.frame(educ=mean(wage1$educ),exper=mean(wage1$exper),tenure=mean(wage1$tenure))

predict(wage1reg,newdata = wagemeans)

#output
#     1 
# 1.623268 
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