R语言update函数（模型修正函数）

toothpaste<-data.frame(
  X1=c(-0.05, 0.25,0.60,0, 0.25,0.20, 0.15,0.05,-0.15, 0.15,
       0.20, 0.10,0.40,0.45,0.35,0.30, 0.50,0.50, 0.40,-0.05,
       -0.05,-0.10,0.20,0.10,0.50,0.60,-0.05,0, 0.05, 0.55),
  X2=c( 5.50,6.75,7.25,5.50,7.00,6.50,6.75,5.25,5.25,6.00,
        6.50,6.25,7.00,6.90,6.80,6.80,7.10,7.00,6.80,6.50,
        6.25,6.00,6.50,7.00,6.80,6.80,6.50,5.75,5.80,6.80),
  Y =c( 7.38,8.51,9.52,7.50,9.33,8.28,8.75,7.87,7.10,8.00,
        7.89,8.15,9.10,8.86,8.90,8.87,9.26,9.00,8.75,7.95,
        7.65,7.27,8.00,8.50,8.75,9.21,8.27,7.67,7.93,9.26)
)
lm.sol<-lm(Y~X1+X2, data=toothpaste)
summary(lm.sol)
attach(toothpaste)
plot(Y~X1)
abline(lm(Y~X1)) #绘制x1与Y的散点图
lm2.sol<-lm(Y~X2+I(X2^2))
x<-seq(min(X2), max(X2), len=200)
y<-predict(lm2.sol, data.frame(X2=x))
plot(Y~X2)
lines(x,y)
lm.new<-update(lm.sol, .~.+I(X2^2)) #修正拟合模型
summary(lm.new)
lm2.new<-update(lm.new, .~.-X2)#去掉x2的一次项
lm3.new<-update(lm.new, .~.+X1*X2) #考虑x1与x2的交互作用---------------------本文来自 lijinxiu123 的CSDN 博客 ，全文地址请点击：https://blog.csdn.net/lijinxiu123/article/details/50939299?utm_source=copy