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R 《回归分析与线性统计模型》page140,5.1

时间:2019-11-03 01:36:58      阅读:101      评论:0      收藏:0      [点我收藏+]
rm(list = ls())
library(car)
library(MASS)
library(openxlsx)
A = read.xlsx("data140.xlsx")
head(A)
attach(A)

  

fm = lm(y~x1+x2+x3 , data=A) #建立模型
vif(fm)                      #查看模型是否存在共线性

> vif(fm) #查看模型是否存在共线性
x1 x2 x3
21.631451 21.894402 1.334751

结果显示存在共线性

summary(fm)

 结果:

> summary(fm)

Call:
lm(formula = y ~ x1 + x2 + x3, data = A)

Residuals:
Min 1Q Median 3Q Max
-2.89129 -0.78230 0.00544 0.93147 2.45478

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 3.2242 3.4598 0.932 0.361983
x1 0.9626 0.2422 3.974 0.000692 ***
x2 -2.6290 3.9000 -0.674 0.507606
x3 -0.1560 3.8838 -0.040 0.968338
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.446 on 21 degrees of freedom
Multiple R-squared: 0.9186, Adjusted R-squared: 0.907
F-statistic: 78.99 on 3 and 21 DF, p-value: 1.328e-11

esti_ling = lm.ridge(y~x1+x2+x3 , data=A,lambda = seq(0,1,0.1)) #岭回归
plot(esti_ling)

  技术分享图片

 

 选取k = 0.6

k = 0.6
X = cbind(1,as.matrix(A[,2:4]))
y = A[,5]
B_ = solve((t(X)%*%X) + k*diag(4))%*%t(X)%*%y
B_

  

回归系数:

> B_
         [,1]
    1.6188146
x1  0.8262986
x2 -0.3076330
x3  1.0780444

  

 

R 《回归分析与线性统计模型》page140,5.1

原文:https://www.cnblogs.com/jiaxinwei/p/11784767.html

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