• R 《回归分析与线性统计模型》page140,5.1


    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)
    

      

     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
    

      

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