前文我們講到線性迴歸建模會有共線性的問題，嶺迴歸和lasso演算法都能一定程度上消除共線性問題。

嶺迴歸

> #########正則化方法消除共線性
> ###嶺迴歸
> ###glmnet只能處理矩陣
> library(glmnet)
> library(mice)
> creditcard_exp<-creditcard_exp[complete.cases(creditcard_exp),]
> x<-as.matrix(creditcard_exp[,c(6,7,10,11)])
> y<-as.matrix(creditcard_exp[,3])
 
> #看一下嶺脊圖
> r1<-glmnet(x=x,y=y,family = "gaussian",alpha = 0)#alpha = 0表示嶺迴歸,x,y不能有缺失值             
> plot(r1,xvar="lambda")

> r1cv<-cv.glmnet(x=x,y=y,family="gaussian",alpha=0,nfolds = 10)#用交叉驗證得到lambda
> plot(r1cv)

> rimin<-glmnet(x=x,y=y,family = "gaussian",alpha = 0,lambda
 
 = r1cv$lambda.min)#取誤差平方和最小時的λ
> coef(rimin)
5 x 1 sparse Matrix of class "dgCMatrix"
                         s0
(Intercept)     106.5467017
Age               0.9156047
Income           19.6903291
dist_home_val     1.7357213
dist_avg_income  71.5765458

我們可以看到這次模型的收入和支出是正相關了。

lasso演算法

#####Lasson演算法：有變數篩選功效
 

r1l<-cv.glmnet(x=x,y=y,family="gaussian",alpha=1,nfolds = 10)
plot(r1l)

> r1l1<-glmnet(x=x,y=y,family = "gaussian",alpha = 1,lambda = r1l$lambda.min)#取λ最小值看建模情況
> coef(r1l1)
5 x 1 sparse Matrix of class "dgCMatrix"
                         s0
(Intercept)      -27.169039
Age                1.314711
Income          -160.195837
dist_home_val      1.538823
dist_avg_income  255.395751

看模型資料，我們得知並沒有解決income為負相關的情況，而且並沒有篩選變數，那麼我們嘗試取lambda.1se*0.5的值

> r1l2<-glmnet(x=x,y=y,family = "gaussian",alpha = 1,lambda = r1l$lambda.1se*0.5)#0.5倍標準誤差的λ
> coef(r1l2)
5 x 1 sparse Matrix of class "dgCMatrix"
                         s0
(Intercept)     267.0510318
Age               .        
Income            .        
dist_home_val     0.6249539
dist_avg_income  83.6952253

看結果，可知把一些變數刪去了，消除共線性的問題，接下來我們看看lambda.1se的值

1 > r1l3<-glmnet(x=x,y=y,alpha = 1,family = "gaussian",lambda = r1l$lambda.1se)
2 > coef(r1l3)
3 5 x 1 sparse Matrix of class "dgCMatrix"
4                        s0
5 (Intercept)     432.00684
6 Age               .      
7 Income            .      
8 dist_home_val     .      
9 dist_avg_income  68.90894

這次結果只留了一個變數，由此可知當lambda越大，變數保留的越少，一般我們在誤差最小和一倍標準差內選擇合適的λ。