试图找出岭回归的测试和训练误差作为样本大小的函数

问题描述

我在 R 中使用 Hitters 数据集。目前我拟合了一个线性回归，从所有其他协变量预测工资，样本大小从 20 到 75 不等，我计算了平均测试/训练错误：

data("Hitters",package = 'ISLR')
Hitters = na.omit(Hitters)
set.seed(1)
train.idx = sample(1:nrow(Hitters),75,replace=FALSE)
train = Hitters[train.idx,-20]
test = Hitters[-train.idx,-20]

errs <- rep(NA,56)
for (ii in 20:75){
  train.idx = sample(1:nrow(Hitters),ii,replace=FALSE)
  train = Hitters[train.idx,-20]
  test = Hitters[-train.idx,-20]
  train.lm <- lm(Salary ~.,- Salary,data = train)
  train.pred <- predict(train.lm,train)
  test.pred <- predict(train.lm,data = test)
  errs[ii-19] <-  mean((test.pred - train$Salary)^2)
}
errs

现在，我尝试使用我之前创建的正则化参数为 20 的样本对 Ridge 回归执行相同的操作。我尝试过：

x_train = model.matrix(Salary~.,train)[,-1]
x_test = model.matrix(Salary~.,test)[,-1]

y_train = train$Salary
y_test = test$Salary

#cv.out = cv.glmnet(x_train,y_train,alpha = 0)
#lam = cv.out$lambda.min


errs.train <- rep(NA,56)
for (ii in 20:75){
  ridge_mod = glmnet(x_train,alpha=0,lambda = 20)
  ridge_pred = predict(ridge_mod,newx = x_test)
  #errs.test[ii] <- mean((ridge_pred - y_test)^2)
  errs.train[ii-19] <- mean((ridge_pred - y_train)^2)
}

errs.train

但是所有的错误都是一样的。我该如何解决这个问题？

解决方法

lm 的第一部分代码存在一些错误。它应该是 predict(train.lm,newdata = test) 而不是 predict(train.lm,data = test) 。如果您不确定输入，请执行 ?predict.lm。其次，如果您对测试集中的错误感兴趣，您应该用 test 和 test$Salary 中的值减去 train 的预测。像下面这样的东西应该可以工作：

data("Hitters",package = 'ISLR')
Hitters = na.omit(Hitters)
set.seed(1)

sample_size = 20:75
errs = vector("numeric",length(sample_size))
for (ii in seq_along(sample_size)){
  train.idx = sample(1:nrow(Hitters),sample_size[ii],replace=FALSE)
  train = Hitters[train.idx,-20]
  test = Hitters[-train.idx,-20]
  train.lm <- lm(Salary ~.,data = train)
  test.pred <- predict(train.lm,newdata = test)
  errs[ii] <-  mean((test.pred - test$Salary)^2)
}

现在对于岭，唯一的区别是您在每次迭代时创建模型矩阵和子集：

errs.test = vector("numeric",length(sample_size))

x_data = model.matrix(Salary~.,Hitters)[,-1]
y_data = Hitters$Salary

for (ii in seq_along(sample_size)){

  train.idx = sample(1:nrow(x_data),replace=FALSE)

  x_train = x_data[train.idx,]
  x_test = x_data[-train.idx,]
  y_train = y_data[train.idx]
  y_test = y_data[-train.idx]

  ridge_mod = glmnet(x_train,y_train,alpha=0,lambda = 20)
  ridge_pred = predict(ridge_mod,newx = x_test)

  errs.test[ii] <- mean((ridge_pred - y_test)^2)
}