R:使用cv.glm计算弹性网预测误差

Question

library(glmnet)
library(boot)
data(iris)
x <- model.matrix(Sepal.Length~., iris)[,-1]
y <- iris$Sepal.Length
m <- cv.glmnet(x, y)
> cv.glm(x, m, K = 10)
Error in UseMethod("predict") : 
  no applicable method for 'predict' applied to an object of class "c('matrix', 'double', 'numeric')"

或

bestLambda = m$lambda.min
m2 <- glmnet(x, y, family = "gaussian", lambda = bestLambda)
>cv.glm(x, m2, K = 10)


 Error in glmnet(x = x, y = y, family = "gaussian", lambda = bestLambda,  : 
  unused argument (data = c(3.5, 3, 3.2, 3.1, 3.6,

引用this问题，我正在尝试使用 cv.glm 获取我的弹性网络模型的 K 倍交叉验证预测误差，但是，由于该错误，我似乎无法这样做。我不太确定 cv.glm 函数是否可用于计算类 cv.glm 或 glmnet 对象的预测误差。

Answer 1

我认为您将 glm 与 glmnet (弹性网，带有套索和山脊惩罚)混淆了。 cv.glm 需要 glm 模型，而不是 glmnet 模型。

尝试以下任一方法:

1. 使用 glmnet 与 cv.glmnet 计算 k 倍交叉验证错误，如下所示:

   library(glmnet)
   library(boot)
   data(iris)
   x <- model.matrix(Sepal.Length~., iris)[,-1]
   y <- iris$Sepal.Length
   m <- cv.glmnet(x, y, nfolds=10)
   m$lambda.min
   #[1] 0.0003839539
   m$lambda.1se
   #[1] 0.009078549
   plot(m$lambda, m$cvm,type='l', xlab=expression(lambda), ylab='CV errors', main=expression(paste('CV error for different ', lambda)))
   lines(m$lambda, m$cvup, col='red')
   lines(m$lambda, m$cvlo, col='red')
   

   

[已编辑]

训练数据集的预测误差:

mean((y-predict(m, newx=x))^2)
# [1] 0.1037433

拟合 glm 模型并使用 cv.glm 计算交叉验证误差增量(无需正则化)。根据cv.glm的文档:

delta: A vector of length two. The first component is the raw cross-validation estimate of prediction error. The second component is the adjusted cross-validation estimate. The adjustment is designed to compensate for the bias introduced by not using leave-one-out cross-validation.

df <- cbind.data.frame(x, y)
m <- glm(y~., df, family='gaussian')
cv.glm(df, m, K = 10)$delta 
# [1] 0.09992177 0.09940190