lin 的留一法 CV 实现。回归
Leave-One-Out CV implementation for lin. regression
我正在 cafe 数据集上构建线性回归,我想通过计算留一法交叉验证来验证结果。
我为此编写了自己的函数,如果我对所有数据都拟合 lm()
,该函数就可以工作,但是当我使用列的子集(来自逐步回归)时,我收到了一个错误。考虑以下代码:
cafe <- read.table("C:/.../cafedata.txt", header=T)
cafe$Date <- as.Date(cafe$Date, format="%d/%m/%Y")
#Delete row 34
cafe <- cafe[-c(34), ]
#wont need date
cafe <- cafe[,-1]
library(DAAG)
#center the data
cafe.c <- data.frame(lapply(cafe[,2:15], function(x) scale(x, center = FALSE, scale = max(x, na.rm = TRUE))))
cafe.c$Day.of.Week <- cafe$Day.of.Week
cafe.c$Sales <- cafe$Sales
#Leave-One-Out CrossValidation function
LOOCV <- function(fit, dataset){
# Attributes:
#------------------------------
# fit: Fit of the model
# dataset: Dataset to be used
# -----------------------------
# Returns mean of squared errors for each fold - MSE
MSEP_=c()
for (idx in 1:nrow(dataset)){
train <- dataset[-c(idx),]
test <- dataset[idx,]
MSEP_[idx]<-(predict(fit, newdata = test) - dataset[idx,]$Sales)^2
}
return(mean(MSEP_))
}
然后当我拟合简单线性模型并调用函数时,它起作用了:
#----------------Simple Linear regression with all attributes-----------------
fit.all.c <- lm(cafe.c$Sales ~., data=cafe.c)
#MSE:258
LOOCV(fit.all.c, cafe.c)
然而,当我仅对列的子集拟合相同的 lm()
时,我得到一个错误:
#-------------------------Linear Stepwise regression--------------------------
step <- stepAIC(fit.all.c, direction="both")
fit.step <- lm(cafe.c$Sales ~ cafe.c$Bread.Sand.Sold + cafe.c$Bread.Sand.Waste
+ cafe.c$Wraps.Waste + cafe.c$Muffins.Sold
+ cafe.c$Muffins.Waste + cafe.c$Fruit.Cup.Sold
+ cafe.c$Chips + cafe.c$Sodas + cafe.c$Coffees
+ cafe.c$Day.of.Week,data=cafe.c)
LOOCV(fit.step, cafe.c)
5495.069
There were 50 or more warnings (use warnings() to see the first 50)
如果我仔细看:
idx <- 1
train <- cafe.c[-c(idx)]
test <- cafe.c[idx]
(predict(fit.step, newdata = test) -cafe.c[idx]$Sales)^2
我得到所有行的 MSE 和一个错误:
Warning message:
'newdata' had 1 row but variables found have 47 rows
编辑
我发现了 this 关于错误的问题,它说当我给列指定不同的名称时会发生此错误,但事实并非如此。
像下面这样更改您的代码:
fit.step <- lm(Sales ~ Bread.Sand.Sold + Bread.Sand.Waste
+ Wraps.Waste + Muffins.Sold
+ Muffins.Waste + Fruit.Cup.Sold
+ Chips + Sodas + Coffees
+ Day.of.Week,data=cafe.c)
LOOCV(fit.step, cafe.c)
# [1] 278.8984
idx <- 1
train <- cafe.c[-c(idx),]
test <- cafe.c[idx,] # need to select the row, not the column
(predict(fit.step, newdata = test) -cafe.c[idx,]$Sales)^2
# 1
# 51.8022
此外,您 LOOCV
实施不正确。每次都必须在留一折叠的训练数据集上拟合一个新模型。现在你在整个数据集上训练模型一次,并使用相同的 single 模型来计算每个留一折叠的保留测试数据集的 MSE,但理想情况下你应该在不同的训练数据集上训练不同的模型。
我正在 cafe 数据集上构建线性回归,我想通过计算留一法交叉验证来验证结果。
我为此编写了自己的函数,如果我对所有数据都拟合 lm()
,该函数就可以工作,但是当我使用列的子集(来自逐步回归)时,我收到了一个错误。考虑以下代码:
cafe <- read.table("C:/.../cafedata.txt", header=T)
cafe$Date <- as.Date(cafe$Date, format="%d/%m/%Y")
#Delete row 34
cafe <- cafe[-c(34), ]
#wont need date
cafe <- cafe[,-1]
library(DAAG)
#center the data
cafe.c <- data.frame(lapply(cafe[,2:15], function(x) scale(x, center = FALSE, scale = max(x, na.rm = TRUE))))
cafe.c$Day.of.Week <- cafe$Day.of.Week
cafe.c$Sales <- cafe$Sales
#Leave-One-Out CrossValidation function
LOOCV <- function(fit, dataset){
# Attributes:
#------------------------------
# fit: Fit of the model
# dataset: Dataset to be used
# -----------------------------
# Returns mean of squared errors for each fold - MSE
MSEP_=c()
for (idx in 1:nrow(dataset)){
train <- dataset[-c(idx),]
test <- dataset[idx,]
MSEP_[idx]<-(predict(fit, newdata = test) - dataset[idx,]$Sales)^2
}
return(mean(MSEP_))
}
然后当我拟合简单线性模型并调用函数时,它起作用了:
#----------------Simple Linear regression with all attributes-----------------
fit.all.c <- lm(cafe.c$Sales ~., data=cafe.c)
#MSE:258
LOOCV(fit.all.c, cafe.c)
然而,当我仅对列的子集拟合相同的 lm()
时,我得到一个错误:
#-------------------------Linear Stepwise regression--------------------------
step <- stepAIC(fit.all.c, direction="both")
fit.step <- lm(cafe.c$Sales ~ cafe.c$Bread.Sand.Sold + cafe.c$Bread.Sand.Waste
+ cafe.c$Wraps.Waste + cafe.c$Muffins.Sold
+ cafe.c$Muffins.Waste + cafe.c$Fruit.Cup.Sold
+ cafe.c$Chips + cafe.c$Sodas + cafe.c$Coffees
+ cafe.c$Day.of.Week,data=cafe.c)
LOOCV(fit.step, cafe.c)
5495.069
There were 50 or more warnings (use warnings() to see the first 50)
如果我仔细看:
idx <- 1
train <- cafe.c[-c(idx)]
test <- cafe.c[idx]
(predict(fit.step, newdata = test) -cafe.c[idx]$Sales)^2
我得到所有行的 MSE 和一个错误:
Warning message: 'newdata' had 1 row but variables found have 47 rows
编辑
我发现了 this 关于错误的问题,它说当我给列指定不同的名称时会发生此错误,但事实并非如此。
像下面这样更改您的代码:
fit.step <- lm(Sales ~ Bread.Sand.Sold + Bread.Sand.Waste
+ Wraps.Waste + Muffins.Sold
+ Muffins.Waste + Fruit.Cup.Sold
+ Chips + Sodas + Coffees
+ Day.of.Week,data=cafe.c)
LOOCV(fit.step, cafe.c)
# [1] 278.8984
idx <- 1
train <- cafe.c[-c(idx),]
test <- cafe.c[idx,] # need to select the row, not the column
(predict(fit.step, newdata = test) -cafe.c[idx,]$Sales)^2
# 1
# 51.8022
此外,您 LOOCV
实施不正确。每次都必须在留一折叠的训练数据集上拟合一个新模型。现在你在整个数据集上训练模型一次,并使用相同的 single 模型来计算每个留一折叠的保留测试数据集的 MSE,但理想情况下你应该在不同的训练数据集上训练不同的模型。