为什么在套索回归中计算MSE会产生不同的输出？

2023年2月12日 256次阅读

我试图从lasso2包中对前列腺癌数据运行不同的回归模型.当我使用Lasso时,我看到了两种不同的方法来计算均方误差.但它们确实给了我不同的结果,所以我想知道我做错了什么,或者它只是意味着一种方法比另一种更好？

# Needs the following R packages.
library(lasso2)
library(glmnet)

# Gets the prostate cancer dataset
data(Prostate)

# Defines the Mean Square Error function 
mse = function(x,y) { mean((x-y)^2)}

# 75% of the sample size.
smp_size = floor(0.75 * nrow(Prostate))

# Sets the seed to make the partition reproductible.
set.seed(907)
train_ind = sample(seq_len(nrow(Prostate)), size = smp_size)

# Training set
train = Prostate[train_ind, ]

# Test set
test = Prostate[-train_ind, ]

# Creates matrices for independent and dependent variables.
xtrain = model.matrix(lpsa~. -1, data = train)
ytrain = train$lpsa
xtest = model.matrix(lpsa~. -1, data = test)
ytest = test$lpsa

# Fitting a linear model by Lasso regression on the "train" data set
pr.lasso = cv.glmnet(xtrain,ytrain,type.measure='mse',alpha=1)
lambda.lasso = pr.lasso$lambda.min

# Getting predictions on the "test" data set and calculating the mean     square error
lasso.pred = predict(pr.lasso, s = lambda.lasso, newx = xtest) 

# Calculating MSE via the mse function defined above
mse.1 = mse(lasso.pred,ytest)
cat("MSE (method 1): ", mse.1, "\n")

# Calculating MSE via the cvm attribute inside the pr.lasso object
mse.2 = pr.lasso$cvm[pr.lasso$lambda == lambda.lasso]
cat("MSE (method 2): ", mse.2, "\n")

所以这些是我为MSE获得的输出：

MSE (method 1): 0.4609978 
MSE (method 2): 0.5654089

而且他们完全不同.有谁知道为什么？
非常感谢您的帮助！

塞缪尔

最佳答案 正如@alistaire指出的那样,在第一种情况下,您使用测试数据来计算MSE,在第二种情况下,报告来自交叉验证(训练)折叠的MSE,因此它不是苹果对苹果的比较.

我们可以做类似以下的事情来进行苹果与苹果的比较(通过在训练折叠处保持拟合值)并且我们可以看到mse.1和mse.2在相同的训练折叠上计算时完全相等(尽管值我的桌面R版本3.1.2,x86_64-w64-mingw32,windows 10)与您的版本略有不同：

# Needs the following R packages.
library(lasso2)
library(glmnet)

# Gets the prostate cancer dataset
data(Prostate)

# Defines the Mean Square Error function 
mse = function(x,y) { mean((x-y)^2)}

# 75% of the sample size.
smp_size = floor(0.75 * nrow(Prostate))

# Sets the seed to make the partition reproductible.
set.seed(907)
train_ind = sample(seq_len(nrow(Prostate)), size = smp_size)

# Training set
train = Prostate[train_ind, ]

# Test set
test = Prostate[-train_ind, ]

# Creates matrices for independent and dependent variables.
xtrain = model.matrix(lpsa~. -1, data = train)
ytrain = train$lpsa
xtest = model.matrix(lpsa~. -1, data = test)
ytest = test$lpsa

# Fitting a linear model by Lasso regression on the "train" data set
# keep the fitted values on the training folds
pr.lasso = cv.glmnet(xtrain,ytrain,type.measure='mse', keep=TRUE, alpha=1)
lambda.lasso = pr.lasso$lambda.min
lambda.id <- which(pr.lasso$lambda == pr.lasso$lambda.min)

# get the predicted values on the training folds with lambda.min (not from test data)
mse.1 = mse(pr.lasso$fit[,lambda.id], ytrain) 
cat("MSE (method 1): ", mse.1, "\n")

MSE (method 1):  0.6044496 

# Calculating MSE via the cvm attribute inside the pr.lasso object
mse.2 = pr.lasso$cvm[pr.lasso$lambda == lambda.lasso]
cat("MSE (method 2): ", mse.2, "\n")

MSE (method 2):  0.6044496 

mse.1 == mse.2
[1] TRUE
点赞