Question

我正在对具有两个预测变量（性别，政治倾向：二进制，连续）的二进制DV进行逻辑回归。我需要帮助以使我的GLM在交叉验证中运行！尽管多次将变量重新分类，但我的代码无法运行。我不确定发生了什么。

这是我的代码：

`

#######################################################
#     Cross-Validation of the Logistic Regression
#######################################################


gen <- as.numeric(choicelife.data$gender)
lnc <- as.numeric(choicelife.data$lc)
procprol <-as.numeric(choicelife.data$views)

# This code could be useful
nCV <- 50
MSE_1 <- numeric(nCV)
MSE_2 <- numeric(nCV)

folds <- cut(sample(n),breaks=nCV,labels=FALSE)

#Perform n.folds fold cross validation
i <- 1
for(i in 1:nCV){

  #Segement your data by fold using the which() function 
  testIndexes <- which(folds==i,arr.ind=TRUE)
  testData <- choicelife.data[testIndexes, ]
  trainData <- choicelife.data[-testIndexes, ]

  # Models
  mod1<- glm(views ~ gen,
             family=binomial(link=logit), data=trainData)

  mod2<- glm(views ~ gen + lnc,
             family=binomial(link=logit), data=trainData)

  # Get predictions
  pred_1 <- predict(mod1, newdata = testData)
  pred_2 <- predict(mod2, newdata = testData)

  # Calculate MSE
  MSE_1[i] <- mean((testData$views - pred_1)^2)
  MSE_2[i] <- mean((testData$views - pred_2)^2)
}
warnings()

# mean MSEs
mean(MSE_1) 
mean(MSE_2) 

# get differences
diffs <- MSE_1 - MSE_2

# get 95% CIs
meandiff <- mean(diffs) 
sddiff <- sd(diffs) 
c(meandiff-2*sddiff, meandiff+2*sddiff) # 95% Confidence interval (n, n)

Answer 1

您将某些变量转换为数字，但未将其放在data.frame中。在通过nCV进行迭代的过程中，子集数据帧不包含数字变量，因此将不起作用。

首先，我模拟一些看起来像您的数据框的选择寿命：

choicelife.data = data.frame(
lc=sample(1:10,100,replace=TRUE),
gender=sample(c("M","F"),100,replace=TRUE),
views = sample(c("Pro","Against"),100,replace=TRUE)
)

有关建议的编辑，请参见下文

choicelife.data$gen <- as.numeric(choicelife.data$gender)
choicelife.data$lnc <- as.numeric(choicelife.data$lc)
# make this 0 or 1
choicelife.data$procprol <-as.numeric(choicelife.data$views)-1

# This code could be useful
nCV <- 5
MSE_1 <- numeric(nCV)
MSE_2 <- numeric(nCV)

folds <- cut(sample(1:nrow(choicelife.data)),breaks=nCV,labels=FALSE)

for(i in 1:nCV){

  testIndexes <- which(folds==i,arr.ind=TRUE)
  testData <- choicelife.data[testIndexes, ]
  trainData <- choicelife.data[-testIndexes, ]

  # Models
  mod1<- glm(procprol ~ gen,
             family=binomial(link=logit), data=trainData)

  mod2<- glm(procprol ~ gen + lnc,
             family=binomial(link=logit), data=trainData)

  # Get predictions
  pred_1 <- predict(mod1, newdata = testData,type="response")
  pred_2 <- predict(mod2, newdata = testData,type="response")

  # Calculate MSE
  MSE_1[i] <- mean((testData$procprol - pred_1)^2)
  MSE_2[i] <- mean((testData$procprol - pred_2)^2)
}

在R中交叉验证逻辑回归

1 个答案: