如何拆分数据集以进行10倍交叉验证

现在我有了一个R数据框（训练），有人可以告诉我如何随机拆分此数据集以进行10倍交叉验证吗？

caret 具有以下功能：

require(caret)
flds <- createFolds(y, k = 10, list = TRUE, returnTrain = FALSE)
names(flds)[1] <- "train"

然后每个元素 flds是每个数据集的索引列表。如果您的数据集称为dat，那么dat[flds$train,]您将获得训练集，dat[ flds[[2]], ]并获得第二个折叠集，依此类推。

这是不使用软件包执行10折的简单方法：

#Randomly shuffle the data
yourData<-yourData[sample(nrow(yourData)),]

#Create 10 equally size folds
folds <- cut(seq(1,nrow(yourData)),breaks=10,labels=FALSE)

#Perform 10 fold cross validation
for(i in 1:10){
    #Segement your data by fold using the which() function 
    testIndexes <- which(folds==i,arr.ind=TRUE)
    testData <- yourData[testIndexes, ]
    trainData <- yourData[-testIndexes, ]
    #Use the test and train data partitions however you desire...
}

可能不是最好的方法，但这是一种方法。我很确定当我编写此代码时，是从这里的另一个答案中借来了一个窍门，但我找不到它的链接。

# Generate some test data
x <- runif(100)*10 #Random values between 0 and 10
y <- x+rnorm(100)*.1 #y~x+error
dataset <- data.frame(x,y) #Create data frame
plot(dataset$x,dataset$y) #Plot the data

#install.packages("cvTools")
library(cvTools) #run the above line if you don't have this library

k <- 10 #the number of folds

folds <- cvFolds(NROW(dataset), K=k)
dataset$holdoutpred <- rep(0,nrow(dataset))

for(i in 1:k){
  train <- dataset[folds$subsets[folds$which != i], ] #Set the training set
  validation <- dataset[folds$subsets[folds$which == i], ] #Set the validation set

  newlm <- lm(y~x,data=train) #Get your new linear model (just fit on the train data)
  newpred <- predict(newlm,newdata=validation) #Get the predicitons for the validation set (from the model just fit on the train data)

  dataset[folds$subsets[folds$which == i], ]$holdoutpred <- newpred #Put the hold out prediction in the data set for later use
}

dataset$holdoutpred #do whatever you want with these predictions

请在下面找到我使用的其他代码（从其他来源借来并改编）。从我刚用过的脚本直接复制到rpart例程中。可能最感兴趣的部分是褶皱的线条。或者，您可以使用bootstrap包中的crossval函数。

#define error matrix
err <- matrix(NA,nrow=1,ncol=10)
errcv=err

#creation of folds
for(c in 1:10){

n=nrow(df);K=10; sizeblock= n%/%K;alea=runif(n);rang=rank(alea);bloc=(rang-1)%/%sizeblock+1;bloc[bloc==K+1]=K;bloc=factor(bloc); bloc=as.factor(bloc);print(summary(bloc))

for(k in 1:10){

#rpart
fit=rpart(type~., data=df[bloc!=k,],xval=0) ; (predict(fit,df[bloc==k,]))
answers=(predict(fit,df[bloc==k,],type="class")==resp[bloc==k])
err[1,k]=1-(sum(answers)/length(answers))

}

err
errcv[,c]=rowMeans(err, na.rm = FALSE, dims = 1)

}
errcv

# Evaluate models uses k-fold cross-validation
install.packages("DAAG")
library("DAAG")

cv.lm(data=dat, form.lm=mod1, m= 10, plotit = F)

只需一行代码即可为您完成所有工作！

?cv.lm for information on input and output

因为我不在此列表中，所以我想为不想安装软件包以进行快速交叉验证的人提供另一种选择

# get the data from somewhere and specify number of folds
data <- read.csv('my_data.csv')
nrFolds <- 10

# generate array containing fold-number for each sample (row)
folds <- rep_len(1:nrFolds, nrow(data))

# actual cross validation
for(k in 1:nrFolds) {
    # actual split of the data
    fold <- which(folds == k)
    data.train <- data[-fold,]
    data.test <- data[fold,]

    # train and test your model with data.train and data.test
}

请注意，上面的代码假定数据已经被改组。如果不是这种情况，您可以考虑添加类似

folds <- sample(folds, nrow(data))