使用R中的拆分功能不会发生精确拆分

Question

https://www.dropbox.com/s/35w66sri5rauv5d/FlightDelays.csv?dl=0

我正在从上面的链接中读取一个包含2201行的数据集。使用分割函数，我给出的比率为0.6。然后我应该得到两个数据集分别包含1320和881个数据集。最初它工作正常，但现在当我分裂时，而不是0.6，分裂发生在0.53，即使我在分裂中指定0.6作为我的比率。这可能是这个突然变化的问题。如何解决这个问题。下面给出了代码。

library(caTools)
originaldata.df<-read.csv("use csv from the link given above")
split<-sample.split(originaldata.df,SplitRatio = 0.6)
Trainingdataset<-subset(originaldata.df,split == "TRUE")
Testingdataset<-subset(originaldata.df,split == "FALSE")

ExpectedOutput:
1320(2201*60/100)
881(2201*40/100)
Actualoutput:
1186
1015

Answer 1

基础-R：

您可以使用索引并按分割比率分配它们;

indexes = sample(1:nrow(originaldata.df), 
size=0.6*nrow(originaldata.df))

Trainingdataset <- originaldata.df[indexes,]
Testingdataset <- originaldata.df[-indexes,]

这将是输出：

> dim(Testingdataset)
# [1] 881  13
> dim(Trainingdataset)
# [1] 1320   13

caTools包：

library(caTools)
#It should be applied on one of column of the data.frame otherwise samples over rows;
split<-sample.split(originaldata.df$schedtime,SplitRatio = 0.6) 

Trainingdataset<-subset(originaldata.df,split == "TRUE")
Testingdataset<-subset(originaldata.df,split == "FALSE")

子集的大小（不完全符合您的预期;）

> dim(Trainingdataset)
# [1] 1323   13
> dim(Testingdataset)
# [1] 878  13

Answer 2

这是一个自定义的分割函数，它将根据给定的比例导出两个rownumber子集：

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使用函数派生训练和测试集：

splitFactor <- function(rows, prop){
  a <- sample(seq(rows), ceiling(rows*prop))
  b <- sample(seq(rows), floor(rows*(1-prop)))
  list(a[order(a)],b[order(b)])
}


sp.53 <- splitFactor(nrow(iris), .53)
lapply(sp.53, length)

# [[1]]
# [1] 80

# [[2]]
# [1] 70