Question

提前致谢。我有一个行程数据框，表示起始位置，完成位置和每个位置之间的距离组合。像这样：

Start = c("Johns House", "Mikes House", "Franks House")
Finish = c("Mikes House", "Johns House", "Lisas House")
Distance = c(1000,1000,500) 
myDF = data.frame(Start, Finish, Distance)

我想返回一个处理两次John / Mike旅行的新数据框，但将它们视为一个独特的组合。具体来说，我想返回每个组合的总出行次数和位置之间的距离 - 因此输出将是：

newStart = c("Johns House", "Franks House")
newFinish = c("Mikes House", "Lisas House")
newDistance = c(1000,500)
Count = c(2,1)
newDF = data.frame(newStart, newFinish, newDistance, Count)

再次感谢。

Answer 1

我认为使用SQL会更容易。在R中安装SQL包，如“sqldf”。

首先，您可以计算每个元组的开始次数 - 以任一方式完成：

library(sqldf)

sqldf("select distinct 
      min(a.Start, a.Finish) Start, 
      max(a.Start, a.Finish) Finish, 
      a.Distance, 
      count(*) Count
      from myDF a, myDF b
      where (a.Start = b.Start and  a.Finish = b.Finish)
      or (a.Start = b.Finish and a.Finish = b.Start)
      group by a.Start")

#          Start      Finish Distance Count
# 1 Franks House Lisas House      500     1
# 2  Johns House Mikes House     1000     2

Answer 2

library(data.table)
myDT <- data.table(myDF)
x <- paste(myDT$Start, myDT$Finish, sep = "|")
myDT$v <- vapply(x, function(xi) paste(sort(strsplit(xi, "[|]")[[1]]), collapse=''), '')
myDT[, Count := length(Distance), by = v]
myDT <- myDT[!duplicated(v), ]
myDT

#          Start      Finish Distance           v Count
#1:  Johns House Mikes House     1000  JohnsMikes     2
#2: Franks House Lisas House      500 FranksLisas     1

我用@ Tommy的答案How to sort letters in a string?对字符串进行排序。

总结两列给出的独特组合

2 个答案: