Question

我有这样的数据结构。

    structure(list(id = c("4031", "1040;2040;3040", "4040", 
    "1050;2050;3050"), description = c("Sentence A", 
    "Sentence B", "Sentence C", 
    "Sentence D")), row.names = 1:4, class = "data.frame")

              id description
1           4031  Sentence A
2 1040;2040;3040  Sentence B
3           4040  Sentence C
4 1050;2050;3050  Sentence D

我想对数据进行重组，以使ID带有“;”分为不同的行-我想要这样：

structure(list(id = c("4031", "1040","2040","3040", "4040", 
"1050","2050","3050"), description = c("Sentence A", 
"Sentence B","Sentence B","Sentence B", "Sentence C", 
"Sentence D","Sentence D","Sentence D")), row.names = 1:8, class = "data.frame")

   id description
1 4031  Sentence A
2 1040  Sentence B
3 2040  Sentence B
4 3040  Sentence B
5 4040  Sentence C
6 1050  Sentence D
7 2050  Sentence D
8 3050  Sentence D

我知道我可以使用strsplit拆分id列，但是无法找出一种有效的方法来将其转换为没有循环的行

strsplit( as.character( a$id ) , ";" )

Answer 1

使用R base：

> IDs <- strsplit(df$id, ";")
> data.frame(ID=unlist(IDs), Description=rep(df$description, lengths(IDs)))
    ID Description
1 4031  Sentence A
2 1040  Sentence B
3 2040  Sentence B
4 3040  Sentence B
5 4040  Sentence C
6 1050  Sentence D
7 2050  Sentence D
8 3050  Sentence D

Answer 2

tidyr的一种非常方便的可能性是：

separate_rows(df, id)

    id description
1 4031  Sentence A
2 1040  Sentence B
3 2040  Sentence B
4 3040  Sentence B
5 4040  Sentence C
6 1050  Sentence D
7 2050  Sentence D
8 3050  Sentence D

根据strsplit将Data.Frame重组为多行

2 个答案: