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
我想重组数据,以便 ids 带有“;”分成单独的行 - 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
我知道我可以用 strsplit
分割 id 列但无法找到一种有效的方法将其转换为没有循环的行
strsplit( as.character( a$id ) , ";" )
最佳答案
使用 R 基础:
> IDs <- strsplit(df$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
关于基于 strsplit 将 Data.Frame 重组为多行,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/56484560/