这是一个基本的 R 解决方案。拆分PolId
现场使用strplit
对于每个这样的分割字段 cbind 它与相应的Description
。这给出了我们的矩阵列表rbind
一起。最后设置列名称。
out <- do.call(rbind, Map(cbind, strsplit(DF$PolId, ";"), DF$Description))
colnames(out) <- colnames(DF)
giving:
> out
PolId Description
[1,] "ABC123" "TEST1"
[2,] "ABC456" "TEST1"
[3,] "ABC789" "TEST1"
[4,] "ABC123" "TEST1"
[5,] "ABC456" "TEST1"
[6,] "ABC789" "TEST1"
[7,] "ABC123" "TEST1"
[8,] "ABC456" "TEST1"
[9,] "ABC789" "TEST1"
[10,] "AAA123" "TEST1"
[11,] "AAA123" "TEST2"
[12,] "ABB123" "TEST3"
[13,] "ABC123" "TEST3"
[14,] "ABB123" "TEST3"
[15,] "ABC123" "TEST3"
Note:我们用它作为输入:
DF <-
structure(list(PolId = c("ABC123;ABC456;ABC789;", "ABC123;ABC456;ABC789;",
"ABC123;ABC456;ABC789;", "AAA123;", "AAA123;", "ABB123;ABC123;",
"ABB123;ABC123;"), Description = c("TEST1", "TEST1", "TEST1",
"TEST1", "TEST2", "TEST3", "TEST3")), .Names = c("PolId", "Description"
), class = "data.frame", row.names = c(NA, -7L))