我有data.table DT，如下所示。
DT <- structure(list(ID = c("Bats", "HL", "JL", "Spidey", "Supes",
"X"), List1 = c("Morrison, Brubaker, Daniel, Loeb", "David, Bryne, Lee",
"", "Loeb, Lee", "Moore, Siegel, Millar", "Bendis, Whendon"),
List2 = c("Rucka, Kane, Morrison", "Lee, Mantlo, Bryne",
"Meltzer, Sekowsky, Morrison", "Waid, Yost, Kirby, Lee",
"", "Claremont, Whendon, Morrison")), .Names = c("ID", "List1",
"List2"), row.names = c(NA, -6L), class = c("data.table", "data.frame"
), .internal.selfref = NULL, sorted = "ID")
DT
ID List1 List2
1: Bats Morrison, Brubaker, Daniel, Loeb Rucka, Kane, Morrison
2: HL David, Bryne, Lee Lee, Mantlo, Bryne
3: JL Meltzer, Sekowsky, Morrison
4: Spidey Loeb, Lee Waid, Yost, Kirby, Lee
5: Supes Moore, Siegel, Millar
6: X Bendis, Whendon Claremont, Whendon, Morrison
我想在列DT$List1和DT$List2中逐行合并两个列表而不重复。
我可以使用apply执行此操作，如下所示。
DT$merged <- apply(DT,1,function(vec){
paste(unique(strsplit(paste(vec[2],vec[3],sep=", "),", ")[[1]]),collapse=", ")
})
DT
ID List1 List2
1: Bats Morrison, Brubaker, Daniel, Loeb Rucka, Kane, Morrison
2: HL David, Bryne, Lee Lee, Mantlo, Bryne
3: JL Meltzer, Sekowsky, Morrison
4: Spidey Loeb, Lee Waid, Yost, Kirby, Lee
5: Supes Moore, Siegel, Millar
6: X Bendis, Whendon Claremont, Whendon, Morrison
merged
1: Morrison, Brubaker, Daniel, Loeb, Rucka, Kane
2: David, Bryne, Lee, Mantlo
3: , Meltzer, Sekowsky, Morrison
4: Loeb, Lee, Waid, Yost, Kirby
5: Moore, Siegel, Millar
6: Bendis, Whendon, Claremont, Morrison
如何在不使用＆＃34;，＆＃34; data.table的情况下有效地获得相同的结果由于空单元格在开始和结束？
答案 0 :(得分：4)
DT[, merged := toString(unique(c(strsplit(List1, split = ", ")[[1]],
strsplit(List2, split = ", ")[[1]]))), by = ID][]
# ID List1 List2
#1: Bats Morrison, Brubaker, Daniel, Loeb Rucka, Kane, Morrison
#2: HL David, Bryne, Lee Lee, Mantlo, Bryne
#3: JL Meltzer, Sekowsky, Morrison
#4: Spidey Loeb, Lee Waid, Yost, Kirby, Lee
#5: Supes Moore, Siegel, Millar
#6: X Bendis, Whendon Claremont, Whendon, Morrison
# merged
#1: Morrison, Brubaker, Daniel, Loeb, Rucka, Kane
#2: David, Bryne, Lee, Mantlo
#3: Meltzer, Sekowsky, Morrison
#4: Loeb, Lee, Waid, Yost, Kirby
#5: Moore, Siegel, Millar
#6: Bendis, Whendon, Claremont, Morrison
如果您的1:nrow(DT)列不是唯一的，请将{by'替换为ID。
答案 1 :(得分：3)
尝试
library(data.table)#v1.9.5+
DT[, merged := do.call(paste, c(.SD, sep=", ")), .SDcols= List1:List2
][, merged:=unlist(lapply(strsplit(merged, ", "),
function(x) toString(unique(x))))]
# ID List1 List2
#1: Bats Morrison, Brubaker, Daniel, Loeb Rucka, Kane, Morrison
#2: HL David, Bryne, Lee Lee, Mantlo, Bryne
#3: JL Meltzer, Sekowsky, Morrison
#4: Spidey Loeb, Lee Waid, Yost, Kirby, Lee
#5: Supes Moore, Siegel, Millar
#6: X Bendis, Whendon Claremont, Whendon, Morrison
# merged
#1: Morrison, Brubaker, Daniel, Loeb, Rucka, Kane
#2: David, Bryne, Lee, Mantlo
#3: , Meltzer, Sekowsky, Morrison
#4: Loeb, Lee, Waid, Yost, Kirby
#5: Moore, Siegel, Millar
#6: Bendis, Whendon, Claremont, Morrison
或者我们可以在regex之后使用do.call(paste删除重复的元素
DT[, merged := gsub('^,\\s*|(\\b\\S+\\b)(?=.*\\b\\1\\b.*),\\s*|,\\s*$', '',
do.call(paste, c(.SD, sep=", ")), perl=TRUE), .SDcols = List1:List2]
# ID List1 List2
#1: Bats Morrison, Brubaker, Daniel, Loeb Rucka, Kane, Morrison
#2: HL David, Bryne, Lee Lee, Mantlo, Bryne
#3: JL Meltzer, Sekowsky, Morrison
#4: Spidey Loeb, Lee Waid, Yost, Kirby, Lee
#5: Supes Moore, Siegel, Millar
#6: X Bendis, Whendon Claremont, Whendon, Morrison
# merged
#1: Brubaker, Daniel, Loeb, Rucka, Kane, Morrison
#2: David, Lee, Mantlo, Bryne
#3: Meltzer, Sekowsky, Morrison
#4: Loeb, Waid, Yost, Kirby, Lee
#5: Moore, Siegel, Millar
#6: Bendis, Claremont, Whendon, Morrison
答案 2 :(得分：3)
这可能是另一种方式
DT[, merged:= paste(union(unlist(strsplit(List1, ', ')),
unlist(strsplit(List2, ', '))), collapse = ', '), by = ID]
#> DT
# ID List1 List2
#1: Bats Morrison, Brubaker, Daniel, Loeb Rucka, Kane, Morrison
#2: HL David, Bryne, Lee Lee, Mantlo, Bryne
#3: JL Meltzer, Sekowsky, Morrison
#4: Spidey Loeb, Lee Waid, Yost, Kirby, Lee
#5: Supes Moore, Siegel, Millar
#6: X Bendis, Whendon Claremont, Whendon, Morrison
# merged
#1: Morrison, Brubaker, Daniel, Loeb, Rucka, Kane
#2: David, Bryne, Lee, Mantlo
#3: Meltzer, Sekowsky, Morrison
#4: Loeb, Lee, Waid, Yost, Kirby
#5: Moore, Siegel, Millar
#6: Bendis, Whendon, Claremont, Morrison
答案 3 :(得分：0)
显示另一种方法的简单示例。函数trimComma只是在开头和结尾处删除逗号。它还修剪重复的逗号，可以轻松地与paste
trimComma <-function(x)
{
gsub(",,",",",gsub("^,+|,+$|","",x))
}