我有示例数据框：
date item1 item2 item3
17-11-2016 2A hp cnf
12-11-2016 1A bp wl
13-11-2016 3A sp dl
14-11-2016 1A hp cnf
16-11-2016 2A bp cnf
10-11-2016 1A sp wl
17-11-2016 2A hp wl
我想基于所有列date，item1和item2对数据进行分组，以便具有相同值的特定列将来到同一个组。
预期产出：
date item1 item2 item3 grp
17-11-2016 2A hp cnf 1
17-11-2016 2A hp wl 1
12-11-2016 1A bp wl 2
13-11-2016 3A sp dl 3
14-11-2016 1A hp cnf 4
16-11-2016 2A bp cnf 5
10-11-2016 1A sp wl 6
答案 0 :(得分：1)
你可以这样做：
DF <- data.frame(date = c("17-11-2016","12-11-2016","13-11-2016","14-11-2016",
"16-11-2016","10-11-2016","17-11-2016"),
item1 = c("2A","1A","3A","1A","2A","1A","2A"),
item2 = c("hp","bp","sp","hp","bp","sp","hp"),
item3 = c("cnf","wl","dl","cnf","cnf","wl","wl")
)
DF$grp <- as.numeric(factor(paste(DF$date,DF$item1,DF$item2)))
DF[order(DF$grp),]
date item1 item2 item3 grp
6 10-11-2016 1A sp wl 1
2 12-11-2016 1A bp wl 2
3 13-11-2016 3A sp dl 3
4 14-11-2016 1A hp cnf 4
5 16-11-2016 2A bp cnf 5
7 17-11-2016 2A hp wl 6
1 17-11-2016 2A hp cnf 6
答案 1 :(得分：0)
我们可以X509KeyManager前三列，转换为paste，将factor指定为'v1'中的levels元素并将其强制转换为{{1} }}
unique或使用integer
v1 <- do.call(paste, df1[1:3])
df1$grp <- as.integer(factor(v1, levels = unique(v1)))
df1$grp
#[1] 1 1 2 3 4 5 6