R 数据表中处理的复杂分组
抱歉描述不清楚，但我不认为一行可以解释我的要求 我有一个data.tableR 数据表中处理的复杂分组,r,group-by,data.table,R,Group By,Data.table,抱歉描述不清楚，但我不认为一行可以解释我的要求 我有一个data.tabledt1，如下所示： id pg pd dt capp vt 1: 1111 hm <NA> 20-10-2020 21:07:54 NA 5 2: 1111 abc abc 20-10-2020 21:07:53 1234 5 3: 1111 hm <NA> 20-10-2020 16:07:56 NA 4 4: 11
dt1 id pg pd dt capp vt
1: 1111 hm <NA> 20-10-2020 21:07:54 NA 5
2: 1111 abc abc 20-10-2020 21:07:53 1234 5
3: 1111 hm <NA> 20-10-2020 16:07:56 NA 4
4: 1111 cde <NA> 20-10-2020 16:06:57 NA 4
5: 1111 cde <NA> 20-10-2020 16:05:58 NA 4
6: 1111 def def 20-07-2020 12:07:59 345 3
7: 1111 abc <NA> 20-06-2020 22:07:59 NA 2
8: 1111 def <NA> 20-06-2020 22:07:58 NA 2
9: 1111 abc <NA> 20-05-2020 21:07:59 NA 1
10: 1112 hm <NA> 20-10-2020 21:07:52 NA 4
11: 1112 cde cde 20-10-2020 21:07:39 456 4
12: 1112 hm <NA> 20-10-2020 16:07:56 NA 3
13: 1112 abc <NA> 20-10-2020 16:06:57 NA 3
14: 1112 abc <NA> 20-07-2020 16:05:58 NA 2
15: 1112 def abc 20-07-2020 16:04:59 234 2
16: 1112 cde <NA> 20-06-2020 22:07:59 NA 1
17: 1112 def <NA> 20-06-2020 21:07:59 NA 1
18: 1112 cde <NA> 20-05-2020 21:07:59 NA 0
dt1structure(list(id = c(1111L, 1111L, 1111L, 1111L, 1111L, 1111L,
1111L, 1111L, 1111L, 1112L, 1112L, 1112L, 1112L, 1112L, 1112L,
1112L, 1112L, 1112L), pg = c("hm", "abc", "hm", "cde", "cde",
"def", "abc", "def", "abc", "hm", "cde", "hm", "abc", "abc",
"def", "cde", "def", "cde"), pd = c(NA, "abc", NA, NA, NA, "def",
NA, NA, NA, NA, "cde", NA, NA, NA, "abc", NA, NA, NA), dt = c("20-10-2020 21:07:54",
"20-10-2020 21:07:53", "20-10-2020 16:07:56", "20-10-2020 16:06:57",
"20-10-2020 16:05:58", "20-07-2020 12:07:59", "20-06-2020 22:07:59",
"20-06-2020 22:07:58", "20-05-2020 21:07:59", "20-10-2020 21:07:52",
"20-10-2020 21:07:39", "20-10-2020 16:07:56", "20-10-2020 16:06:57",
"20-07-2020 16:05:58", "20-07-2020 16:04:59", "20-06-2020 22:07:59",
"20-06-2020 21:07:59", "20-05-2020 21:07:59"), capp = c(NA, 1234L,
NA, NA, NA, 345L, NA, NA, NA, NA, 456L, NA, NA, NA, 234L, NA,
NA, NA), vt = c(5L, 5L, 4L, 4L, 4L, 3L, 2L, 2L, 1L, 4L, 4L, 3L,
3L, 2L, 2L, 1L, 1L, 0L)), .Names = c("id", "pg", "pd", "dt",
"capp", "vt"), row.names = c(NA, -18L), class = c("data.table",
"data.frame"), .internal.selfref = <pointer: 0x0000000002650788>)
dt1[,
prev := with(.SD, vapply(
seq_along(vt),
function(i) {tmp <- vt[vt < vt[[i]] & pg == pd[[i]] & !is.na(capp[[i]])]; if (length(tmp) < 1L) NA_real_ else max(tmp)},
numeric(1L)
)),
by = id
]
dt1[,，
prev:=带（.SD，vapply）(
顺时针（vt），
函数（i）{tmp这里是另一个选项，对非空capp的每一行使用非等联接，然后通过引用进行更新：
dt1[!is.na(capp), prev :=
dt1[.SD, on=.(id, pg=pd, vt<vt), max(x.vt), by=.EACHI]$V1
]
dt1[！is.na（capp），prev:=
dt1[.SD，on=。（ID，PG= PD，VTAS，我可以说，你从来没有考虑过代码< >代码> CAP >代码> No>代码，所有的值都是从<代码> VT < /代码>列中得到的。<代码>第二行> <代码> >代码> > <代码> CAPP > <代码> VT <代码> 2 < /代码>id=1111
？@ekoam：我知道你是从哪里来的。vt小于对应于非空capp的vt值，实际上，我正在寻找vt
（其中capp
可以为空）的值，但小于capp
非空行中vt
的值（ergo对应于非空capp）这很有效，但速度非常慢。谢谢你的帮助。你给了我一些食物来咀嚼。谢谢你的帮助。奇怪的是，如果我需要添加另一个条件，比如pg！=“cc”
，它会进入On（）
？dt1[！is.na（capp），prev:=dt1[.SD，On=（id，pg=pd，vt
id pg pd dt capp vt prev
1: 1111 hm <NA> 20-10-2020 21:07:54 NA 5 NA
2: 1111 abc abc 20-10-2020 21:07:53 1234 5 2
3: 1111 hm <NA> 20-10-2020 16:07:56 NA 4 NA
4: 1111 cde <NA> 20-10-2020 16:06:57 NA 4 NA
5: 1111 cde <NA> 20-10-2020 16:05:58 NA 4 NA
6: 1111 def def 20-07-2020 12:07:59 345 3 2
7: 1111 abc <NA> 20-06-2020 22:07:59 NA 2 NA
8: 1111 def <NA> 20-06-2020 22:07:58 NA 2 NA
9: 1111 abc <NA> 20-05-2020 21:07:59 NA 1 NA
10: 1112 hm <NA> 20-10-2020 21:07:52 NA 4 NA
11: 1112 cde cde 20-10-2020 21:07:39 456 4 1
12: 1112 hm <NA> 20-10-2020 16:07:56 NA 3 NA
13: 1112 abc <NA> 20-10-2020 16:06:57 NA 3 NA
14: 1112 abc <NA> 20-07-2020 16:05:58 NA 2 NA
15: 1112 def abc 20-07-2020 16:04:59 234 2 NA
16: 1112 cde <NA> 20-06-2020 22:07:59 NA 1 NA
17: 1112 def <NA> 20-06-2020 21:07:59 NA 1 NA
18: 1112 cde <NA> 20-05-2020 21:07:59 NA 0 NA
dt1[!is.na(capp), prev :=
dt1[.SD, on=.(id, pg=pd, vt<vt), max(x.vt), by=.EACHI]$V1
]