5

我有一个如下所示的数据集：

structure(list(A = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = c("1", 
"2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13", 
"14", "15", "16", "17", "18", "19", "20", "21", "22", "23", "24", 
"25"), class = "factor"), T = c(0.04, 0.08, 0.12, 0.16, 0.2, 
0.24), X = c(464.4, 464.4, 464.4, 464.4, 464.4, 464.4), Y = c(418.5, 
418.5, 418.5, 418.5, 418.5, 418.5), V = c(0, 0, 0, 0, 0, 0), 
    GD = c(0, 0, 0, 0, 0, 0), ND = c(NA, 0, 0, 0, 0, 0), ND2 = c(NA, 
    0, 0, 0, 0, 0), TID = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = c("t1", 
    "t10", "t11", "t12", "t13", "t14", "t15", "t16", "t17", "t18", 
    "t19", "t2", "t20", "t21", "t22", "t23", "t24", "t25", "t3", 
    "t4", "t5", "t6", "t7", "t8", "t9"), class = "factor")), .Names = c("A", 
"T", "X", "Y", "V", "GD", "ND", "ND2", "TID"), row.names = c(NA, 
6L), class = "data.frame")

我想为每个 TID 选择所有变量的前 80 个观察值。到目前为止，我只能使用以下代码对第一个 TID 执行此操作：

sub.data1<-NM[1:80, ]

我怎样才能为所有其他 TID 做到这一点？

谢谢！

4

4 回答 4

7

我会做：

lapply(split(dat, dat$TID), head, 80)

它返回一个包含 80 行（或更少）行的 data.frames 列表。相反，如果您希望将所有内容都放在一个 data.frame 中：

do.call(rbind, lapply(split(dat, dat$TID), head, 80))
于 2013-05-23T19:46:15.377 回答
5

使用您的功能ddply()可以plyr按 TID 拆分数据，然后选择 forst 80 ，head()然后将所有内容再次放入一个数据帧中，

library(plyr)
ddply(NM, .(TID), head, n = 80)
于 2013-05-23T19:48:34.667 回答
3

使用数据表，我制作了一个较短的示例，其中仅包含 TID t1 和 t2，它返回 t1 和 t2 的前 2 行。它可以根据您的数据进行调整。

library(data.table)
data<-structure(list(A = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = c("1", 
                "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13", 
                "14", "15", "16", "17", "18", "19", "20", "21", "22", "23", "24", 
                "25"), class = "factor"), T = c(0.04, 0.08, 0.12, 0.16, 0.2, 
                0.24), X = c(464.4, 464.4, 464.4, 464.4, 464.4, 464.4), Y = c(418.5, 
                        418.5, 418.5, 418.5, 418.5, 418.5), V = c(0, 0, 0, 0, 0, 0), 
                GD = c(0, 0, 0, 0, 0, 0), ND = c(NA, 0, 0, 0, 0, 0), ND2 = c(NA, 
                        0, 0, 0, 0, 0), TID = c("t1","t1","t1","t2","t2","t2")), .Names = c("A", 
                "T", "X", "Y", "V", "GD", "ND", "ND2", "TID"), row.names = c(NA, 
                6L), class = "data.frame")
dt<-data.table(data)
dt[,head(.SD,2),by=TID]

这导致：

   TID A    T     X     Y V GD ND ND2
1:  t1 1 0.04 464.4 418.5 0  0 NA  NA
2:  t1 1 0.08 464.4 418.5 0  0  0   0
3:  t2 1 0.16 464.4 418.5 0  0  0   0
4:  t2 1 0.20 464.4 418.5 0  0  0   0

如果需要，可以通过将最后一行更改为

as.data.frame(dt[,head(.SD,2),by=TID])
于 2013-05-23T20:05:17.653 回答
2

这是base中的另一个解决方案：

do.call(rbind, by(NM, NM$TID, head, 80))
于 2013-05-23T20:14:43.933 回答