6

我有一个数据表：

> (mydt <- data.table(id=c(1,1,1,1,2,2),
                      time=1:6,
                      v1=letters[1:6],
                      v2=LETTERS[1:6],
                      key=c("id","time")))
   id time v1 v2
1:  1    1  a  A
2:  1    2  b  B
3:  1    3  c  C
4:  1    4  d  D
5:  2    5  e  E
6:  2    6  f  F

我想“卷起来”（这是正确的术语吗？），到“更改”表：对象1更改了 3 次（从时间戳 1 到 2、2 到 3 和 3 到 4）对象2更改了一次（时间5至6）；我对初始 v1和最终 v2感兴趣。所以，结果应该是：

> (res <- data.table(beg.time=c(1,2,3,5),
                     end.time=c(2,3,4,6),
                     v1=c('a','b','c','e'),
                     v2=c('B','C','D','F'),
                     key=c("beg.time","end.time")))
   beg.time end.time v1 v2
1:        1        2  a  B
2:        2        3  b  C
3:        3        4  c  D
4:        5        6  e  F
4

1 回答 1

8

感谢您提供可重复的示例！这是一个镜头。

首先，请注意，您可以使用以下头尾成语将相距一定距离的向量条目彼此相邻放置：

x <- letters[1:5]
cbind(head(x, -1), tail(x, -1))
     # [,1] [,2]
# [1,] "a"  "b" 
# [2,] "b"  "c" 
# [3,] "c"  "d" 
# [4,] "d"  "e" 
cbind(head(x, -2), tail(x, -2))
     # [,1] [,2]
# [1,] "a"  "c" 
# [2,] "b"  "d" 
# [3,] "c"  "e"

然后，我们可以使用 的by功能data.table按组进行此操作。

mydt[,{
    ## if there's just one row in the group of ID's, return nothing
    if (.N == 1) return(NULL) 
    else {
        list(
            ## head and tail take the first and last parts of a vector
            ## this will place an element next to its subsequent element
            beg.time = head(time, -1),
            end.time = tail(time, -1),
            v1 = head(v1, -1),
            v2 = tail(v2, -1)
## group by ID
)}}, by = id]

#    id beg.time end.time v1 v2
# 1:  1        1        2  a  B
# 2:  1        2        3  b  C
# 3:  1        3        4  c  D
# 4:  2        5        6  e  F
于 2013-09-17T15:31:01.413 回答