r - 携带/使用前一组的值

数据:

structure(list(id = c(1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 4, 4, 5),
               ax = c("a", "a", "b", "b", "b", "b", "b", "b", "c", "c",
                      "d", "d", "e"), time = c(1, 3, 0, 2, 4, 5, 6, 8, 7, 9, 10,
                                               11, 12)), .Names = c("id", "ax", "time"), class = c("data.table",
                                                                                                   "data.frame"), row.names = c(NA, -13L))

好像:

    id ax time
 1:  1  a    1
 2:  1  a    3
 3:  2  b    0
 4:  2  b    2
 5:  2  b    4
 6:  2  b    5
 7:  2  b    6
 8:  2  b    8
 9:  3  c    7
10:  3  c    9
11:  4  d   10
12:  4  d   11
13:  5  e   12

我想将上一组的max放在实际组的旁边:
所需的输出:

    id ax time newCol
 1:  1  a    1     NA
 2:  1  a    3     NA
 3:  2  b    0      3
 4:  2  b    2      3
 5:  2  b    4      3
 6:  2  b    5      3
 7:  2  b    6      3
 8:  2  b    8      3
 9:  3  c    7      8
10:  3  c    9      8
11:  4  d   10      9
12:  4  d   11      9
13:  5  e   12     11

是否还可能具有“previous-previous” grp的值？
对baseR，data.table和tidyverse解决方案感兴趣
笔记:
可以按EITHER id或ax分组。这里的示例有点多余。

最佳答案

data.table解决方案:

dtt.max <- dtt[, .(max = max(time)), by = ax]
dtt.max[, max.prev := shift(max)]

dtt[dtt.max, newCol := i.max.prev, on = 'ax']
# > dtt
#     id ax time newCol
#  1:  1  a    1     NA
#  2:  1  a    3     NA
#  3:  2  b    0      3
#  4:  2  b    2      3
#  5:  2  b    4      3
#  6:  2  b    5      3
#  7:  2  b    6      3
#  8:  2  b    8      3
#  9:  3  c    7      8
# 10:  3  c    9      8
# 11:  4  d   10      9
# 12:  4  d   11      9
# 13:  5  e   12     11

关于r - 携带/使用前一组的值，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/49754911/