基于连续字段值的索引

Question

我在 R 中有一个 data.table，用于跟踪系统内项目的移动。我想根据ID 和Location 这两个字段对这些数据进行分组。

library(data.table)
example <- data.table(ID = rep(LETTERS[1:3], each = 6),
                      Location = c(1,2,3,1,2,1,2,2,2,3,3,1,2,3,3,3,1,3))
example
#     ID Location
#  1:  A        1
#  2:  A        2
#  3:  A        3
#  4:  A        1
#  5:  A        2
#  6:  A        1
#  7:  B        2
#  8:  B        2
#  9:  B        2
# 10:  B        3
# 11:  B        3
# 12:  B        1
# 13:  C        2
# 14:  C        3
# 15:  C        3
# 16:  C        3
# 17:  C        1
# 18:  C        3

我希望输出的是一个新列，其编号在每次位置更改时递增，无论新位置是什么（即该位置是否存在于历史记录中的其他位置）。与 this question 相反，它只在组内递增。

expected_output <- data.table(ID = rep(LETTERS[1:3], each = 6),
                     Location = c(1,2,3,1,2,1,2,2,2,3,3,1,2,3,3,3,1,3),
                     Group = c(1,2,3,4,5,6,1,1,1,2,2,3,1,2,2,2,3,4))
expected_output
#     ID Location Group
#  1:  A        1     1
#  2:  A        2     2
#  3:  A        3     3
#  4:  A        1     4
#  5:  A        2     5
#  6:  A        1     6
#  7:  B        2     1
#  8:  B        2     1
#  9:  B        2     1
# 10:  B        3     2
# 11:  B        3     2
# 12:  B        1     3
# 13:  C        2     1
# 14:  C        3     2
# 15:  C        3     2
# 16:  C        3     2
# 17:  C        1     3
# 18:  C        3     4

到目前为止，我已经尝试了几种不同的 by 参数组合，但运气不佳。我似乎能够得到的最接近的是diff，它部分正确地显示了更改发生的时间，但在内部递增。

output <- example
output[, Group := 1:.N, by = paste0(ID, Location, diff(Location))]
output
#     ID Location Group
#  1:  A        1     1 
#  2:  A        2     1 # not incrementing/new group
#  3:  A        3     1 # not incrementing/new group
#  4:  A        1     2
#  5:  A        2     1
#  6:  A        1     3
#  7:  B        2     1
#  8:  B        2     2 # incrementing when shouldn't
#  9:  B        2     1 
# 10:  B        3     1 
# 11:  B        3     1
# 12:  B        1     1
# 13:  C        2     1
# 14:  C        3     1
# 15:  C        3     2
# 16:  C        3     1
# 17:  C        1     1
# 18:  C        3     1

在这一点上，我很迷茫，尽管我确信解决方案正盯着我的脸。

Answer 1

您可以按ID 分组，然后取Location 列的rleid：

example[, Group := rleid(Location), ID]
example
#    ID Location Group
# 1:  A        1     1
# 2:  A        2     2
# 3:  A        3     3
# 4:  A        1     4
# 5:  A        2     5
# 6:  A        1     6
# 7:  B        2     1
# 8:  B        2     1
# 9:  B        2     1
#10:  B        3     2
#11:  B        3     2
#12:  B        1     3
#13:  C        2     1
#14:  C        3     2
#15:  C        3     2
#16:  C        3     2
#17:  C        1     3
#18:  C        3     4

all.equal(example, expected_output)
# [1] TRUE