r - 使用 data.table 识别从 0 到 1 的跳转

Question

我有以下数据表，其中包含感兴趣的变量x。我想创建另一个变量来指示 x 从 0 到 1 的跳跃，这意味着该变量在某一年之前一直为 0，而在此之后的所有年份中都为 1。这应该由 id_d 完成。

是否有一种简单的 data.table 方法可以做到这一点？

原始数据表:

 fullDat <- data.table(id_d = rep(letters[1:3], each=12), 
                  year=rep(1:12, 3), 
                  x = c(rep(0, 5), rep(1, 7), 0,1,0,1,2,2,4, rep(5,5), 1, rep(0, 3), rep(1, 8)))

    id_d year x
  1:    a    1 0
  2:    a    2 0
  3:    a    3 0
  4:    a    4 0
  5:    a    5 0
  6:    a    6 1
  7:    a    7 1
  8:    a    8 1
  9:    a    9 1
 10:    a   10 1
 11:    a   11 1
 12:    a   12 1
 13:    b    1 0
 14:    b    2 1
 15:    b    3 0
 16:    b    4 1
 17:    b    5 2
 18:    b    6 2
 19:    b    7 4
 20:    b    8 5
 21:    b    9 5
 22:    b   10 5
 23:    b   11 5
 24:    b   12 5
 25:    c    1 1
 26:    c    2 0
 27:    c    3 0
 28:    c    4 0
 29:    c    5 1
 30:    c    6 1
 31:    c    7 1
 32:    c    8 1
 33:    c    9 1
 34:    c   10 1
 35:    c   11 1
 36:    c   12 1
id_d year x

结果应该是这样的:

    id_d year x jump
  1:    a    1 0    0
  2:    a    2 0    0
  3:    a    3 0    0
  4:    a    4 0    0
  5:    a    5 0    0
  6:    a    6 1    1
  7:    a    7 1    0
  8:    a    8 1    0
  9:    a    9 1    0
 10:    a   10 1    0
 11:    a   11 1    0
 12:    a   12 1    0
 13:    b    1 0    0
 14:    b    2 1    0
 15:    b    3 0    0
 16:    b    4 1    0
 17:    b    5 2    0
 18:    b    6 2    0
 19:    b    7 4    0
 20:    b    8 5    0
 21:    b    9 5    0
 22:    b   10 5    0
 23:    b   11 5    0
 24:    b   12 5    0
 25:    c    1 1    0
 26:    c    2 0    0
 27:    c    3 0    0
 28:    c    4 0    0
 29:    c    5 1    0
 30:    c    6 1    0
 31:    c    7 1    0
 32:    c    8 1    0
 33:    c    9 1    0
 34:    c   10 1    0
 35:    c   11 1    0
 36:    c   12 1    0
id_d year x jump

Answer 1

the variable has been 0 until a certain year and 1 in all years following

# find rows to assign one
wDT = fullDat[, .(year = year[with(rle(x), 
  if (identical(values, c(0, 1))) first(lengths) + 1L
  else 0L
)]), by=id_d]

# initialize to zero
fullDat[, jump := 0L ]
# update join to assign ones
fullDat[wDT, on=.(id_d, year), jump := 1L ]

不需要制作中间表wDT；将其完整代码写入最终语句也是可行的。事实上，如果需要的话，它可以全部在一行中，例如......

DT[, x := 0L][code_for_wDT, on=on_cols, x := 1L]

---

或者，不使用联接，只需使用 .I 中的行号:

# find rows to assign one
w = fullDat[, with(rle(x), .I[
  if (identical(values, c(0, 1))) first(lengths) + 1L
  else 0L
]), by=id_d]$V1

# initialize to zero
fullDat[, jump := 0L ]
# update to assign ones
fullDat[w, jump := 1L ]