按组使用data.table的shift（）产生意外结果（错误？）

Question

Consider this dataset

Data

dt <- data.table(ID = c(1,8,9,20,32,33), Char = c("A", "A", "B", "B", "C", "C"))
dt
   ID Char
1:  1    A
2:  8    A
3:  9    B
4: 20    B
5: 32    C
6: 33    C

I want to identify "runs" by ID, ie consecutive rows where the ID differs by 1, but I only want to consider runs within the same Char group. I can do this as follows

Correct

dt[, InRun := FALSE]
dt[, DistToAbove := abs(ID - shift(ID, type="lag")), by=Char]
dt[, DistToBelow := abs(ID - shift(ID, type="lead")), by=Char]
dt[DistToAbove <= 1 | DistToBelow <= 1, InRun := TRUE, by=Char]
dt
   ID Char InRun DistToAbove DistToBelow
1:  1    A FALSE          NA           7
2:  8    A FALSE           7          NA
3:  9    B FALSE          NA          11
4: 20    B FALSE          11          NA
5: 32    C  TRUE          NA           1
6: 33    C  TRUE           1          NA

I tried simplifying the above code into the lines below, but the answer differs

Incorrect / Unexpected

dt[, InRun := FALSE]
dt[abs(ID - shift(ID, type="lag")) <= 1 | abs(shift(ID, type="lead") - ID) <= 1, InRun := TRUE, by=Char]
dt
   ID Char InRun DistToAbove DistToBelow
1:  1    A FALSE          NA           7
2:  8    A  TRUE           7          NA
3:  9    B  TRUE          NA          11
4: 20    B FALSE          11          NA
5: 32    C  TRUE          NA           1
6: 33    C  TRUE           1          NA

What gives? (Note I'm using data.table v1.9.7)

Answer 1

I want to identify "runs" by ID, ie consecutive rows where the ID differs by 1, but I only want to consider runs within the same Char group.

Here's how I'd approach it:

dt[, run_id := cumsum(
  ( ID != shift(ID, fill = ID[1L]) + 1L )
  |
  ( Char != shift(Char, fill = Char[1L]) )
)]
dt[, in_run := .N > 1L, by=.(Char, run_id)]

   ID Char run_id in_run
1:  1    A      1  FALSE
2:  8    A      2  FALSE
3:  9    B      3  FALSE
4: 20    B      4  FALSE
5: 32    C      5   TRUE
6: 33    C      5   TRUE

This code identifies all runs (including those with length of one) and then tests for length greater than one (the OP's definition).

---

Regarding the OP's approach:

dt[abs(ID - shift(ID, type="lag")) <= 1 | abs(shift(ID, type="lead") - ID) <= 1, # i
  InRun := TRUE # j
  , by=Char] # by

In DT[i,j,by] the steps are: filter using i , then group with by , then calculate j . You can't do by-group calculations in i in the way attempted here.