如何通过唯一的组/子组对找到不等式的总和？

Question

Suppose I am working with the following data.table :

dta <- setDT(
  data.frame(
    id = c("A","A","A","B","B","C","C","C"),
    subid = c(1,1,2,1,2,1,1,1),
    x1 = c(1,1,3,1,2,3,3,3),
    x2 = c(3,3,1,1,1,3,3,3)
  )
)

> dta
   id subid x1 x2
1:  A     1  1  3
2:  A     1  1  3
3:  A     2  3  1
4:  B     1  1  1
5:  B     2  2  1
6:  C     1  3  3
7:  C     1  3  3
8:  C     1  3  3

For each unique id - subid pairing, I would like to find the total number of times that x1<x2 and the total number of times that x1>=x2 , and have those counts be added to the data.table as new columns/variables but aggregated to the id level.

The outcome would look something like:

   id subid x1 x2 lt gt
1:  A     1  1  3  1  1
2:  A     1  1  3  1  1
3:  A     2  3  1  1  1
4:  B     1  1  1  0  2
5:  B     2  2  1  0  2
6:  C     1  3  3  0  1
7:  C     1  3  3  0  1
8:  C     1  3  3  0  1

For example, of the two unique id-subid parings for id="A" , one has x1<x2 and one has x1>x2 , which means that for A the variable for "less-than" has a value of 1 (ie dta$lt[dta$id==A] <- 1 ), and the same for "greater-than" ( dta$gt[dta$id==A] <- 1 ).

I have been searching for a solution to this but have not had much luck. I have found solutions to similar problems (eg counting number of unique observations by unique pairings), but have not been able to modify them to suit my needs. In particular, I am struggling to aggregate the count from the id-subid level to the id level. (It could be that I'm not exactly sure how to search for -- or even word -- this question.)

I've been able to do this using nested loops on a data frame, but I suspect there is a more efficient way of doing it. In particular, I am curious about doing this using data.table .

Answer 1

A possible solution:

dta[, c('lt', 'gt') := unique(.SD)[, .(sum(x1 < x2), sum(x1 >= x2))], by = .(id)]

which gives:

 > dta id subid x1 x2 lt gt 1: A 1 1 3 1 1 2: A 1 1 3 1 1 3: A 2 3 1 1 1 4: B 1 1 1 0 2 5: B 2 2 1 0 2 6: C 1 3 3 0 1 7: C 1 3 3 0 1 8: C 1 3 3 0 1