How to count occurrences combinations in data.table in R

Question

I have two data.tables. I would like to count the number of rows matching a combination of a table in another table. I have checked the data.table documentation but I have not found my answer. I am using data.table 1.9.2.

DT1 <- data.table(a=c(3,2), b=c(8,3))
DT2 <- data.table(w=c(3,3,3,2,3), x=c(8,8,8,3,7), z=c(2,6,7,2,2))
DT1
#    a b
# 1: 3 8
# 2: 2 3

DT2
#    w x z
# 1: 3 8 2
# 2: 3 8 6
# 3: 3 8 7
# 4: 2 3 2
# 5: 3 7 2

Now I would like to count the number of (3, 8) pairs and (2, 3) pairs in DT2.

setkey(DT2, w, x)
nrow(DT2[J(3, 8), nomatch=0])
# [1] 3    ## OK !

nrow(DT2[J(2, 3), nomatch=0])
# [1] 1    ## OK !

DT1[,count_combination_in_dt2 := nrow(DT2[J(a, b), nomatch=0])]
DT1
#    a b count_combination_in_dt2
# 1: 3 8                        4 ## not ok.
# 2: 2 3                        4 ## not ok.

Expected result:

#    a b count_combination_in_dt2
# 1: 3 8                        3 
# 2: 2 3                        1 

Pointtopoint

Answer 1

You just need to add by=list(a,b).

DT1[,count_combination_in_dt2:=nrow(DT2[J(a,b),nomatch=0]), by=list(a,b)]
DT1
## 
##    a b count_combination_in_dt2
## 1: 3 8                        3
## 2: 2 3                        1

EDIT: Some more details: In your original version, you used DT2[DT1, nomatch=0] (because you used all a, b combinations. If you want to use J(a,b) for each a, b combination separately, you need to use the by argument. The data.table is then grouped by a, b and the nrow(...) is evaluated within each group.

Jaddo

Answer 2

setkey(DT2, w, x)

DT2[DT1, .N, by = .EACHI]
#   w x N
#1: 3 8 3
#2: 2 3 1

# In versions <= 1.9.2, use DT2[DT1, .N] instead

The above simply does the merge and counts the number of rows for each group defined by the i-expression, thus by = .EACHI.

Adamo

Thanks! Is is possible to assign the result column in DT1? I have tried DT2[DT1, count_combination_in_dt2:=.N] but it is not working. – Pointtopoint

DT1[,count:=DT2[DT1, .N][,N]] works and is very fast compare to the first solution. – Pointtopoint