With a 2-column data.table
, I\'d like to summarize the pairwise relationships in column 1 by summing the number of shared elements in column 2. In other words,
You already have solution written in SQL so I suggest R package sqldf
Here's code:
library(sqldf)
result <- sqldf("SELECT A.X, B.X, COUNT(A.Y) as N FROM test as A JOIN test as B WHERE A.Y==B.Y GROUP BY A.X, B.X")
If you can split your Y
's into groups that don't have a large intersection of X
's, you could do the computation by those groups first, resulting in a smaller intermediate table:
d[, grp := Y <= 3] # this particular split works best for OP data
d[, .SD[.SD, allow = T][, .N, by = .(X, i.X)], by = grp][,
.(N = sum(N)), by = .(X, i.X)]
The intermediate table above has only 16 rows, as opposed to 26. Unfortunately I can't think of an easy way to create such grouping automatically.
How about this one using foverlaps()
. The more consecutive values of Y
you've for each X
, the lesser number of rows this'll produce compared to a cartesian join.
d = data.table(X=c(1,1,1,2,2,2,2,3,3,3,4,4), Y=c(1,2,3,1,2,3,4,1,5,6,4,5))
setorder(d, X)
d[, id := cumsum(c(0L, diff(Y)) != 1L), by=X]
dd = d[, .(start=Y[1L], end=Y[.N]), by=.(X,id)][, id := NULL][]
ans <- foverlaps(dd, setkey(dd, start, end))
ans[, count := pmin(abs(i.end-start+1L), abs(end-i.start+1L),
abs(i.end-i.start+1L), abs(end-start+1L))]
ans[, .(count = sum(count)), by=.(X, i.X)][order(i.X, X)]
# X i.X count
# 1: 1 1 3
# 2: 2 1 3
# 3: 3 1 1
# 4: 1 2 3
# 5: 2 2 4
# 6: 3 2 1
# 7: 4 2 1
# 8: 1 3 1
# 9: 2 3 1
# 10: 3 3 3
# 11: 4 3 1
# 12: 2 4 1
# 13: 3 4 1
# 14: 4 4 2
Note: make sure
X
andY
are integers for faster results. This is because joins on integer types are faster than on double types (foverlaps
performs binary joins internally).
You can make this more memory efficient by using which=TRUE
in foverlaps()
and using the indices to generate count
in the next step.