fast subsetting in data.table in R - r

Question

Given a data.table, I would like to subset the items in there quickly. For example:
dt = data.table(a=1:10, key="a")
dt[a > 3 & a <= 7]
This is pretty slow still. I know I can do joins to get individual rows but is there a way to fact that the data.table is sorted to get quick subsets of this kind?
This is what I'm doing:
dt1 = data.table(id = 1, ym = c(199001, 199006, 199009, 199012), last_ym = c(NA, 199001, 199006, 199009), v = 1:4, key=c("id", "ym"))
dt2 = data.table(id = 1, ym = c(199001, 199002, 199003, 199004, 199005, 199006, 199007, 199008, 199009, 199010, 199011, 199012), v2 = 1:12, key=c("id","ym"))
For each id, here there is only 1, and ym in dt1, I would like to sum the values of v2 between current ym in dt1 and the last ym in dt1. That is, for ym == 199006 in dt1 I would like to return list(v2 = 2 + 3 + 4 + 5 + 6). These are the values of v2 in dt2 that are equal to or less than the current ym (excluding the previous ym). In code:
expr = expression({ #browser();
cur_id = id;
cur_ym = ym;
cur_dtb = dt2[J(cur_id)][ym <= cur_ym & ym > last_ym];
setkey(cur_dtb , ym);
list(r = sum(cur_dtb$v2))
})
dt1[,eval(expr ),by=list(id, ym)]

Answer 1

To avoid the logical condition, perform a rolling join of dt1 and dt2. Then shift ym forward by one position within id. Finally, sum over v2 by id and ym:
setkey(dt1, id, last_ym)
setkey(dt2, id, ym)
dt1[dt2,, roll = TRUE][
, list(v2 = v2, ym = c(last_ym[1], head(ym, -1))), by = id][
, list(v2 = sum(v2)), by = list(id, ym)]
Note that we want to sum everything since the last_ym so the key on dt1 must be last_ym rather than ym.
The result is:
id ym v2
1: 1 199001 1
2: 1 199006 20
3: 1 199009 24
4: 1 199012 33
UPDATE: correction

Answer 2

Regardless of the fact that data.table is sorted, you will be limited to the amount of time it takes to evaluate a > 3 & a <= 7 in the first place:
> dt = data.table(a=1:10000000, key="a")
> system.time(dt$a > 3 & dt$a <= 7)
user system elapsed
0.18 0.01 0.20
> system.time(dt[,a > 3 & a <= 7])
user system elapsed
0.18 0.05 0.24
> system.time(dt[a > 3 & a <= 7])
user system elapsed
0.25 0.07 0.31
Alternative approach:
> system.time({Indices = dt$a > 3 & dt$a <= 7 ; dt[Indices]})
user system elapsed
0.28 0.03 0.31
Multiple Subsets
There can be a speed issue here if you break up factors on an ad hoc basis rather than doing it all at once first:
> dt <- data.table(A=sample(letters, 10000, replace=T))
> system.time(for(i in unique(dt$A)) dt[A==i])
user system elapsed
5.16 0.42 5.59
> system.time(dt[,.SD,by=A])
user system elapsed
0.32 0.03 0.36