Fast top N by count by group in data.table - r

Question

I'd like to know the preferred way to frank subgroups on the count of their appearances by group.
For example, I have customers who belong to segments and who have postal codes. I would like to know the most common 3 postal codes for each segment.
library(data.table)
set.seed(123)
n <- 1e6
df <- data.table( cust_id = 1:n,
cust_segment = sample(LETTERS, size=n, replace=T),
cust_postal = sample(as.character(5e4:7e4),size=n, replace=T)
)
This chain (inside the dcast() below) produces the desired output but requires two passes, the first to count by group-subgroup and the second to rank the counts by group.
dcast(
df[,.(.N),
by = .(cust_segment, cust_postal)
][,.(cust_postal,
postal_rank = frankv(x=N, order=-1, ties.method = 'first')
), keyby=cust_segment
][postal_rank<=3],
cust_segment ~ paste0('postcode_rank_',postal_rank), value.var = 'cust_postal'
)
# desired output:
# cust_segment postcode_rank_1 postcode_rank_2 postcode_rank_3
# A 51274 64588 59212
# B 63590 69477 50380
# C 60619 66249 53494 ...etc...
Is that the best there is, or is there a single-pass approach?

Answer 1

Taking the answer from Frank out of the comments:
Using forder instead of frankv and using keyby as this is faster than just using by
df[, .N,
keyby = .(cust_segment, cust_postal)
][order(-N), r := rowid(cust_segment)
][r <= 3, dcast(.SD, cust_segment ~ r, value.var ="cust_postal")]
cust_segment 1 2 3
1: A 51274 53440 55754
2: B 63590 69477 50380
3: C 60619 66249 52122
4: D 68107 50824 59305
5: E 51832 65249 52366
6: F 51401 55410 65046
microbenchmark time:
library(microbenchmark)
microbenchmark(C8H10N4O2 = dcast(
df[,.(.N),
by = .(cust_segment, cust_postal)
][,.(cust_postal,
postal_rank = frankv(x=N, order=-1, ties.method = 'first')
), keyby=cust_segment
][postal_rank<=3],
cust_segment ~ paste0('postcode_rank_',postal_rank), value.var = 'cust_postal'
),
frank = df[, .N,
keyby = .(cust_segment, cust_postal)
][order(-N), r := rowid(cust_segment)
][r <= 3, dcast(.SD, cust_segment ~ r, value.var ="cust_postal")])
Unit: milliseconds
expr min lq mean median uq max neval
C8H10N4O2 136.3318 140.8096 156.2095 145.6099 170.4862 205.8457 100
frank 102.2789 110.0140 118.2148 112.6940 119.2105 192.2464 100
Frank's answer is about 25% faster.