I need to compute the ranks of dates by group. There are many small groups.
library(data.table)
library(lubridate)
library(microbenchmark)
set.seed(1)
NN <- 1000000
EE <- 10
# Just an example.
todo <- data.table(id=paste0("ID",rep(1:NN, each=EE)),
val=dmy("1/1/1980") + sample(1:14000,NN*EE,replace=T))
# I want to benchmark this:
todo[,ord := frank(val, ties.method="first"), by=id]
In order to compare it you can try with smaller NN, the timing is linear.
For NN = 1 million it takes 560 seconds.
Is there any way to do it faster?
I've been using lubridate but I can use any library you suggest.
In my real problem the number of rows within each ID is not constant.
I believe it is due to the overhead of calling frank multiple times for many small groups (the memory usage below should give you a hint on the bottleneck). Here is another option:
DT1[order(id, val), ord := rowid(id)]
timing code:
library(data.table)
set.seed(1L)
NN <- 1e6
EE <- 10
todo <- data.table(id=paste0("ID",rep(1:NN, each=EE)),
val=as.IDate("1980-01-01") + sample(1:14000,NN*EE,replace=T))
DT0 <- copy(todo)
DT1 <- copy(todo)
bench::mark(
todo[, ord := frank(val, ties.method="first"), by=id],
DT0[, ord := rank(unclass(val), ties.method = "first"), by = id],
DT1[order(id, val), ord := rowid(id)])
all.equal(todo$ord, DT0$ord)
# [1] TRUE
all.equal(todo$ord, DT1$ord)
# [1] TRUE
timings:
expression min median `itr/sec` mem_alloc `gc/sec` n_itr n_gc total_time result memory time
<bch:expr> <bch> <bch:> <dbl> <bch:byt> <dbl> <int> <dbl> <bch:tm> <list> <list> <lis>
1 todo[, `:=`(ord, frank(val, ties.method = "first")), by = id] 6.32m 6.32m 0.00264 15.7GB 0.177 1 67 6.32m <df[,~ <df[,~ <bch~
2 DT0[, `:=`(ord, rank(unclass(val), ties.method = "first")), by = id] 1.12m 1.12m 0.0149 99.3MB 0.969 1 65 1.12m <df[,~ <df[,~ <bch~
3 DT1[order(id, val), `:=`(ord, rowid(id))] 7.85s 7.85s 0.127 236.8MB 0 1 0 7.85s <df[,~ <df[,~ <bch~
It can be even faster if we remove id in order:
DT1[order(val), ord := rowid(id)]
timing code:
bench::mark(DT0[order(id, val), ord := rowid(id)],
DT1[order(val), ord := rowid(id)])
all.equal(DT0$ord, DT1$ord)
# [1] TRUE
timings:
# A tibble: 2 x 13
expression min median `itr/sec` mem_alloc `gc/sec` n_itr n_gc total_time result memory time gc
<bch:expr> <bch:tm> <bch:tm> <dbl> <bch:byt> <dbl> <int> <dbl> <bch:tm> <list> <list> <list> <list>
1 DT0[order(id, val), `:=`(ord, rowid(id))] 7.44s 7.44s 0.134 237MB 0 1 0 7.44s <df[,3] [10,000,000 x 3]> <df[,3] [15 x 3]> <bch:tm> <tibble [1 x 3]>
2 DT1[order(val), `:=`(ord, rowid(id))] 4.66s 4.66s 0.215 237MB 0 1 0 4.66s <df[,3] [10,000,000 x 3]> <df[,3] [14 x 3]> <bch:tm> <tibble [1 x 3]>