CODEWITHSUNDEEP
rdata.table
Beasterfield
Beasterfield

Reputation: 7113

Equivalent for dlply in data.table

I try to achieve the same what dlply does with data.table. So just as a very simple example:

library(plyr)
library(data.table)
dt <- data.table( p = c("A", "B"), q = 1:2 )

dlply( dt, "p", identity )
$A
  p q
1 A 1

$B
  p q
1 B 2

dt[ , identity(.SD), by = p ]
   p q
1: A 1
2: B 2

foo <- function(x) as.list(x)
dt[ , foo(.SD), by = p ]
   p q
1: A 1
2: B 2

Obviously the return values of foo are collapsed to one data.table. And I don't want to use dlply because it passes the split data.tables as data.frames to foo which makes further data.table operations within foo inefficient.

Upvotes: 7

Views: 649

Answers (3)

eddi
eddi

Reputation: 49448

Here's a more data.table oriented approach:

setkey(dt, p)
dt[, list(list(dt[J(.BY[[1]])])), by = p]$V1
#[[1]]
#   p q
#1: A 1
#
#[[2]]
#   p q
#1: B 2

There are more data.table style alternatives to the above but that seems to be the fastest - here's a comparison with lapply:

dt <- data.table( p = rep( LETTERS[1:25], 1E6), q = 25*1E6, key = "p" )
microbenchmark(dt[, list(list(dt[J(.BY[[1]])])), by = p]$V1, lapply(unique(dt$p), function(x) dt[x]), times = 10)
#Unit: seconds
#                                        expr      min       lq   median       uq      max neval
#dt[, list(list(dt[J(.BY[[1]])])), by = p]$V1 1.111385 1.508594 1.717357 1.966694 2.108188    10
#     lapply(unique(dt$p), function(x) dt[x]) 1.871054 1.934865 2.216192 2.282428 2.367505    10

Upvotes: 3

Beasterfield
Beasterfield

Reputation: 7113

Regarding G. Grothendieck's answer I was curious how well split performs:

dt <- data.table( p = rep( LETTERS[1:25], 1E6), q = 25*1E6, key = "p" )

system.time(
  ll <- split(dt, dt[ ,p ] )
)
  user  system elapsed 
  5.237   1.340   6.563 

system.time(
  ll <- lapply( unique(dt[,p]), function(x) dt[x] )
)
  user  system elapsed 
  1.179   0.363   1.541

So if there is no better answer, I'd stick with

lapply( unique(dt[,p]), function(x) dt[x] )

Upvotes: 2

G. Grothendieck
G. Grothendieck

Reputation: 269501

Try this:

> split(dt, dt[["p"]])
$A
   p q
1: A 1

$B
   p q
1: B 2

Upvotes: 2

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