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Subset by group with data.table
(1 answer)
Closed 2 years ago.
What is the idiomatic way to do the action below in data.table
?
library(dplyr)
df %>%
group_by(b) %>%
slice(1:10)
I can do
library(data.table)
df[, .SD[1:10]
, by = b]
but that appears much slower. Is there a better way?
set.seed(0)
df <- rep(1:500, sample(500:1000, 500, T)) %>%
data.table(a = runif(length(.))
,b = .)
f1 <- function(df){
df %>%
group_by(b) %>%
slice(1:10)
}
f2 <- function(df){
df[, .SD[1:10]
, by = b]
}
library(microbenchmark)
microbenchmark(f1(df), f2(df))
#Unit: milliseconds
# expr min lq mean median uq max neval
# f1(df) 17.67435 19.50381 22.06026 20.50166 21.42668 78.3318 100
# f2(df) 69.69554 79.43387 119.67845 88.25585 106.38661 581.3067 100
========== Benchmarks with suggested methods ==========
set.seed(0)
df <- rep(1:500, sample(500:1000, 500, T)) %>%
data.table(a = runif(length(.))
,b = .)
use.slice <- function(df){
df %>%
group_by(b) %>%
slice(1:10)
}
IndexSD <- function(df){
df[, .SD[1:10]
, by = b]
}
Index.I <- function(df) {
df[df[, .I[seq_len(10)], by = b]$V1]
}
use.head <- function(df){
df[, head(.SD, 10)
, by = b]
}
library(microbenchmark)
microbenchmark(use.slice(df)
, IndexSD(df)
, Index.I(df)
, use.head(df)
, unit = "relative"
, times = 100L)
#Unit: relative
# expr min lq mean median uq max neval
# use.slice(df) 9.804549 10.269234 9.167413 8.900060 8.782862 6.520270 100
# IndexSD(df) 38.881793 42.548555 39.044095 38.636523 39.942621 18.981748 100
# Index.I(df) 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 100
# use.head(df) 3.666898 4.033038 3.728299 3.408249 3.545258 3.951565 100