dplyr - filter by group size
Comparing the answers timewise:
require(dplyr)
require(data.table)
n <- 1e5
x <- rnorm(n)
# Category size ranging each from 1 to 5
cat <- rep(seq_len(n/3), sample(1:5, n/3, replace = TRUE))[1:n]
dat <- data.frame(x = x, cat = cat)
# second data set for the dt approch
dat2 <- data.frame(x = x, cat = cat)
sol_floo0 <- function(dat){
dat <- group_by(dat, cat)
all_ind <- rep(seq_len(n_groups(dat)), group_size(dat))
take_only <- which(group_size(dat) == 5L)
dat[all_ind %in% take_only, ]
}
sol_floo0_v2 <- function(dat){
g <- group_by(dat, cat) %>% group_size()
ind <- rep(g == 5, g)
dat[ind, ]
}
sol_docendo_discimus <- function(dat){
dat <- group_by(dat, cat)
semi_join(dat, count(dat, cat) %>% filter(n == 5), by = "cat")
}
sol_akrun <- function(dat2){
setDT(dat2)[dat2[, .I[.N==5], by = cat]$V1]
}
sol_sotos <- function(dat2){
setDT(dat2)[, if(.N == 5) .SD, by = cat]
}
sol_chirayu_chamoli <- function(dat){
rle_ <- rle(dat$cat)
dat[dat$cat %in% rle_$values[rle_$lengths==5], ]
}
microbenchmark::microbenchmark(times = 20,
sol_floo0(dat),
sol_floo0_v2(dat),
sol_docendo_discimus(dat),
sol_akrun(dat2),
sol_sotos(dat2),
sol_chirayu_chamoli(dat))
Results in:
Unit: milliseconds
expr min lq mean median uq max neval cld
sol_floo0(dat) 58.00439 65.28063 93.54014 69.82658 82.79997 280.23114 20 cd
sol_floo0_v2(dat) 42.27791 50.27953 72.51729 58.63931 67.62540 238.97413 20 bc
sol_docendo_discimus(dat) 100.54095 113.15476 126.74142 121.69013 132.62533 183.05818 20 d
sol_akrun(dat2) 26.88369 34.01925 41.04378 37.07957 45.44784 63.95430 20 ab
sol_sotos(dat2) 16.10177 19.78403 24.04375 23.06900 28.05470 35.83611 20 a
sol_chirayu_chamoli(dat) 20.67951 24.18100 38.01172 27.61618 31.97834 230.51026 20 ab
You can do it more concisely with n()
:
library(dplyr)
dat %>% group_by(cat) %>% filter(n() == 5)
I know you asked for a dplyr
solution but if you combine it with some purrr
you can get it in one line without specifying any new functions. (A little slower though.)
library(dplyr)
library(purrr)
library(tidyr)
dat %>%
group_by(cat) %>%
nest() %>%
mutate(n = map(data, n_distinct)) %>%
unnest(n = n) %>%
filter(n == 5) %>%
select(cat, n)
Here's another dplyr approach you can try
semi_join(dat, count(dat, cat) %>% filter(n == 5), by = "cat")
--
Here's another approach based on OP's original approach with a little modification:
n <- 1e5
x <- rnorm(n)
# Category size ranging each from 1 to 5
cat <- rep(seq_len(n/3), sample(1:5, n/3, replace = TRUE))[1:n]
dat <- data.frame(x = x, cat = cat)
# second data set for the dt approch
dat2 <- data.frame(x = x, cat = cat)
sol_floo0 <- function(dat){
dat <- group_by(dat, cat)
all_ind <- rep(seq_len(n_groups(dat)), group_size(dat))
take_only <- which(group_size(dat) == 5L)
dat[all_ind %in% take_only, ]
}
sol_floo0_v2 <- function(dat){
g <- group_by(dat, cat) %>% group_size()
ind <- rep(g == 5, g)
dat[ind, ]
}
microbenchmark::microbenchmark(times = 10,
sol_floo0(dat),
sol_floo0_v2(dat2))
#Unit: milliseconds
# expr min lq mean median uq max neval cld
# sol_floo0(dat) 43.72903 44.89957 45.71121 45.10773 46.59019 48.64595 10 b
# sol_floo0_v2(dat2) 29.83724 30.56719 32.92777 31.97169 34.10451 38.31037 10 a
all.equal(sol_floo0(dat), sol_floo0_v2(dat2))
#[1] TRUE