dplyr - filter by group size

Question

What is the best way to filter a data.frame to only get groups of say size 5?

So my data looks as follows:

require(dplyr)
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)

The dplyr way i could come up with was

dat <- group_by(dat, cat)

system.time({
  out1 <- dat %>% filter(n() == 5L)
})
#    user  system elapsed 
#   1.157   0.218   1.497

But this is very slow... Is there a better way in dplyr?

So far my workaround solutions looks as follows:

system.time({
  all_ind <- rep(seq_len(n_groups(dat)), group_size(dat))
  take_only <- which(group_size(dat) == 5L)
  out2 <- dat[all_ind %in% take_only, ]
})
#    user  system elapsed 
#   0.026   0.008   0.036
all.equal(out1, out2) # TRUE

But this doesn't feel very dplyr like...

Answer 1

You can do it more concisely with n():

library(dplyr)
dat %>% group_by(cat) %>% filter(n() == 5)

Answer 2

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

Answer 3

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)

Answer 4

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  

Answer 5

I generalised the function written by docendo discimus, to use it alongside existing dplyr functions:

#' inherit dplyr::filter
#' @param min minimal group size, use \code{min = NULL} to filter on maximal group size only
#' @param max maximal group size, use \code{max = NULL} to filter on minimal group size only
#' @export
#' @source Stack Overflow answer by docendo discimus, \url{https://stackoverflow.com/a/43110620/4575331}
filter_group_size <- function(.data, min = NULL, max = min) {
  g <- dplyr::group_size(.data)
  if (is.null(min) & is.null(max)) {
    stop('`min` and `max` cannot both be NULL.')
  }
  if (is.null(max)) {
    max <- base::max(g, na.rm = TRUE)
  }
  ind <- base::rep(g >= min & g <= max, g)
  .data[ind, ]
}

Let's check it for a minimal group size of 5:

dat2 %>%
  group_by(cat) %>%
  filter_group_size(5, NULL) %>%
  summarise(n = n()) %>%
  arrange(desc(n))

# # A tibble: 6,634 x 2
#      cat     n
#    <int> <int>
#  1    NA    19
#  2     1     5
#  3     2     5
#  4     6     5
#  5    15     5
#  6    17     5
#  7    21     5
#  8    27     5
#  9    33     5
# 10    37     5
# # ... with 6,624 more rows

Great, now check for the OP's question; a group size of exactly 5:

dat2 %>%
  group_by(cat) %>%
  filter_group_size(5) %>%
  summarise(n = n()) %>%
  pull(n) %>%
  unique()
# [1] 5

Hooray.

Answer 6

A very simple way of accelerating the dplyr-way n() filter is to store the result in a new column. The initial time of calculating the group size is amortised if there are multiple filters later on.

library(dplyr)

prep_group <- function(dat) {
    dat %>%
        group_by(cat) %>%
        mutate(
            Occurrences = n()
        ) %>%
        ungroup()
}

# Create a new data frame with the `Occurrences` column:
# dat_prepped <- dat %>% prep_group

Filtering the Occurrences field is much faster than the workaround solution:

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_cached <- function(dat) {
    out <- filter(dat, Occurrences == 5L)
}

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)

dat_prepped <- prep_group(dat)

microbenchmark::microbenchmark(times=50, sol_floo0(dat), sol_floo0_v2(dat), sol_cached(dat_prepped))

Unit: microseconds
                    expr       min        lq      mean    median        uq        max neval cld
          sol_floo0(dat) 33345.764 35603.446 42430.441 37994.477 41379.411 144103.471    50   c
       sol_floo0_v2(dat) 26180.539 27842.927 29694.203 29089.672 30997.411  37412.899    50  b 
 sol_cached(dat_prepped)   801.402   930.025  1342.348  1098.843  1328.192   5049.895    50 a  

---

The preparation can be further accelerated by using count() -> left_join():

prep_join <- function(dat) {
    dat %>%
        left_join(
            dat %>%
                count(cat, name="Occurrences")
        )
}

microbenchmark::microbenchmark(times=10, prep_group(dat), prep_join(dat))

Unit: milliseconds
            expr      min       lq     mean   median       uq      max neval cld
 prep_group(dat) 45.67805 47.68100 48.98929 49.11258 50.08214 52.44737    10   b
  prep_join(dat) 35.01945 36.20857 37.96460 36.86776 38.71056 45.59041    10  a 

Answer 7

dat %>%
  dplyr::group_by(cat) %>%
  dplyr::add_tally() %>%
  dplyr::filter(n == 5)