过滤data.frame以获取大小为5的组的最佳方法是什么？
所以我的数据如下：
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)
我能提出的dplyr方式是
dat <- group_by(dat, cat)
system.time({
out1 <- dat %>% filter(n() == 5L)
})
# user system elapsed
# 1.157 0.218 1.497
但这很慢......在dplyr中有更好的方法吗？
到目前为止，我的解决方案解决方案如下所示：
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
但是这并没有像...那样感觉非常明显。
答案 0 :(得分：5)
这是您可以尝试的另一种dplyr方法
semi_join(dat, count(dat, cat) %>% filter(n == 5), by = "cat")
-
这是另一种基于OP原创方法的方法，稍作修改：
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
答案 1 :(得分：2)
我知道您要求提供dplyr解决方案，但如果将其与某些purrr结合使用，您可以在一行中获取该解决方案，而无需指定任何新功能。 （虽然有点慢。）
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)
答案 2 :(得分：1)
按时间比较答案：
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))
结果：
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
答案 3 :(得分：1)
我概括了docendo discimus编写的函数，并将其与现有的dplyr函数一起使用：
#' 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, ]
}
让我们检查一下5的最小群组大小：
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
太好了，现在检查一下OP的问题;群组大小恰好为5：
dat2 %>%
group_by(cat) %>%
filter_group_size(5) %>%
summarise(n = n()) %>%
pull(n) %>%
unique()
# [1] 5
万岁。
答案 4 :(得分：0)
您可以使用n()来做得更简洁：
library(dplyr)
dat %>% group_by(cat) %>% filter(n() == 5)
答案 5 :(得分：0)
加速dplyr-way n()过滤器的一种非常简单的方法是将结果存储在新列中。如果以后有多个filter，则摊销计算组大小的初始时间。
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
过滤Occurrences字段比解决方法快得多：
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
使用count()-> left_join()可进一步加快准备工作：
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
答案 6 :(得分：0)
dat %>%
dplyr::group_by(cat) %>%
dplyr::add_tally() %>%
dplyr::filter(n == 5)