Ren*_*rop 6 r subset filter dataframe dplyr
过滤data.frame以获得大小为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)
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我能想到的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
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但这很慢...... dplyr有更好的方法吗?
到目前为止,我的解决方案解决方案如下:
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
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但这并不像是......
Joe*_*Joe 14
您可以更简洁地使用n():
library(dplyr)
dat %>% group_by(cat) %>% filter(n() == 5)
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这是你可以尝试的另一种dplyr方法
semi_join(dat, count(dat, cat) %>% filter(n == 5), by = "cat")
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-
这是基于OP原始方法的另一种方法,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
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我知道你需要一个dplyr解决方案,但如果你将它与一些解决方案结合起来,purrr你可以在一行中得到它,而无需指定任何新功能。(虽然慢了一点。)
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)
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