r - 当它实际上是一个因素时，是否有更快的方法来重新编码字符数据？

Question

我经常处理需要重新编码的字符数据。一种常见的情况是，正在记录的字符向量本质上是一个因素，但不一定属于类。例如，考虑一个 chr 向量，例如以下 vec:

set.seed(2021)
vec <- sample(rep(c("animal_dog_xyz", "animal_cat_abc", "animal_alligator_tyl"), 10))
vec
#>  [1] "animal_dog_xyz"       "animal_alligator_tyl" "animal_cat_abc"      
#>  [4] "animal_cat_abc"       "animal_alligator_tyl" "animal_alligator_tyl"
#>  [7] "animal_cat_abc"       "animal_cat_abc"       "animal_cat_abc"      
#> [10] "animal_dog_xyz"       "animal_dog_xyz"       "animal_cat_abc"      
#> [13] "animal_alligator_tyl" "animal_alligator_tyl" "animal_alligator_tyl"
#> [16] "animal_cat_abc"       "animal_dog_xyz"       "animal_alligator_tyl"
#> [19] "animal_alligator_tyl" "animal_cat_abc"       "animal_dog_xyz"      
#> [22] "animal_cat_abc"       "animal_cat_abc"       "animal_dog_xyz"      
#> [25] "animal_dog_xyz"       "animal_dog_xyz"       "animal_dog_xyz"      
#> [28] "animal_dog_xyz"       "animal_alligator_tyl" "animal_alligator_tyl"

^{由 reprex package 创建于 2021-07-19 (v2.0.0)}

如果我想重新编码这个向量并只提取动物名称，我会选择适合字符数据的解决方案:

library(stringr)

sapply(str_split(vec, "_",  n = 3), `[`, 2)
#>  [1] "dog"       "alligator" "cat"       "cat"       "alligator" "alligator"
#>  [7] "cat"       "cat"       "cat"       "dog"       "dog"       "cat"      
#> [13] "alligator" "alligator" "alligator" "cat"       "dog"       "alligator"
#> [19] "alligator" "cat"       "dog"       "cat"       "cat"       "dog"      
#> [25] "dog"       "dog"       "dog"       "dog"       "alligator" "alligator"

问题

如果向量很长，这样的重新编码过程需要很长时间。 R 将遍历每个向量元素并应用该过程。鉴于向量中只有 3 个唯一值，这似乎效率低下。换句话说，我们不需要一个一个地检查元素并找出重新编码的值应该是什么。

此处，vec_long 的长度为 30000。这是在我的机器上重新编码所需的时间。

vec_long <- sample(rep(c("animal_dog_xyz", "animal_cat_abc", "animal_alligator_tyl"), 10000))
length(vec_long)
#> [1] 30000

library(microbenchmark)

microbenchmark(sapply(str_split(vec_long, "_",  n = 3), `[`, 2))
#> Unit: milliseconds
#>                                             expr     min       lq     mean
#>  sapply(str_split(vec_long, "_", n = 3), `[`, 2) 51.6972 52.66918 57.42299
#>    median      uq     max neval
#>  54.47867 58.7653 115.754   100

有没有办法利用这个向量实际上是一个因素这一事实？从而识别唯一值(“级别”)，重新编码它们，并重新部署到整个矢量长度？是否有这样的程序可以加快处理时间？

谢谢!

---

编辑

---

我只想根据@GKi 的回答、@ThomasIsCoding 的回答和@user20650 的评论总结我的测试。

## The Data
set.seed(2021)

unique_vals <- c("animal_dog_xyz", "animal_cat_abc", "animal_alligator_tyl")

vec <- sample(rep(unique_vals, 10))
vec_long <- sample(rep(unique_vals, 1000))
vec_very_long <- sample(unique_vals, 100000))

## The functions

## function #1 -- as @user20650 proposed
via_fac_levels <- function(x) {
  x_factor <- factor(x)
  levels(x_factor) <- sapply(str_split(levels(x_factor), "_",  n = 3), `[`, 2)
  as.character(x_factor)
}
####################

## function #2 --  as @GKi proposed
via_fac_no_levels <- function(x) {
  x_factor <- as.factor(x)
  x_factor <- sapply(strsplit(levels(x_factor), "_", TRUE), `[`, 2)[x_factor]
  as.character(x_factor)
}
####################

## function #3 -- the original slow method shown in the question
via_chr_only <- function(x) {
  sapply(str_split(x, "_",  n = 3), `[`, 2)
}

####################

## function #4 -- as @ThomasIsCoding proposed
via_read_table <- function(x) {
  read.table(text = paste0(x, collapse = "\n"), sep = "_", header = FALSE)$V2
}

###################

## function #5 -- forcats::fct_relabel()
via_fct_relabel <- function(x) {
  x_factor <- as.factor(x)
  x_factor <- fct_relabel(x_factor, ~sapply(str_split(.x, "_",  n = 3), `[`, 2))
  as.character(x_factor)
}

## Performance assessment
### I ran it on Rstudio cloud
bm_short <- bench::mark(fac_levels = via_fac_levels(vec),
                  fac_no_levels = via_fac_no_levels(vec),
                  chr = via_chr_only(vec), 
                  read_t = via_read_table(vec),
                  fct_relabel = via_fct_relabel(vec),
                  iterations = 1000)

bm_long <- bench::mark(fac_levels = via_fac_levels(vec_long),
                        fac_no_levels = via_fac_no_levels(vec_long),
                        chr = via_chr_only(vec_long), 
                       read_t = via_read_table(vec_long),
                       fct_relabel = via_fct_relabel(vec_long),
                       iterations = 1000)

bm_very_long <- bench::mark(fac_levels = via_fac_levels(vec_very_long),
                  fac_no_levels = via_fac_no_levels(vec_very_long),
                  chr = via_chr_only(vec_very_long),
                  read_t = via_read_table(vec_very_long),
                  fct_relabel = via_fct_relabel(vec_very_long),
                  iterations = 1000)

## visualize
library(ggplot2)
library(tidyr)
library(ggbeeswarm)
library(beeswarm)

autoplot(bm_short) + ggtitle("data of length 30")
autoplot(bm_long) + ggtitle("data of length 3000")
autoplot(bm_very_long) + ggtitle("data of length 300000")

## verify all functions give the same output
v1 <- via_fac_levels(vec_long)
v2 <- via_fac_no_levels(vec_long)
v3 <- via_chr_only(vec_long)
v4 <- via_read_table(vec_long)
v5 <- via_fct_relabel(vec_long)

all(sapply(list(v1, v2, v3, v4), FUN = identical, v5)) # https://stackoverflow.com/a/30850654/6105259
## [1] TRUE

Answer 1

如果您创建一个因素，这也需要一些时间，它可能看起来像:

vec_fac <- as.factor(vec_long)
sapply(strsplit(levels(vec_fac), "_", TRUE), `[`, 2)[vec_fac]

如果数据也需要重新编码:

levels(vec_fac) <- sapply(strsplit(levels(vec_fac), "_", TRUE), `[`, 2)
vec_fac

基准:

bench::mark(
         Question = sapply(strsplit(vec_long, "_",  TRUE), `[`, 2)
       , HaveFactor = sapply(strsplit(levels(vec_fac), "_", TRUE), `[`, 2)[vec_fac]
       , CreateFactor = {vec_fac <- as.factor(vec_long)
         sapply(strsplit(levels(vec_fac), "_", TRUE), `[`, 2)[vec_fac]}
       )
#  expression        min   median `itr/sec` mem_alloc `gc/sec` n_itr  n_gc
#  <bch:expr>   <bch:tm> <bch:tm>     <dbl> <bch:byt>    <dbl> <int> <dbl>
#1 Question       16.2ms   16.4ms      59.6    1.17MB     51.7    15    13
#2 HaveFactor    233.9µs    251µs    3757.   234.42KB     16.8  1788     8
#3 CreateFactor  882.8µs  920.2µs    1073.   959.36KB     19.2   503     9