r - 取消列出/取消数据框中字符向量的嵌套？

Question

是否有使用 dplyr 在数据框中取消列出或取消嵌套矢量的功能或方法？我有以下示例。

library(tidyverse)
df <- tibble(x = 
      c("c(\"Police\", \"Pensions\")",
        "c(\"potato\", \"sweetpotato\")"))
df
# A tibble: 2 x 1
  x                               
  <chr>                           
1 "c(\"Police\", \"Pensions\")"   
2 "c(\"potato\", \"sweetpotato\")"

我想将该数据框列转换成这样的格式。

> df
# A tibble: 4 x 1
  x          
  <chr>      
1 Police     
2 Pensions   
3 Potato     
4 Sweetpotato

Answer 1

一个选项是separate_rows

library(tidyverse)
df %>% 
    separate_rows(x) %>%
    filter(!x %in% c('c', ''))
# A tibble: 4 x 1
#  x          
#  <chr>      
#1 Police     
#2 Pensions   
#3 potato     
#4 sweetpotato

注意:分离和过滤

会更快/更有效

---

或者另一种选择是提取引号之间的单词，然后 unnest

df %>% 
   mutate(x = str_extract_all(x, '(?<=")[A-Za-z]+')) %>%
   unnest
# A tibble: 4 x 1
#  x          
#  <chr>      
#1 Police     
#2 Pensions   
#3 potato     
#4 sweetpotato

基准

在稍微大一点的数据上，

df1 <- df[rep(1:nrow(df), each = 1e5), ]

system.time({
df1 %>% 
    separate_rows(x) %>%
    filter(!x %in% c('c', ''))


})
#. user  system elapsed 
#  0.916   0.033   0.939 


system.time({
df1 %>%
 mutate(x = str_extract_all(x, '(?<=")[A-Za-z]+')) %>%
   unnest

})
#  user  system elapsed 
#  0.763   0.015   0.773 


system.time({

df1 %>% 
  mutate(x = map(x,~eval(parse(text=.)))) %>% 
  unnest

})
#user  system elapsed 
# 15.643   1.813  17.375