使用 enquo() 以 dplyr 语法编写可管道函数，不返回预期输出

Question

library(tidyverse)
library(stringr)
library(janitor) 

word_count <- function(data, char_col) {
  char_col <- enquo(char_col)

  data %>% 
    select(!!char_col) %>% 
    mutate(char_col = str_remove_all(!!char_col, '[[:punct:]]')) %>% 
    mutate(char_col = str_split(!!char_col, ' ')) %>% 
    separate(char_col, into = paste0('col', 1:30), fill = 'right') %>% 
    select(-col1) %>% 
    gather(value = word) %>% 
    select(word) %>% 
    remove_empty(c('rows')) %>%
    filter(word != '') %>% 
    mutate(word = str_to_lower(word)) %>% 
    group_by(word) %>% 
    summarize(freq = n()) %>% 
   arrange(desc(freq))
}

iris %>% 
  as.tibble() %>% 
  mutate(Species = str_c(Species, ' species')) %>% 
  word_count(Species)

此代码在函数外部按预期工作，但当我在函数内部使用它时，它将 return 每个单词和每个 'non-split' 字符串的频率。

我认为这是我放置“!!”的方式的问题运营商，但我无法通过与他们一起试错来解决这个问题。这也可能是一个懒惰的问题，我不确定如何解决。

我希望函数的输出与下面代码的输出匹配。

iris %>% 
  as.tibble() %>% 
  mutate(Species = str_c(Species, ' species')) %>% 
  select(Species) %>% 
  mutate(Species = str_remove_all(Species, '[[:punct:]]')) %>% 
  mutate(Species = str_split(Species, ' ')) %>% 
  separate(Species, into = paste0('col', 1:30), fill = 'right') %>% 
  select(-col1) %>% 
  gather(value = word) %>% 
  select(word) %>% 
  remove_empty(c('rows')) %>%
  filter(word != '') %>% 
  mutate(word = str_to_lower(word)) %>% 
  group_by(word) %>% 
  summarize(freq = n()) %>% 
  arrange(desc(freq))

Answer 1

我们不能将 char_col 对象作为列名分配给 mutate。它需要被评估。 char_col 是一个 quosure 对象，可以将其转换为 character (quo_name(char_col)) 或 symbol，在评估时 (!!) 将分配 ( :=) 正确的列名

 word_count <- function(data, char_col) {
  char_col <- enquo(char_col)
  char_colC <- quo_name(char_col)

  data %>% 
    select(!!char_col) %>% 
    mutate(!!char_colC := str_remove_all(!!char_col, '[[:punct:]]')) %>% 
    mutate(!!char_colC := str_split(!!char_col, ' ')) %>% 
    separate(char_colC, into = paste0('col', 1:30), fill = 'right') %>%
    select(-col1) %>% 
    gather(value = word) %>% 
    select(word) %>% 
    remove_empty(c('rows')) %>%
    filter(word != '') %>% 
    mutate(word = str_to_lower(word)) %>% 
    group_by(word) %>% 
    summarize(freq = n()) %>% 
    arrange(desc(freq))
}

out2 <- iris %>%
           as.tibble() %>%
           mutate(Species =str_c(Species, ' species')) %>%
           word_count(Species)

-不使用函数

检查输出

out1 <- iris %>% 
          as.tibble() %>% 
          mutate(Species = str_c(Species, ' species')) %>% 
          select(Species) %>% 
          mutate(Species = str_remove_all(Species, '[[:punct:]]')) %>% 
          mutate(Species = str_split(Species, ' ')) %>% 
          separate(Species, into = paste0('col', 1:30), fill = 'right') %>% 
          select(-col1) %>% 
          gather(value = word) %>% 
          select(word) %>% 
          remove_empty(c('rows')) %>%
          filter(word != '') %>% 
          mutate(word = str_to_lower(word)) %>% 
          group_by(word) %>% 
          summarize(freq = n()) %>% 
          arrange(desc(freq)) 


identical(out1, out2)
#[1] TRUE

使用 enquo() 以 dplyr 语法编写可管道函数，不返回预期输出

Writing pipeable function in dplyr syntax, using enquo(), not returning intended output

r

lazy-evaluation

dplyr

tidyr