Har*_*y M 0 r dplyr tidyverse tidyeval
我正在尝试编写一个自定义 case_when 函数以在 dplyr 内部使用。我一直在阅读其他问题中发布的 tidyeval 示例,但仍然不知道如何使其工作。这是一个代表:
\n\ndf1 <- data.frame(animal_1 = c("Horse", "Pig", "Chicken", "Cow", "Sheep"),\n animal_2 = c(NA, NA, "Horse", "Sheep", "Chicken"))\n\n\ntranslate_title <- function(data, input_col, output_col) {\n mutate(data, \n !!output_col := case_when(\n input_col == "Horse" ~ "Cheval",\n input_col == "Pig" ~ "\xd0\xa0orc",\n input_col == "Chicken" ~ "Poulet",\n TRUE ~ NA)\n )\n}\n\n\ndf1 %>% \n translate_title("animal_1", "animaux_1") %>% \n translate_title("animal_2", "animaux_2")\n
Run Code Online (Sandbox Code Playgroud)\n\n当我尝试运行它时,我得到\nError in mutate_impl(.data, dots) : Evaluation error: must be type string, not logical.
另外,我实际上想重写该函数,以便可以像这样使用它:
\n\ndf1 %>% \n mutate(animaux_1 = translate_title(animal_1),\n animaux_2 = translate_title(animal_2)\n )\n
Run Code Online (Sandbox Code Playgroud)\n\n但不确定如何。
\n根据您希望如何将输入传递给函数,您可以通过两种方式解决它:
\n\n1)使用不带引号的方式传递输入{{}}
library(dplyr)\n\ntranslate_title <- function(data, input_col, output_col) {\n\n mutate(data, \n !!output_col := case_when(\n {{input_col}} == "Horse" ~ "Cheval",\n {{input_col}} == "Pig" ~ "\xd0\xa0orc",\n {{input_col}} == "Chicken" ~ "Poulet",\n TRUE ~ NA_character_)\n )\n}\n\ndf1 %>% \n translate_title(animal_1, "animaux_1") %>%\n translate_title(animal_2, "animaux_2")\n\n# animal_1 animal_2 animaux_1 animaux_2\n#1 Horse <NA> Cheval <NA>\n#2 Pig <NA> \xd0\xa0orc <NA>\n#3 Chicken Horse Poulet Cheval\n#4 Cow Sheep <NA> <NA>\n#5 Sheep Chicken <NA> Poulet\n
Run Code Online (Sandbox Code Playgroud)\n\n2) 使用sym
and传递引用的输入!!
translate_title <- function(data, input_col, output_col) {\n mutate(data, \n !!output_col := case_when(\n !!sym(input_col) == "Horse" ~ "Cheval",\n !!sym(input_col) == "Pig" ~ "\xd0\xa0orc",\n !!sym(input_col) == "Chicken" ~ "Poulet",\n TRUE ~ NA_character_)\n )\n}\n\ndf1 %>% \n translate_title("animal_1", "animaux_1") %>%\n translate_title("animal_2", "animaux_2")\n
Run Code Online (Sandbox Code Playgroud)\n