如何在dplyr::mutate中传递字符串作为列名实现标签映射？

有一个包含调查选项列choices与选择索引列choice的dataframe，示例数据如下：

df <- tibble(
  record_id = 1:9,
  choices = c(rep("1, A | 2, B | 3, C", 3), 
              rep("1, Apple | 2, Banana | 3, Cherry", 3),
              rep("1, America | 2, Belgium | 3, China", 3)),
  choice = sample(1:3, size = 9, replace = T)
)

目标是根据choices列中的标签映射生成label列。已编写make_key函数，单独调用可正常运行：

make_key <- function(.str) {
    
  lstr <- str_split(.str, pattern = " \\| ")
  
  out <- map(lstr, ~str_remove(.x, pattern = "^([0-9]+), ")) %>% as_vector()
  
  out_names <- map(lstr, ~str_extract(.x, pattern = "^([0-9]+)")) %>% as_vector()
  
  names(out) <- out_names
  
  return(out)
}

# 正常运行示例：
my_string <- c("1, A | 2, B | 3, C")
recode(1, !!!make_key(my_string))
# [1] "A"

但在rowwise管道中调用dplyr::mutate时出现报错：

rowwise(df) %>%
  mutate(label = recode(choice, !!!make_key(choices))
)

# Error in stri_split_regex(string, pattern, n = n, simplify = simplify, : 
# object 'choices' not found

尝试过添加双大括号{{}}及rlang相关函数，均未解决问题，寻求可行解决方案。

方法1：用purrr::map2逐行配对处理 #

无需依赖rowwise，直接用map2_chr将choice和choices逐组传入，结合原make_key函数完成匹配：

library(dplyr)
library(purrr)
library(stringr)

df <- df %>%
  mutate(label = map2_chr(choice, choices, function(ch, str) {
    key <- make_key(str)
    recode(ch, !!!key)
  }))

方法2：整合匹配逻辑到函数，适配rowwise #

修改函数逻辑，让它直接接收索引和选项字符串并返回对应标签，避免非标准求值的环境问题：

get_label <- function(choice_idx, choices_str) {
  split_str <- str_split(choices_str, " \\| ")[[1]]
  idx <- str_extract(split_str, "^\\d+") %>% as.integer()
  labels <- str_remove(split_str, "^\\d+, ")
  labels[match(choice_idx, idx)]
}

df <- df %>%
  rowwise() %>%
  mutate(label = get_label(choice, choices)) %>%
  ungroup()

方法3：用tidyr拆分后匹配（无自定义函数） #

先拆分choices列提取索引与标签，再通过关联匹配生成结果，逻辑更直观：

library(tidyr)

df_label <- df %>%
  select(record_id, choices) %>%
  separate_rows(choices, sep = " \\| ") %>%
  mutate(
    choice_idx = str_extract(choices, "^\\d+") %>% as.integer(),
    label = str_remove(choices, "^\\d+, ")
  ) %>%
  select(-choices)

df <- df %>%
  left_join(df_label, by = c("record_id", "choice" = "choice_idx"))

内容的提问来源于stack exchange，提问作者sometimes_sci