我正在尝试编写一个使用pivot_longer的函数，并且希望将我的函数对象用作pivot_longer中的names_to参数的对象。

record <- c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
x214532 <- c("shirts, shoes",
"shoes, purses, hats",
"shirts, shoes, hats, heavy machinery",
"sponges, shoes",
"hats, heavy machinery",
"",
"heavy machinery, purses, shirts",
"heavy machinery, shoes, sponges",
"sponges",
"shoes")
screening_data_responses_char <- data.frame(record, x214532)

record                              x214532
1       1                        shirts, shoes
2       2                  shoes, purses, hats
3       3 shirts, shoes, hats, heavy machinery
4       4                       sponges, shoes
5       5                hats, heavy machinery
6       6                                     
7       7      heavy machinery, purses, shirts
8       8      heavy machinery, shoes, sponges
9       9                              sponges
10     10                                shoes

最后，我尝试取消连接列x214532并创建一个长数据集，将数据分离到列中列出的项中，然后创建一个长数据集，如下所示：

record         x214532
1       1          shirts
2       1           shoes
3       2           shoes
4       2          purses
5       2            hats
6       3          shirts
7       3           shoes
8       3            hats
9       3 heavy machinery
10      4         sponges
11      4           shoes
12      5            hats
13      5 heavy machinery
14      6                
15      7 heavy machinery
16      7          purses
17      7          shirts
18      8 heavy machinery
19      8           shoes
20      8         sponges
21      9         sponges
22     10           shoes

型
我希望包含数据的列仍然被称为x214532，但我在通过pivot_longer的names_to传递它时遇到了问题。下面是我得到的结果：

remove_col_prefix <- function(x) {
  pattern <- "^[^_]+_"
  stringr::str_remove(x, pattern)
}

deconcatenate <- function(questionID) {
  screening_data_responses_questionID <- cSplit_e(data=screening_data_responses_char,split.col=questionID,sep=",",type="character")
  screening_data_responses_questionID <- screening_data_responses_questionID %>% 
    select(-questionID) %>% 
    pivot_longer(cols=c(starts_with(questionID)), 
                 names_to="questionID", 
                 values_to="questionID_resp") %>% 
    drop_na(questionID_resp) %>% 
    select(-questionID_resp) %>% 
    mutate(questionID=remove_col_prefix(questionID)) %>%
    select(c(deid_pat_id, questionID))
  
  screening_data_responses_char <- screening_data_responses_char %>% 
    select(-questionID)
  
  screening_data_responses_char <-merge(screening_data_responses_char,screening_data_responses_questionID,by="deid_pat_id",all=TRUE)
  }

screening_data_responses_char <- deconcatenate(questionID="x214532")

型

我尝试过的事情：

-{{}}和！！运算符（Using function arguments as column names）
-enquo

• 这个水泥函数：（https://adv-r.hadley.nz/quasiquotation.html）
  -deparse（subsitute（x））

我得到的东西：

• 列从输出中完全消失
• 列名为questionID，而不是x214532
• 该列名为questionID，所有文本都转换为x214532
  我肯定有一些地方我做错了，或者可能是我在pivot_longer中做对了一些事情，但还需要进一步修改语法，但我不能完全弄清楚。任何帮助都将不胜感激！