在R中的purrr：：pmap中使用带有子集的变量名的正确方法

我有一个叫description的tibble:

description <- structure(list(col1 = "age", col2 = "> 7 months", col3 = "<= 7 months"), row.names = c(NA, 
-1L), class = c("tbl_df", "tbl", "data.frame"))

> description
# A tibble: 1 × 3
  col1  col2       col3       
  <chr> <chr>      <chr>      
1 age   > 7 months <= 7 months

和一个称为my_df的数据帧:

my_df <- structure(list(ID = c("ID1", "ID2", "ID3", "ID4", "ID5", "ID6"
), age = structure(c(1L, 2L, 1L, 2L, 2L, 2L), .Label = c("<= 7 months", 
"> 7 months"), class = "factor")), row.names = c("ID1", "ID2", 
"ID3", "ID4", "ID5", "ID6"), class = "data.frame")

> my_df
     ID         age
ID1 ID1 <= 7 months
ID2 ID2  > 7 months
ID3 ID3 <= 7 months
ID4 ID4  > 7 months
ID5 ID5  > 7 months
ID6 ID6  > 7 months

我目前有以下功能:

updated_df <- purrr::pmap(description, function(col1, col2, col3) {
        subset(
            my_df,
            age == col2
        )
})

这将产生:

[[1]]
     ID        age
ID2 ID2 > 7 months
ID4 ID4 > 7 months
ID5 ID5 > 7 months
ID6 ID6 > 7 months

但是我想使用变量col1而不是age.我try 了以下方法，但都不起作用:

updated_df <- purrr::pmap(description, function(col1, col2, col3) {
        subset(
            my_df,
            col1 == col2
        )
})


updated_df <- purrr::pmap(description, function(col1, col2, col3) {
        subset(
            my_df,
            !!as.name(col1) == col2
        )
})

updated_df <- purrr::pmap(description, function(col1, col2, col3) {
        subset(
            my_df,
            !!col1 == col2
        )
})

我哪里做错了/用col1代替age的正确方法是什么？