how to rename a variable using a dynamic name and dplyr?

Question

I want to rename a column inside a function with a name passed as an argument for this function. Basically, I have a function

produce_data_frame <- function(name) {
  return(iris)
}

And I want that this function change the Sepal.length column name with 'name' (with name taking the value of name) I tried different things such as

produce_data_frame <- function(name) {
  name <- enquo(name)
  iris %>%
    rename((!!name) = Sepal.Length) %>%
    return()
}

And when calling

produce_data_frame("newName")

I would like to get back the iris data.frame with the Sepal.Length column named newName. But my understanding of NSE is still very basic, and it doesn't even compile.

not sure what was wrong with my question, I tried to develop my exemple, if this is not the problem, could some one explain what is the problem with it ? Thanks, — Malta, Apr 06 '18 at 05:30

Roman · Accepted Answer · 2021-09-08T11:39:45.177

You can try

library(tidyverse)

produce_data_frame <- function(name) {
  iris %>%
    as.tibble() %>% 
    select(!!quo_name(name) := Sepal.Length)
}

produce_data_frame("new_name")
#> # A tibble: 150 x 1
#>    new_name
#>       <dbl>
#>  1     5.10
#>  2     4.90
#>  3     4.70
#>  4     4.60
#>  5     5.00
#>  6     5.40
#>  7     4.60
#>  8     5.00
#>  9     4.40
#> 10     4.90
#> # ... with 140 more rows

Created on 2018-04-04 by the reprex package (v0.2.0).

Or you can use{{...}}

produce_data_frame <- function(name) {
  iris %>%
    as_tibble() %>% 
    select({{name}} := Sepal.Length)
}

score 22 · Answer 2 · answered Apr 04 '18 at 12:20

22

From Different input and output variable in the Programming with dplyr vignette, we can use the := operator:

library(dplyr)
library(rlang)


produce_data_frame <- function(name) {
  name = quo_name(name)
  iris %>%
    rename(!!name := Sepal.Length)
}

produce_data_frame('test') %>% colnames()
#> [1] "test"         "Sepal.Width"  "Petal.Length" "Petal.Width" 
#> [5] "Species"

answered Apr 04 '18 at 12:20

GGamba

13,140
3
38
47

1

This is clean and beautiful! – Helene Sep 25 '21 at 20:47
1

I had no idea that dplyr has a walrus operator. Thanks for sharing! – philiporlando Mar 21 '22 at 23:40

Shique · Answer 3 · 2018-04-06T07:19:07.957

6

You can use base R names() instead.

produce_data_frame <- function(name) {
  temp_df <- iris
  names(temp_df)[names(temp_df) == "Sepal.Length"] <- name
  return(temp_df)
}

produce_data_frame("newName")
#        newName Sepal.Width Petal.Length Petal.Width    Species
#1           5.1         3.5          1.4         0.2     setosa
#2           4.9         3.0          1.4         0.2     setosa

edited Apr 06 '18 at 07:19

answered Apr 04 '18 at 12:14

Shique

724
3
18

score 4 · Answer 4 · answered May 21 '21 at 23:32

Using dplyr::rename and glue there is little need to write such a function, as you can always do the following:

library(glue)
library(dplyr)
name <- "new_name"
rename(iris,"{name}" := Sepal.Length)

If a function is needed, there are many ways to implement it. The following is a generic function to rename a named vector or list (which includes data frames):

rename_named_obj <- function(named_obj,new_names) {
  setNames(named_obj,
    purrr::map2_chr(names(named_obj),
                    names(new_names)[match(names(named_obj),new_names)],
      ~ {
        ifelse(is.na(.y),.x,.y)
      })
  )
}

Giving the following results:

> new_iris_names <- list("new_name1"="Sepal.Length","new_name2"="Petal.Length")
> rename_named_obj(iris,new_iris_names) %>% colnames()
[1] "new_name1"   "Sepal.Width" "new_name2"   "Petal.Width" "Species"    
> rename_named_obj(setNames(1:5,colnames(df)),new_iris_names)
  new_name1 Sepal.Width   new_name2 Petal.Width     Species 
          1           2           3           4           5

how to rename a variable using a dynamic name and dplyr?

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