Passing column name into function

Question

I have a simple problem with non-standard evaluation: passing a variable name as an argument into a function.

As a reproducible example, here's a simple thing: taking the mean of one variable, mpg from the mtcars dataset. My end goal is to have a function where I can input the dataset and the variable, and get the mean.

So without a function:

library(tidyverse)
mtcars %>% summarise(mean = mean(mpg))

#>       mean
#> 1 20.09062

I've tried to use get() for non-standard evaluation, but I'm getting errors:

library(tidyverse)
summary_stats <- function(variable, dataframe){
  dataframe %>% summarise(mean = get(variable))
}

summary_stats(mpg, mtcars)

#> Error: Problem with `summarise()` input `mean`.
#> x invalid first argument
#> ℹ Input `mean` is `get(variable)`.

^{Created on 2020-09-19 by the reprex package (v0.3.0)}

Edit:

I also had one additional follow-up question.

I also need the variable argument as a char string, I tried the code below, but I'm still missing how to do that:

library(tidyverse)
summary_stats <- function(variable, dataframe){
  dataframe %>% summarise(mean = mean({{variable}}))
  print(as.character({{variable}}))
}

summary_stats(disp, mtcars)
#> Error in print(as.character({: object 'disp' not found

^{Created on 2020-09-19 by the reprex package (v0.3.0)}

Here's the dplyr vignette that you'll find helpful for stuff like tidyeval => https://cran.r-project.org/web/packages/dplyr/vignettes/programming.html — theairbend3r, Sep 19 '20 at 03:56
For anyone reading this question in the future, @theairbend3r 's link is exactly the sort of documentation that steps through the answer, and is just what I was looking for. — Jeremy K., Sep 19 '20 at 03:59

Ronak Shah · Accepted Answer · 2020-09-19T04:58:10.877

1

You could use the curly-curly ({{}}) operator to pass column name as unquoted variable.

To get variables passed as character value we can use deparse, substitute.

library(dplyr)
library(rlang)

summary_stats <- function(variable, dataframe){
  print(deparse(substitute(variable)))
  dataframe %>% summarise(mean = mean({{variable}}))
}

#[1] "mpg"

#      mean
#1 20.09062

edited Sep 19 '20 at 04:58

answered Sep 19 '20 at 03:36

Ronak Shah

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I also had a follow up question (in the edit above) if you're able to help. Thank you so much. – Jeremy K. Sep 19 '20 at 04:55
1

We can use `deparse`, `substitute` for that. See updated answer. – Ronak Shah Sep 19 '20 at 04:58

Passing column name into function

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