R - Passing arguments to `dplyr::select()` inside a wrapper

Question

I have a wrapper function which uses dplyr::select(), however when I attempt to use it to select columns by name, it throws an object * not found error. I know it has something to do with the way select() uses quasiquotation but I don't know exactly why.

Here is my attempt using as_mapper:

fun1 = as_mapper(~select(.x, .y)) fun1(iris, Species) Error in .f(.x[[i]], ...) : object 'Species' not found

Using base function notation:

fun2 = function(dat, x) {select(substitute(dat), substitute(x))} fun2(iris, Species:Sepal.Length) Error in UseMethod("select_") : no applicable method for 'select_' applied to an object of class "name"

I would be grateful if someone could shed some light on why these errors are occurring.

as_mapper expects a character vector (or numeric for number-based indexing). Calling fun1(iris, "Species") or fun1(iris, 5) works, but not when Species is supplied as a name of an object (that might not exist in the environment). — teofil
– teofil, Commented Nov 7, 2019 at 5:21
With the latest version of dplyr you would use fun3 = function(dat, x) {select({{dat}}, {{x}})}; fun3(iris, Species:Sepal.Length). The problem with substitute is it happens too late. The problem with as_mapper is it doesn't support non-standard evaluation. — MrFlick
– MrFlick, Commented Nov 7, 2019 at 5:32

al-obrien · Accepted Answer · 2019-11-07 05:44:07Z

I believe this is related to non-standard evaluation (NSE) in R. Why dont you try using rlang to work with dplyr functions, as referenced here.

library(magrittr) fun2 <- function(dat, x) { x <- rlang::enquo(x) dplyr::select(dat,!!x) } fun2(iris, Species:Sepal.Length) %>% tibble::tibble() # A tibble: 150 x 1 .$Species $Petal.Width $Petal.Length $Sepal.Width $Sepal.Length <fct> <dbl> <dbl> <dbl> <dbl> 1 setosa 0.2 1.4 3.5 5.1 2 setosa 0.2 1.4 3 4.9 3 setosa 0.2 1.3 3.2 4.7 4 setosa 0.2 1.5 3.1 4.6 5 setosa 0.2 1.4 3.6 5 6 setosa 0.4 1.7 3.9 5.4 7 setosa 0.3 1.4 3.4 4.6 8 setosa 0.2 1.5 3.4 5 9 setosa 0.2 1.4 2.9 4.4 10 setosa 0.1 1.5 3.1 4.9 # ... with 140 more rows

I would also like to point out that @MrFlick is also correct and the new interpolation method through the curly-curly operator is a nice shortcut

Collectives™ on Stack Overflow

R - Passing arguments to `dplyr::select()` inside a wrapper

1 Answer 1

Comments

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Related