Check if object is Null or undefined

Question

I have a function that includes an optional variable parameter. By default, I set the variable to NULL, but if it isn't NULL I'd like my function to do some stuff. I need a way to check if the variable is not null. This is complicated because I am using tidyeval, and just using is.null(var) throws an object not found error. I've found a hacky solution using try, but am hoping there's a better way.

library(dplyr)
dat <- data.frame(value = 1:8, 
                         char1 = c(rep("a", 4), rep("b", 4)), 
                         char2 = rep(c(rep("c", 2), rep("d", 2)), 2))

myfun <- function(dat, group = NULL, var = NULL) {
    
    x <- dat %>% 
        group_by({{group}}, {{var}}) %>% 
        summarize(mean = mean(value), 
                            .groups = "drop")
    
    # if(!is.null(var)) { # Throws object not found error if not null

    null_var <- try(is.null(var), silent = TRUE)
    null_var <- null_var == TRUE
    if(!null_var)   {
        print("do something with `var`")
    }
    x
}
myfun(dat)
myfun(dat, char1)
myfun(dat, char1, char2)

Answer 1

{{ is the combination of enquo() and !! in one step. To inspect the contents of var, you need to decompose these two steps. enquo() defuses the argument and returns a quosure that you can inspect. !! injects the quosure into other calls.

Below we enquo() the argument at the start of the function. Inspect it with quo_is_null() (you could also use quo_get_expr() to see what's inside it). Then inject it inside group_by() with !!:

myfun <- function(dat, group = NULL, var = NULL) {
    var <- enquo(var)
    
    if (!quo_is_null(var)) {
        print("It works!")
    }

    # Use `!!` instead of `{{` because we have decomposed the
    # `enquo()` and `!!` steps that `{{` bundles
    dat %>% 
        group_by({{ group }}, !!var) %>% 
        summarize(mean = mean(value), .groups = "drop")
}