Passing strings as arguments in dplyr verbs

Question

I would like to be able to define arguments for dplyr verbs

condition <- \"dist > 50\"

and then use these strings in

Answer 1

Since these 2014 answers, two new ways are possible using rlang's quasiquotation.

Conventional hard-coded filter statement. For the sake of comparison, the statement dist > 50 is included directly in dplyr::filter().

library(magrittr)

# The filter statement is hard-coded inside the function.
cars_subset_0 <- function( ) {
  cars %>%
    dplyr::filter(dist > 50)
}
cars_subset_0()

results:

   speed dist
1     14   60
2     14   80
3     15   54
4     18   56
...
17    25   85

rlang approach with NSE (nonstandard evaluation). As described in the Programming with dplyr vignette, the statement dist > 50 is processed by rlang::enquo(), which "uses some dark magic to look at the argument, see what the user typed, and return that value as a quosure". Then rlang's !! unquotes the input "so that it’s evaluated immediately in the surrounding context".

# The filter statement is evaluated with NSE.
cars_subset_1 <- function( filter_statement ) {
  filter_statement_en <- rlang::enquo(filter_statement)
  message("filter statement: `", filter_statement_en, "`.")

  cars %>%
    dplyr::filter(!!filter_statement_en)
}
cars_subset_1(dist > 50)

results:

filter statement: `~dist > 50`.

expr: ^dist > 50
env:  global
   speed dist
1     14   60
2     14   80
3     15   54
4     18   56
17    25   85

rlang approach passing a string. The statement "dist > 50" is passed to the function as an explicit string, and parsed as an expression by rlang::parse_expr(), then unquoted by !!.

# The filter statement is passed a string.
cars_subset_2 <- function( filter_statement ) {
  filter_statement_expr <- rlang::parse_expr(filter_statement)
  message("filter statement: `", filter_statement_expr, "`.")

  cars %>%
    dplyr::filter(!!filter_statement_expr)
}
cars_subset_2("dist > 50")

results:

filter statement: `>dist50`.
   speed dist
1     14   60
2     14   80
3     15   54
4     18   56
...
17    25   85

Things are simpler with dplyr::select(). Explicit strings need only !!.

# The select statement is passed a string.
cars_subset_2b <- function( select_statement ) {
  cars %>%
    dplyr::select(!!select_statement)
}
cars_subset_2b("dist")