What is the tidyeval way of using dplyr::filter?

Question

Call the function below using foo(c("b")). The outputs are shown inline.

What is the right way of writing df %>% filter(!!x > (!!x))?

I have included an example of using mutate in tidyeval style to contrast it with filter.

foo <- function(variables) {

  x <- rlang::sym(variables[[1]])

  print(x)
  #> b

  print(typeof(x))
  #> [1] "symbol"

  df <- data_frame(a = 1, b = 2)

  print(df %>% mutate(!!x := 100 + !!x))

  #> # A tibble: 1 x 2
  #>         a     b
  #>       <dbl> <dbl>
  #>   1     1   102  

  print(df %>% filter(!!x  > (!!x)))

  #> Error in !x : invalid argument type

  print(df %>% filter(magrittr::is_greater_than(!!x, !!x)))

  #> # A tibble: 0 x 2
  #> # ... with 2 variables: a <dbl>, b <dbl>

}

Answer 1

You are most of the way there except for a minor typo, the round brackets in your filter statement should be on the variable and not the value.

print(df %>% filter((!!x) > !!x))

#> # A tibble: 0 x 2
#> # ... with 2 variables: a <dbl>, b <dbl>

Answer 2

Edit: All of this no longer applies. The precedence tree is reorganised so that !!x + !!y etc do the right thing by default. The parentheses are no longer necessary since rlang 0.2.0.

---

The ! operator has really low precedence. This means that it will apply to most of the expression appearing on its right.

!! x > 3

is implicitly equivalent to:

(!! x > 3)

So you have to help R figure out the right precedence with explicit parentheses:

(!! x) > 3

Note that in most cases if you're unquoting on both sides of an operator, you technically don't have to apply the parentheses on the last one:

(!! x) + (!! y) + z

However that will vary according to often mysterious rules of precedence, so I suggest to always enclose in parentheses when operators are involved:

(!! x ) + (!! y) + (!! z)

Answer 3

You can use filter_at

oof <- function(variables) {
  x <- rlang::sym(variables[[1]])
  df <- data.frame(a = 1, b = 2)
  print(df %>% filter_at(vars(!!x), any_vars(. == !!x)))
  print(df %>% filter(magrittr::equals(!!x, !!x)))
}

I use magrittr::equals to show the magrittr style works as well

oof(c("b"))

#   a b
# 1 1 2
#   a b
# 1 1 2

Answer 4

This is a very generic way of handling any field value condition

data%>%
    filter(!!quo((!!as.name (field1)) > (!!myVal)))