Why is enquo + !! preferable to substitute + eval

Question

In the following example, why should we favour using f1 over f2? Is it more efficient in some sense? For someone used to base R, it seems more natu

Answer 1

I want to give an answer that is independent of dplyr, because there is a very clear advantage to using enquo over substitute. Both look in the calling environment of a function to identify the expression that was given to that function. The difference is that substitute() does it only once, while !!enquo() will correctly walk up the entire calling stack.

Consider a simple function that uses substitute():

f <- function( myExpr ) {
  eval( substitute(myExpr), list(a=2, b=3) )
}

f(a+b)   # 5
f(a*b)   # 6

This functionality breaks when the call is nested inside another function:

g <- function( myExpr ) {
  val <- f( substitute(myExpr) )
  ## Do some stuff
  val
}

g(a+b)
# myExpr     <-- OOPS

Now consider the same functions re-written using enquo():

library( rlang )

f2 <- function( myExpr ) {
  eval_tidy( enquo(myExpr), list(a=2, b=3) )
}

g2 <- function( myExpr ) {
  val <- f2( !!enquo(myExpr) )
  val
}

g2( a+b )    # 5
g2( b/a )    # 1.5

And that is why enquo() + !! is preferable to substitute() + eval(). dplyr simply takes full advantage of this property to build a coherent set of NSE functions.

UPDATE: rlang 0.4.0 introduced a new operator {{ (pronounced "curly curly"), which is effectively a short hand for !!enquo(). This allows us to simplify the definition of g2 to

g2 <- function( myExpr ) {
  val <- f2( {{myExpr}} )
  val
}