R - How to extract object names from expression

Question

Given an rlang expression:

expr1 <- rlang::expr({
  d <- a + b
})

How to retrieve the names of the objects refered to within the expression ?

> extractObjects(expr1)
[1] "d" "a" "b"

Better yet, how to retrieve the object names and categorise them by "required"(input) and "created"(output) ?

> extractObjects(expr1)
$created
[1] "d"

$required
[1] "a" "b"

Answer 1

The base function all.vars does this:

〉all.vars(expr1)
[1] "d" "a" "b"

Alternatively, you can use all.names to get all names in the expression rather than just those that aren’t used as calls or operators:

〉all.names(expr1)
[1] "{"  "<-" "d"  "+"  "a"  "b"

Don’t be misled: this result is correct! All of these appear in the expression, not just a, b and d.

But it may not be what you want.

In fact, I’m assuming what you want corresponds to the leaf tokens in the abstract syntax tree (AST) — in other words, everything except function calls (and operators, which are also function calls).

The syntax tree for your expression looks as follows:¹

   {
   |
   <-
   /\
  d  +
    / \
   a   b

Getting this information means walking the AST:

leaf_nodes = function (expr) {
    if(is.call(expr)) {
        unlist(lapply(as.list(expr)[-1L], leaf_nodes))
    } else {
        as.character(expr)
    }
}

〉leaf_nodes(expr1)
[1] "d" "a" "b"

Thanks to the AST representation we can also find inputs and outputs:

is_assignment = function (expr) {
    is.call(expr) && as.character(expr[[1L]]) %in% c('=', '<-', '<<-', 'assign')
}

vars_in_assign = function (expr) {
    if (is.call(expr) && identical(expr[[1L]], quote(`{`))) {
        vars_in_assign(expr[[2L]])
    } else if (is_assignment(expr)) {
        list(created = all.vars(expr[[2L]]), required = all.vars(expr[[3L]]))
    } else {
        stop('Expression is not an assignment')
    }
}

 〉vars_in_assign(expr1)
$created
[1] "d"

$required
[1] "a" "b"

Note that this function does not handle complex assignments (i.e. stuff like d[x] <- a + b or f(d) <- a + b very well.

---

_{¹ lobstr::ast shows the syntax tree differently, namely as}

█─`{`
└─█─`<-`
  ├─d
  └─█─`+`
    ├─a
    └─b

_{… but the above representation is more conventional outside R, and I find it more intuitive.}

Answer 2

Another solution is to extract the abstract symbolic tree explicitly:

getAST <- function(ee) purrr::map_if(as.list(ee), is.call, getAST)

str(getAST(expr1))
#  List of 2
#   $ : symbol {
#   $ :List of 3
#    ..$ : symbol <-
#    ..$ : symbol d
#    ..$ :List of 3
#    .. ..$ : symbol +
#    .. ..$ : symbol a
#    .. ..$ : symbol b

Then traverse the AST to find the assignment(s):

extractObjects <- function(ast)
{
    ## Ensure that there is at least one node
    if( length(ast) == 0 ) stop("Provide an AST")

    ## If we are working with the assigment
    if( identical(ast[[1]], as.name("<-")) ) {
        ## Separate the LHS and RHS
        list(created = as.character(ast[[2]]),
             required = sapply(unlist(ast[[3]]), as.character))
    } else {
        ## Otherwise recurse to find all assignments
        rc <- purrr::map(ast[-1], extractObjects)

        ## If there was only one assignment, simplify reporting
        if( length(rc) == 1 ) purrr::flatten(rc)
        else rc
    }
}

extractObjects( getAST(expr1) )
# $created
# [1] "d"
#
# $required
# [1] "+" "a" "b"

You may then filter math operators out, if needed.

Answer 3

This is an interesting one. I think that conceptually, it might not be clear in ALL possible expressions what exactly is input and output. If you look at the so called abstract syntax tree (AST), which you can visualize with lobstr::ast(), it looks like this.

So in simple cases when you always have LHS <- operations on RHS variables, if you iterate over the AST, you will always get the LST right after the <- operator. If you assign z <- rlang::expr(d <- a+b), then z behaves like a list and you can for example do the following:

z <- rlang::expr(d <- a+b)

for (i in 1:length(z)) {
  if (is.symbol(z[[i]])) {
    print(paste("Element", i, "of z:", z[[i]], "is of type", typeof(z[[i]])))
    if (grepl("[[:alnum:]]", z[[i]])) {print(paste("Seems like", z[[i]], "is a variable"))}
  } else {
    for (j in 1:length(z[[i]])){
      print(paste("Element", j, paste0("of z[[",i,"]]:"), z[[i]][[j]], "is of type", typeof(z[[i]][[j]])))
      if (grepl("[[:alnum:]]", z[[i]][[j]])) {print(paste("Seems like", z[[i]][[j]], "is a variable"))}
    }
  }
}
#> [1] "Element 1 of z: <- is of type symbol"
#> [1] "Element 2 of z: d is of type symbol"
#> [1] "Seems like d is a variable"
#> [1] "Element 1 of z[[3]]: + is of type symbol"
#> [1] "Element 2 of z[[3]]: a is of type symbol"
#> [1] "Seems like a is a variable"
#> [1] "Element 3 of z[[3]]: b is of type symbol"
#> [1] "Seems like b is a variable"

Created on 2020-09-03 by the reprex package (v0.3.0)

As you can see these trees can quickly get complicated and nested. So in a simple case like in your example, assuming that variables are using alphanumeric representations, we can kind of identify what the "objects" (as you call them) are and what are operators (which don't match the [[:alnum:]] regex). As you can see the type cannot be used to distinguish between objects and operators since it is always symbol (btw z below is a language as is z[[3]] which is why we can condition on whether z[[i]] is a symbol or not and if not, dig a level deeper). You could then (at your peril) try to classify that the objects that appear immediately after a <- are "outputs" and the rest are "inputs" but I don't have too much confidence in this, especially for more complex expressions.

In short, this is all very speculative.