R：替代所有父环境中绑定的变量

Question

The function base::substitute(expr, env) , as per its documentation page,

returns the parse tree for the (unevaluated) expression expr, substituting any variables bound in env.

I am looking for a way of substituting any variables bound not in one specific environment, but in all environments in the current call stack, ie all environments encountered by iterating over parent.frame(i) where i is in seq_len(sys.nframe()) . Additionally, I'd like standard scoping rules to apply.

This is a contradiction: standard scoping in R is lexical, but what I describe here is dynamic scoping (thank you @MikkoMarttila for helping me clear this up). What I actually want is a way of substituting any variables bound not in one specific environment, but in all parent enclosing environments, the set of which can be enumerated by repeatedly applying base::parent.env() .

Consider the following example:

do_something <- function(todo) {
  cat(
    paste(
      deparse(substitute(todo, environment())),
      collapse = "\n"
    )
  )
}

nested_do <- function() {

  var_2 <- "goodbye"

  do_something({
    print(var_1)
    print("world")
    print(var_2)
  })

}

var_1 <- "hello"

nested_do()

Currently this gives

print(var_1)
print("world")
print(var_2)

where I'd like to have

print("hello")
print("world")
print("goodbye")

I have looked at base::bquote() and rlang::enexpr() but for both I have to explicitly mark the variables for substitution/unquoting with .() or !! . I'd rather not have to specify variables manually, but have everything resolved that is found (just like in base::substitute() ). Furthermore, I tried iteratively applying base::substitute() with the respective env arguments and I had a look at oshka::expand() , but nothing I tried, does what I need.

Any help is much appreciated.

Additional context

What I'm trying to achieve is the following: I'm working on a cluster running LSF. This means that I can submit jobs using the submission tool bsub which may take an R file as input. Now I would like to have a script that generates these input files (eg using the function do_something() ).

long_running_fun <- function(x) {
  Sys.sleep(100)
  x / 2
}

var_1 <- 2 + 2
var_2 <- var_1 + 10

do_something({
  print(var_1)
  var_3 <- long_running_fun(var_2)
  print(var_3)
})

I in the above case, want the following (or something equivalent) to be written to a file

print(4)
var_3 <- long_running_fun(14)
print(var_3)

Answer 1

Rather than do that I suggest you just pass the environment like this:

esubstitute <- function(expr, envir) do.call("substitute", list(expr, envir))

do_something <- function(todo, envir = parent.frame()) {
  cat(
    paste(
      deparse(esubstitute(todo, envir)),
      collapse = "\n"
    )
  )
}

nested_do <- function(envir = parent.frame()) {

  var_2 <- "goodbye"

  do_something({
    print(var_1)
    print("world")
    print(var_2)
  }, envir)

}

var_1 <- "hello"

nested_do()

giving:

[1] "hello"
[1] "world"
[1] "goodbye"
"goodbye"> 

You may also want to look at the envnames package.

Answer 2

You can define a function to do such a substitution sequence: that is, take an expression and substitute it in all of the environments in the call stack. Here's one way:

substitute_stack <- function(expr) {
  expr <- substitute(expr)

  # Substitute in all envs in the call stack
  envs <- rev(sys.frames())
  for (e in envs) {
    expr <- substitute_q(expr, e)
  }

  # sys.frames() does not include globalenv() and
  # substitute() doesnt "substitute" there
  e <- as.list(globalenv())
  substitute_q(expr, e)
}

# A helper to substitute() in a pre-quoted expression
substitute_q <- function(expr, env = parent.frame()) {
  eval(substitute(substitute(x, env), list(x = expr)))
}

Let's give it a go:

do_something <- function(todo) {
  cat(
    paste(
      deparse(substitute_stack(todo)),
      collapse = "\n"
    )
  )
}

nested_do <- function() {
  var_2 <- "goodbye"

  do_something({
    print(var_1)
    print("world")
    print(var_2)
  })
}

var_1 <- "hello"

nested_do()
#> {
#>     print("hello")
#>     print("world")
#>     print("goodbye")
#> }

Whether or not actually doing this is a good idea is a whole other question. The approach suggested by @G.Grothendieck is likely to be preferrable.

Created on 2018-07-19 by the reprex package (v0.2.0.9000).

Answer 3

Building on @MikkoMarttila's answer , I think the following does what I requested

do_something <- function(todo) {

  # A helper to substitute() in a pre-quoted expression
  substitute_q <- function(expr, env) {
    eval(substitute(substitute(x, env), list(x = expr)))
  }

  substitute_parents <- function(expr) {
    expr <- substitute(expr)

    # list all parent envs
    envs <- list()
    env <- environment()
    while (!identical(env, globalenv())) {
      envs <- c(envs, env)
      env <- parent.env(env)
    }
    # substitute in all parent envs
    for (e in envs) {
      expr <- substitute_q(expr, e)
    }

    # previously did not include globalenv() and
    # substitute() doesnt "substitute" there
    e <- as.list(globalenv())
    substitute_q(expr, e)
  }

  cat(
    paste(
      deparse(substitute_parents(todo)),
      collapse = "\n"
    )
  )
}

This gives

nested_do <- function() {
  var_2 <- "not_this"

  do_something({
    print(var_1)
    Sys.sleep(100)
    print("world")
    print(var_2)
  })
}

var_1 <- "hello"
var_2 <- "goodbye"

do_something({
  print(var_1)
  Sys.sleep(100)
  print("world")
  print(var_2)
})
#> {
#>     print("hello")
#>     Sys.sleep(100)
#>     print("world")
#>     print("goodbye")
#> }
nested_do()
#> {
#>     print("hello")
#>     Sys.sleep(100)
#>     print("world")
#>     print("goodbye")
#> }