[help] tar_eval() cannot track targets nested in list

[psychelzh]

Help

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Description

Please ask your question and provide a reprex if troubleshooting.

For example:

# with issues
targets::tar_dir({
  targets::tar_script({
    library(targets)
    list(
      tar_target(x, 1),
      tar_target(y, 2),
      tarchetypes::tar_eval(
        tar_target(name, list(val)),
        values = tibble::tribble(
          ~name, ~val,
          rlang::sym("t"), rlang::syms(c("x", "y"))
        )
      )
    )
  })
  targets::tar_visnetwork()
})

The dependency between t and x, y does not work.

For now I can use tar_map():

# workaround?
targets::tar_dir({
  targets::tar_script({
    library(targets)
    list(
      tar_target(x, 1),
      tar_target(y, 2),
      tarchetypes::tar_map(
        values = tibble::tribble(
          ~name, ~val,
          rlang::sym("t"), rlang::syms(c("x", "y"))
        ),
        names = name,
        tar_target(
          tar_name, val
        )
      )
    )
  })
  targets::tar_visnetwork()
})

Is this by design? The tar_eval() works better here because we can specify the names of the targets in values too.

[wlandau]

substitute(list(x), env = list(x = rlang::syms(list("y", "z")))) and substitute(list(x), env = list(x = quote(list(y, z)))) may look the same when you print them, but the returned objects are actually very different. The former has a non-language list object in place of x, whereas the latter is a better language object with true language objects at every level of the abstract syntax tree. tar_eval() is built on substitute(), so it works the same way. The following _targets.R file does what you want.

# _targets.R
library(targets)
list(
  tar_target(x, 1),
  tar_target(y, 2),
  tarchetypes::tar_eval(
    tar_target(name, list(val)),
    values = tibble::tribble(
      ~name, ~val,
      quote(t), quote(list(x, y))
    )
  )
)

[psychelzh]

Thanks! Does it mean that I cannot create this type of dependency based on string input of "x" and "y" (these names are created programatically)?

[psychelzh]

Found it now, I should use bquote() to quasi-quote my expressions.

[psychelzh]

After a second thought, I found bquote() does not always help because I have many such names stored in a character vector. I cannot create such an object. For example: say I have two rows:

tibble::tribble(
  ~name, ~val,
  "x", c("var_x1", "var_x2"),
  "y", c("var_y1", "var_y2"),
)

Each row depends on the two names (in character class) in val column, making bquote() hard to program with.

[psychelzh]

Several days later, I finally make it clear now. Initially, I had a misunderstanding about the abstract syntax tree. If symbols were created beforehand and nested into list as another variable, the abstract syntax tree was ruined.

Now I abandoned the method of nesting those symbols beforehand and separate them into different columns. In this way, it even makes the code cleaner. Thanks again for your great package!

(BTW, see psychelzh/g-factor@d4649f4 for my solution if anyone is interested.)