CODEWITHSUNDEEP
rpipemagrittr
Migwell
Migwell

Reputation: 20107

Piping into `if` returns the pipeline without evaluating it

I am trying to implement a pipeline that has an optional step which consists of a pipeline of several functions. It runs this pipeline based on a condition, and otherwise it just passes through the original value. However I've tried implementing this using if and also purrr::when, but in both cases the positive case simply returns the pipeline instead of executing it.

Here's a simple contrived example. I've simplified the conditional pipeline to only one function, but being able to use magrittr pipes inside the TRUE branch is important here.

library(magrittr)

maybe_round = function(x, round = TRUE){
  x %>%
    `if`(
      round,
      . %>% round(),
      .
    )
}

> maybe_round(5.3, TRUE)
Functional sequence with the following components:

 1. round(.)

Use 'functions' to extract the individual functions. 

> maybe_round(5.3, FALSE)
[1] 5.3

This second case is working correctly; it's returning the original value unchanged. But this first case is not, it's returning the magrittr pipeline, but not actually feeding it with 5. How can I get this to work as I intend? I assume this has something to do with how magrittr rewrites the syntax tree, but I can't quite work it out.

Upvotes: 4

Views: 136

Answers (3)

G. Grothendieck
G. Grothendieck

Reputation: 269491

The syntax . %>% round(.) means function(.) round(.). Any time dot starts a pipeline it defines a function rather than being an ordinary pipeline. Put parentheses around the dot to prevent the dot from starting the inner pipeline.

 maybe_round = function(x, round = TRUE){
   x %>%
     `if`(
       round,
       (.) %>% round(),
       .
     )
 }

maybe_round(5.3, TRUE)
## [1] 5

Another possibility is to just leave it as a function and then evaluate that function at the outer dot like this:

 maybe_round = function(x, round = TRUE){
   x %>%
     `if`(
       round,
       (. %>% round())(.),
       .
     )
 }

Upvotes: 1

Konrad Rudolph
Konrad Rudolph

Reputation: 545578

This is unrelated to `if` — the same issue would manifest with any other function. In fact, the issue is the . %>% … expression, which is not a normal pipeline. Instead, this special syntax creates a lambda (see “Using the dot-place holder as lhs” in the documentation).

If you insist on using a pipe here, you’ll need to first assign . to a different variable name, e.g.:

maybe_round = function(x, round = TRUE){
  x %>%
    `if`(
      round,
      {
          x = .
          x %>% round()
      },
      .
    )
}

… honestly, I would instead use a regular if expression and encapsulate it inside a function that can be piped into, i.e.

maybe_round = function (x, round = TRUE) {
    x %>% maybe_round_impl(round)
}

maybe_round_impl = function (x, round = TRUE) {
    if (round) {
        x %>% round()
    } else {
        x
    }
}

Upvotes: 1

jay.sf
jay.sf

Reputation: 72693

Just don't pipe again in the `if`().

maybe_round = function(x, round = TRUE){
  x %>%
    `if`(
      round,
      round(.),
      .
    )
}

maybe_round(5.3, TRUE)
# [1] 5
maybe_round(5.3, FALSE)
# [1] 5.3

Upvotes: 0

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