Why is plotly() and enquo + !! in conflict?

Question

I am writing a function that uses plot_ly for a pieplot. The tilde (~) within argument labels = ~ is in conflict with the unquote operator !!. Is there a solution for this problem?

pieplotr <- function (df, Property){

Property_Name <- enquo(Property)

Pie <- plot_ly(df,
              labels = ~!!Property_Name,
              type = "pie")

return(Pie)

}

Toy_dataframe <- data.frame(
                 Sex = c("Male", "Female", "Female", "Male", "Male","NA"),
                 Height = c(12,11,7,3,9,NA),
                 Name = c("John", "Dora", "Dora","Dora", "Anna", "John"),
                 Last = c("Henry", "Paul", "House", "Houze", "Henry", "Bill"),
                 Location = c("Chicago", "Chicago", "Portland", "NYC", "NYC", "NYC"),
                 stringsAsFactors = TRUE)

e.g.

pieplotr(df = Toy_dataframe,
          Property = Name)

I expect to return a pieplot, but I am getting the following error message:

Error in as.list.environment(x, all.names = TRUE) :
object 'Name' not found

the `enquo` `!!` works in tidyverse packages/environment. `plotly` is not a tidyverse package — akrun, Nov 07 '19 at 17:50
Thank you for the observation! See added Toy_dataframe for testing. — Soph Carr, Nov 07 '19 at 18:01

akrun · Answer 1 · 2019-11-07T18:37:39.777

2

We can evaluate after creating creating an expression

pieplotr <- function (df, Property){

Property_Name <- rlang::enquo(Property)
rlang::eval_tidy(
    rlang::quo_squash(quo(
    plot_ly(df, labels = ~!!Property_Name,  type = "pie")
    )))


} 




pieplotr(df = Toy_dataframe, Property = Name)

-output

Or remove the ~!!

pieplotr <- function (df, Property){

Property_Name <- rlang::enquo(Property)

    plot_ly(df, labels = Property_Name,  type = "pie")



}

pieplotr(df = Toy_dataframe, Property = Name)

edited Nov 07 '19 at 18:37

answered Nov 07 '19 at 18:10

akrun

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That worked! Thank you so much. Do you mind walking me through your thought process? I would greatly appreciate it so I can have a much better understanding if the solution. – Soph Carr Nov 07 '19 at 18:30
1

@SophCarr It is not a straightforward solution as we are `eval`uating an expression created with `quo`, `quo_squash` flattens the expression and on top of which we are evaluating. As I mentioned in the comments, `plotly` is not a tidyverse compatible package. – akrun Nov 07 '19 at 18:33
1

@SophCarr Noticed that the issue is also with the `~!!` part. Removing it works – akrun Nov 07 '19 at 18:38
Don't really need `quo` / `quo_quash`. You can just use `rlang::expr( plot_ly(df, labels = ~!!Property_name, type="pie") )`, which will create the appropriate expression. (@SophCarr: try returning the output of `rlang::expr` from the function to see what it does). The expression is then evaluated using `eval_tidy`. – Artem Sokolov Nov 07 '19 at 18:38
@ArtemSokolov Not even that is needed – akrun Nov 07 '19 at 18:38
Also, the correct verb here is `ensym()` not `enquo()`, because you are providing the context of the symbol directly to the function as the dataframe. – Artem Sokolov Nov 07 '19 at 18:38

score 2 · Answer 2 · answered Nov 07 '19 at 18:43

An alternative solution is to avoid non-standard-evaluation / quasiquoation entirely and simply modify how you are calling the function

pieplotr <- function (df, Property) {
    plot_ly(df, labels = Property, type = "pie")
}

pieplotr( Toy_dataframe, ~Name )    # Notice the ~

The reason this works here is because the plotly universe uses formulas (defined via ~) to pass around information about variables. Tidyverse uses unevaluated expressions instead of formulas, which is the reason for needing enquo(), etc.

score 0 · Answer 3 · answered Jun 03 '20 at 02:21

I was trying to find a way to use plot_ly inside a function, but also trying to avoid passing unquoted variable names. Turns out passing quoted column names, parsing, and then evaluating within the plot_ly call works nicely, and, at least for me, its easier to read and understand than some of the other solutions:

ply <- function(data, x, y) {

  name_x <- parse(text = x)
  name_y <- parse(text = y)

  plotly::plot_ly(data = data, x = ~eval(name_x), y = ~eval(name_y)) %>%
    plotly::add_trace(type = "scatter")
}

ply(faithful, "eruptions", "waiting")

Why is plotly() and enquo + !! in conflict?

3 Answers3