How to detect if bare variable or string

Question

I am trying to write a plotting function where you can pass bare column names to select which columns are plotted. I would like also to be able to specify a string as the color.

I have found that I need to use shQuote if I want to pass a string to aes_string. Now my problem is to figure out if a bare name or a string was passed. How would I do this?

dat <- data.frame(
    time = factor(c("Lunch","Dinner"), levels=c("Lunch","Dinner")),
    total_bill = c(14.89, 17.23)
)



plot_it <- function(dat, x,y, fill){
    require(rlang)
    require(ggplot2)

    x <- enquo(x)
    y <- enquo(y)
    fill <- enquo(fill)

    xN <- quo_name(x)
    yN <- quo_name(y)
    fillN <- quo_name(fill)

ggplot(data=dat, aes_string(x=xN, y=yN, fill=fillN)) +
    geom_bar(stat="identity")

}

This works:

plot_it(dat, time, total_bill, time)

This does not:

plot_it(dat, time, total_bill, "grey")

Note that this requires the newest versions of rlang and ggplot2.

Answer 1

Based on @akrun's suggestion of how to detect which case we had (was removed) I found something that does what I asked for:

plot_it <- function(dat, x, y, fill) {

    lst <- as.list(match.call())

    if(is.character(lst$fill)){
        fillN <- shQuote(fill)
    } else{
        fillN <- quo_name(enquo(fill))
    }

    x <- enquo(x)
    y <- enquo(y)


    xN <- quo_name(x)
    yN <- quo_name(y)


    p <- ggplot(data=dat, aes_string(x=xN, y=yN, fill=fillN)) +
        geom_bar(stat="identity")

    return(p)
}

Turns out this doesn't actually do what I had in mind since it assigns the quoted value as a factor to assign colors by. Not the actual color.

I came up with this that seems to work but is not really elegant:

plot_it <- function(dat, x, y, fill) {

    lst <- as.list(match.call())

    if(!(type_of(lst$fill)=="symbol" | (type_of(lst$fill)=="string" & length(lst$fill)==1))) stop("Fill must either be a bare name or a vector of length 1.")

    x <- enquo(x)
    y <- enquo(y)

    xN <- quo_name(x)
    yN <- quo_name(y)


    if(is.character(lst$fill)){
        dat[,"fillN"] <- fill
        fillN <- fill

        p <- ggplot(data=dat, aes_string(x=xN, y=yN, fill = shQuote(fillN))) +
             scale_fill_manual(name="fill", values=setNames(fillN,fillN))
    } else{
        fillN <- quo_name(enquo(fill))

        p <- ggplot(data=dat, aes_string(x=xN, y=yN, fill = fillN))
    }



       p <- p + geom_bar(stat="identity")

    return(p)
}

Any idea to make this a bit more elegant?

Answer 2

Working from your answer and Eumenedies' answer/test cases, here is a version that:

• Solves your original problem
• Prevents an unnecessary legend, as Eumenedies did
• Provides ggplot object that can be added to without error
• Uses the more appropriate geom_col, per Eumenedies' suggestion

The main trick is that, if fill isn't a factor in the plot, you want it outside of the aes / aes_string block.

plot_it <- function(dat, x, y, fill) {

  lst <- as.list(match.call())

  xN <- quo_name(enquo(x))
  yN <- quo_name(enquo(y))

  if(is.character(lst$fill)) {
    p <- ggplot(data=dat, aes_string(x=xN, y=yN)) +
      geom_col(fill = fill)
  } else {
    p <- ggplot(data=dat, aes_string(x=xN, y=yN, fill = quo_name(enquo(fill)))) +
      geom_col()
  }

  return(p)
}

plot_it(dat, time, total_bill, time)
plot_it(dat, time, total_bill, "blue")
plot_it(dat, time, total_bill, "blue") + geom_point()

You could make the if block shorter by moving the fill aesthetic in the second case to the geom_col call, but that will extend in a different way if you add more geoms.

Also, once ggplot is updated to support rlang, it would be cleaner to avoid aes_string and quo_name combination and just use !!fill.

Note that, assuming that the fill factor exists, if it's always going to be the same as the x factor, it would probably make more sense to have a version where fill is an optional argument. You would only overwrite the default per-factor color if the argument is included.