converting a quosure to a string in R - r

Question

I've been using quosures with dplyr:
library(dplyr)
library(ggplot2)
thing <- quo(clarity)
diamonds %>% select(!!thing)
print(paste("looking at", thing))
[1] "looking at ~" "looking at clarity"
I really want to print out the string value put into the quo, but can only get the following:
print(thing)
<quosure: global>
~clarity
print(thing[2])
clarity()
substr(thing[2],1, nchar(thing[2]))
[1] "clarity"
is there a simpler way to "unquote" a quo()?

Answer 1

We can use quo_name
print(paste("looking at", quo_name(thing)))

Answer 2

quo_name does not work if the quosure is too long:
> q <- quo(a + b + c + d + e + f + g + h + i + j + k + l + m + n + o + p + q + r + s + t + u + v + w + x + y + z)
> quo_name(q)
[1] "+..."
rlang::quo_text (not exported by dplyr) works better, but introduces line breaks (which can be controlled with parameter width):
> rlang::quo_text(q)
[1] "a + b + c + d + e + f + g + h + i + j + k + l + m + n + o + p + \n q + r + s + t + u + v + w + x + y + z"
Otherwise, as.character can also be used, but returns a vector of length two. The second part is what you want:
> as.character(q)
[1] "~"
[2] "a + b + c + d + e + f + g + h + i + j + k + l + m + n + o + p + q + r + s + t + u + v + w + x + y + z"
> as.character(q)[2]
[1] "a + b + c + d + e + f + g + h + i + j + k + l + m + n + o + p + q + r + s + t + u + v + w + x + y + z"

Answer 3

If you use within a function, you will need to enquo() it first. Note also that with newer versions of rlang, as_name() seems to be preferred!
library(rlang)
fo <- function(arg1= name) {
print(rlang::quo_text(enquo(arg1)))
print(rlang::as_name(enquo(arg1)))
print(rlang::quo_name(enquo(arg1)))
}
fo()
#> [1] "name"
#> [1] "name"
#> [1] "name"