dplyr: deselecting columns given by

Question

How can I deselect columns given in the ... argument of a self written function. (I also need to select the columns at another point, so just specifying the columns with - in ... does not solve my problem.)

Any soltions are apreciated, select-helpers, manipulating quosures or expressions, ...

# very simple example data
test <- data.frame(a=1:3, b=1:3, c=1:3)

# function skeleton
testfun <- function(x, ...){
  y <- select(x, ...)
  z <- select(x, -...) # does of course not work like this
  return(list(y, z))   # just as an example
}

# calling the function to select different columns
testfun(test, a)
testfun(test, a, b)

Lionel Henry uses this example in his webinar https://www.rstudio.com/resources/webinars/?wvideo=zt9kl921rh and solve it with `neg_dots <- map(dots, ~ quo(- !!.x))` — aurelien, Nov 05 '18 at 08:24

score 6 · Accepted Answer · answered Oct 19 '17 at 12:11

These easiest solution would be to select the positive columns, and then compare names to figure out which columns to drop, as in this answer.

To work on the dots directly,

We will capture them in a list of quosures (quos).
Unquote and splice the dots in with UQS for the positive selection.
Do the same thing inside of c() so that we have a vector of selection.
Negate that vector to do the negative selection.

This is the transformation described by (3) and (4).

library(dplyr)
dots <- quos(a, b)
quos(-c(UQS(dots)))
#> [[1]]
#> <quosure: frame>
#> ~-c(~a, ~b)
#> 
#> attr(,"class")
#> [1] "quosures"

The full solution then would be

test <- data.frame(a = 1:3, b = 1:3, c = 1:3)

# function skeleton
testfun <- function(x, ...) {
  dots <- quos(...)
  y <- select(x, UQS(dots))
  z <- select(x, -c(UQS(dots)))
  return(list(y, z))   
}

testfun(test, a)
#> [[1]]
#>   a
#> 1 1
#> 2 2
#> 3 3
#> 
#> [[2]]
#>   b c
#> 1 1 1
#> 2 2 2
#> 3 3 3

testfun(test, a, b)
#> [[1]]
#>   a b
#> 1 1 1
#> 2 2 2
#> 3 3 3
#> 
#> [[2]]
#>   c
#> 1 1
#> 2 2
#> 3 3

Test on selection helpers.

testfun(test, starts_with("b"), one_of("c"))
#> [[1]]
#>   b c
#> 1 1 1
#> 2 2 2
#> 3 3 3
#> 
#> [[2]]
#>   a
#> 1 1
#> 2 2
#> 3 3

it seems `c` is not necessary, and the line for `y` was already working in OP's question, so you could just change the line for `z` into `z <- select(x, -!!!quos(...))` — moodymudskipper, Oct 19 '17 at 12:50
my bad, you need `c` or `testfun(test, a,b)` will fail, the line then becomes `z <- select(x, -c(!!!quos(...)))` — moodymudskipper, Oct 19 '17 at 12:52

moodymudskipper · Answer 2 · 2017-10-19T12:22:36.393

2

You can use this trick with purrr::modify_at

library(purrr)
testfun <- function(x, ...){
  y <- select(x, ...)
  z <- modify_at(x,c(...),~NULL)
  return(list(y, z))   # just as an example
}

testfun(test,"a")
# [[1]]
#   a
# 1 1
# 2 2
# 3 3
# 
# [[2]]
#   b c
# 1 1 1
# 2 2 2
# 3 3 3

edited Oct 19 '17 at 12:22

answered Oct 19 '17 at 12:05

moodymudskipper

46,417
11
121
167

This looks really nice, I'll try this, accepted the other answer, because it was closer to what I had in mind when I asked the question, but this looks more practical. – snaut Oct 20 '17 at 07:16

score 1 · Answer 3 · answered Oct 19 '17 at 11:12

1

What about that one?

testfun <- function(x, ...){
  y <- select(x, ...)
  z <- x[, !names(x) %in% names(y)]
  return(list(y, z))
}

answered Oct 19 '17 at 11:12

Tino

2,091
13
15

score 0 · Answer 4 · answered Oct 19 '17 at 11:15

You can try:

testfun <- function(x, y, z){
  y1 <- select(x, y)
  z1 <- select(x, -one_of(z)) 
  return(list(y1, z1)) 
}
testfun(test, "a", "b")
[[1]]
  a
1 1
2 2
3 3

[[2]]
  a c
1 1 1
2 2 2
3 3 3

You can also specify more variables using c

testfun(test, c("a", "c"), c("b", "a"))

dplyr: deselecting columns given by

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