How to use tidyeval on a column to mutate?

Question

I\'m sorry for the confusion but eventually, the first example I posted (at the bottom of the page), did not help me to figure out how tidyeval works with mutate, so I\'m ad

Answer 1

As the error says "Error: LHS must be a name or string" where LHS means left-hand side and it's specifically referring to the !!col := 1 part. You need to turn the quosure you made into a string or symbol. It's probably easiest to get the string version of the quosure with quo_name

colTo1 <- function(dt, ...) {
    col <- quo_name(quo(...))
    mutate(mtcars, !!col := 1)
}

Answer 2

@MrFlick's solution works for the one column case, but since OP used ... as an argument, I assume OP would also want the function to be able to take in multiple columns. For example, the following would not work:

colTo1 <- function(dt, ...) {
  col <- quo_name(quo(...))
  mutate(dt, !!col := 1)
}

colTo1(mtcars, mpg, cyl)

Error in inherits(x, "quosure") : object 'cyl' not found

What we can do is to use quos instead of quo and mutate_at instead of mutate:

colTo1 <- function(dt, ...) {
  cols <- quos(...)
  mutate_at(dt, vars(!!!cols), function(x) x=1)
}

quos converts each argument from ... into vector of quosures. Using mutate_at's vars syntax and explicit splicing with !!! from rlang, we can unquote each quosure in cols, and mutate on those specified columns.

Now this works as intended:

colTo1(mtcars, mpg, cyl)

Result:

   mpg cyl  disp  hp drat    wt  qsec vs am gear carb
1    1   1 160.0 110 3.90 2.620 16.46  0  1    4    4
2    1   1 160.0 110 3.90 2.875 17.02  0  1    4    4
3    1   1 108.0  93 3.85 2.320 18.61  1  1    4    1
4    1   1 258.0 110 3.08 3.215 19.44  1  0    3    1
5    1   1 360.0 175 3.15 3.440 17.02  0  0    3    2
6    1   1 225.0 105 2.76 3.460 20.22  1  0    3    1
7    1   1 360.0 245 3.21 3.570 15.84  0  0    3    4
8    1   1 146.7  62 3.69 3.190 20.00  1  0    4    2
9    1   1 140.8  95 3.92 3.150 22.90  1  0    4    2
10   1   1 167.6 123 3.92 3.440 18.30  1  0    4    4
...

It's also easy enough to let "1" be another argument to be passed into the function:

colToX <- function(dt, ..., X) {
  cols <- quos(...)
  mutate_at(dt, vars(!!!cols), function(x) x=X)
}

colToX(mtcars, mpg, cyl, X = 2)

Edit: OP changed the question to require that X should be another column in the same dataframe. Below is my new solution:

colToX <- function(dt, ..., X) {
  cols <- quos(...)
  X_quo <- enquo(X)
  mutate_at(dt, vars(!!!cols), funs(.data$`!!`(X_quo)))
}

colToX(mtcars, mpg, cyl, X = hp)

Here, I am using the funs function to create a list of function calls to each column referenced from vars. .data refers to the input dataframe into mutate_at (in this case dt). I used enquo to convert what's called from X into a quosure and unquote it using !!.

Result:

   mpg cyl  disp  hp drat    wt  qsec vs am gear carb
1  110 110 160.0 110 3.90 2.620 16.46  0  1    4    4
2  110 110 160.0 110 3.90 2.875 17.02  0  1    4    4
3   93  93 108.0  93 3.85 2.320 18.61  1  1    4    1
4  110 110 258.0 110 3.08 3.215 19.44  1  0    3    1
5  175 175 360.0 175 3.15 3.440 17.02  0  0    3    2
6  105 105 225.0 105 2.76 3.460 20.22  1  0    3    1
7  245 245 360.0 245 3.21 3.570 15.84  0  0    3    4
8   62  62 146.7  62 3.69 3.190 20.00  1  0    4    2
9   95  95 140.8  95 3.92 3.150 22.90  1  0    4    2
10 123 123 167.6 123 3.92 3.440 18.30  1  0    4    4
...