dplyr/rlang: parse_expr with multiple expressions

Question

dplyr/rlang: parse_expr with multiple expressions

For example if i want to parse some string to mutate i can

e1 = "vs + am"
mtcars %>% mutate(!!parse_expr(e1))

But when i want to parse any text with special characters like "," it will give me an error,

e2 = "vs + am , am +vs"
mtcars %>% mutate(!!parse_expr(e2))

Error in parse(text = x) : <text>:1:9: unexpected ','
1: vs + am ,
           ^

Are there any ways to work around this?

Thanks

Optimism

Answer 1

We can use the triple-bang operator with the plural form parse_exprs and a modified e2 expression to parse multiple expressions (see ?parse_quosures):

Explanation:

1. Multiple expressions in e2 need to be separated either by ; or by new lines.
2. From ?quasiquotation: The !!! operator unquotes and splices its argument. The argument should represents a list or a vector.

e2 = "vs + am ; am +vs";
mtcars %>% mutate(!!!parse_exprs(e2))
#    mpg cyl  disp  hp drat    wt  qsec vs am gear carb vs + am am + vs
#1  21.0   6 160.0 110 3.90 2.620 16.46  0  1    4    4       1       1
#2  21.0   6 160.0 110 3.90 2.875 17.02  0  1    4    4       1       1
#3  22.8   4 108.0  93 3.85 2.320 18.61  1  1    4    1       2       2
#4  21.4   6 258.0 110 3.08 3.215 19.44  1  0    3    1       1       1
#5  18.7   8 360.0 175 3.15 3.440 17.02  0  0    3    2       0       0
#6  18.1   6 225.0 105 2.76 3.460 20.22  1  0    3    1       1       1
#7  14.3   8 360.0 245 3.21 3.570 15.84  0  0    3    4       0       0
#8  24.4   4 146.7  62 3.69 3.190 20.00  1  0    4    2       1       1
#9  22.8   4 140.8  95 3.92 3.150 22.90  1  0    4    2       1       1
#10 19.2   6 167.6 123 3.92 3.440 18.30  1  0    4    4       1       1
#11 17.8   6 167.6 123 3.92 3.440 18.90  1  0    4    4       1       1
#12 16.4   8 275.8 180 3.07 4.070 17.40  0  0    3    3       0       0
#13 17.3   8 275.8 180 3.07 3.730 17.60  0  0    3    3       0       0
#14 15.2   8 275.8 180 3.07 3.780 18.00  0  0    3    3       0       0
#15 10.4   8 472.0 205 2.93 5.250 17.98  0  0    3    4       0       0
#16 10.4   8 460.0 215 3.00 5.424 17.82  0  0    3    4       0       0
#17 14.7   8 440.0 230 3.23 5.345 17.42  0  0    3    4       0       0
#18 32.4   4  78.7  66 4.08 2.200 19.47  1  1    4    1       2       2
#19 30.4   4  75.7  52 4.93 1.615 18.52  1  1    4    2       2       2
#20 33.9   4  71.1  65 4.22 1.835 19.90  1  1    4    1       2       2
#21 21.5   4 120.1  97 3.70 2.465 20.01  1  0    3    1       1       1
#22 15.5   8 318.0 150 2.76 3.520 16.87  0  0    3    2       0       0
#23 15.2   8 304.0 150 3.15 3.435 17.30  0  0    3    2       0       0
#24 13.3   8 350.0 245 3.73 3.840 15.41  0  0    3    4       0       0
#25 19.2   8 400.0 175 3.08 3.845 17.05  0  0    3    2       0       0
#26 27.3   4  79.0  66 4.08 1.935 18.90  1  1    4    1       2       2
#27 26.0   4 120.3  91 4.43 2.140 16.70  0  1    5    2       1       1
#28 30.4   4  95.1 113 3.77 1.513 16.90  1  1    5    2       2       2
#29 15.8   8 351.0 264 4.22 3.170 14.50  0  1    5    4       1       1
#30 19.7   6 145.0 175 3.62 2.770 15.50  0  1    5    6       1       1
#31 15.0   8 301.0 335 3.54 3.570 14.60  0  1    5    8       1       1
#32 21.4   4 121.0 109 4.11 2.780 18.60  1  1    4    2       2       2

Sauternes

Answer 2

You could always split them outside the expressions for example:

e2 = "vs + am"
e3 = "am +vs"
mtcars %>% mutate(!!parse_expr(e2),!!parse_expr(e3))

You can do this with parse_exprs and a semicolon instead of a comma thanks @Maurits Evers. !!! takes a list of elements and splices them into to the current call.

e2 = "vs + am ; am +vs"
mtcars %>% mutate(!!!parse_exprs(e2))

Rechabite

That's not quite correct. Take a look at ?parse_quosures. parse_exprs allows you to parse multiple expressions. – Sauternes

I did not know this. Thank you. – Rechabite

How would you name the result? e.g. "vs_am_sum = vs + am ; am_vs_sum = am +vs" – Drawtube

Answer 3

Here a little trick I use to name variables (as Genom asked)
Exemple with 2 named expressions :

across_funs <- function(x, .fns, .cols) {
  stopifnot(length(.fns) == length(.cols))
  stopifnot(all(sapply(.fns, class) == "call"))

  for (i in 1:length(.fns)) {
    x <- x %>% mutate(!!.cols[i] := !!.fns[[i]])
  }
  return(x)
}

funs = parse_exprs(c("vs+am", "am+vs"))
cols = c("var1", "var2")
mtcars %>% across_funs(.fns = funs, .cols = cols)

Greyso