How to evaluate a constructed string with non-stan

2020-04-10 07:34发布

问题:

I have read several guides on programming with dplyr now and I am still confused about how to solve the problem of evaluating constructed/concatenated strings with non-standard evaluation (NSE). I realize that there are better ways to solve this example than using NSE, but want to learn how to.

t <- tibble( x_01 = c(1, 2, 3), x_02 = c(4, 5, 6))
i <- 1

This is my desired outcome but want the variables in mutate() to be constructed:

t %>% mutate(d_01 = x_01 * 2)
#>   A tibble: 3 x 3
#>   x_01  x_02  d_01
#>   <dbl> <dbl> <dbl>
#> 1  1.00  4.00  2.00
#> 2  2.00  5.00  4.00
#> 3  3.00  6.00  6.00

This is my first attempt, trying to use strings:

new <- sprintf("d_%02d", i)
var <- sprintf("x_%02d", i)
t %>% mutate(new = var * 2)
#> Error in mutate_impl(.data, dots) : 
#> Evaluation error: non-numeric argument to binary operator.

This is my second attempt, trying to use quosures:

new <- rlang::quo(sprintf("d_%02d", i))
var <- rlang::quo(sprintf("x_%02d", i))
t %>% mutate(!!new = !!var * 2)
#> Error: unexpected '=' in "t %>% mutate(!!new ="

This is my third attempt, trying to use quosures and the := operator:

new <- rlang::quo(sprintf("d_%02d", i))
var <- rlang::quo(sprintf("x_%02d", i))
t %>% mutate(!!new := !!var * 2)
#> Error in var * 2 : non-numeric argument to binary operator

回答1:

Use sym and := like this:

library(dplyr)
library(rlang)

t <- tibble( x_01 = c(1, 2, 3), x_02 = c(4, 5, 6))
i <- 1

new <- sym(sprintf("d_%02d", i))
var <- sym(sprintf("x_%02d", i))
t %>% mutate(!!new := (!!var) * 2)

giving:

# A tibble: 3 x 3
   x_01  x_02  d_01
  <dbl> <dbl> <dbl>
1     1     4     2
2     2     5     4
3     3     6     6

Also note that this is trivial in base R:

tdf <- data.frame( x_01 = c(1, 2, 3), x_02 = c(4, 5, 6))
i <- 1

new <- sprintf("d_%02d", i)
var <- sprintf("x_%02d", i)
tdf[[new]] <- 2 * tdf[[var]]


回答2:

You might find package friendlyeval useful while learning tidy eval. It simplifies the API and makes function choice clear in cases like these.

You have two strings you want to use as column names, so you can write:

t <- tibble( x_01 = c(1, 2, 3), x_02 = c(4, 5, 6))
i <- 1

new <- sprintf("d_%02d", i)
var <- sprintf("x_%02d", i)
t %>% mutate(!!treat_string_as_col(new) := 
               !!treat_string_as_col(var) * 2)

You can convert from friendlyeval code to plain tidy eval code at any time using an RStudio addin. This may be helpful given your learning goal.