如何使用 dplyr 编程语法创建和评估变量名称 [英] How to use dplyr programming syntax to create and evaluate variable names

问题描述

我想使用 dplyr 编程语法动态输入变量名，但是，正如许多人所描述的那样，这可能会令人困惑.

I would like to dynamically input a variable name using dplyr programming syntax, however, as many have described this can be quite confusing.

我玩过 quo/enquo 的各种组合！！等无济于事.这是我的代码的最简单形式

I've played around with various combinations of quo/enquo !! etc. to no avail. Here is the simplest form of my code

library(tidyverse)

df <- tibble(
  color1 = c("blue", "blue", "blue", "blue", "blue"),
  color2 = c("black", "black", "black", "black", "black"),
  value = 1:5
)

num <- 2

df %>%
  mutate(color3 = !!(paste0("color", num)))


#> # A tibble: 5 x 4
#>   color1 color2 value color3
#>   <chr>  <chr>  <int> <chr> 
#> 1 blue   black      1 color2
#> 2 blue   black      2 color2
#> 3 blue   black      3 color2
#> 4 blue   black      4 color2
#> 5 blue   black      5 color2

^{由 reprex 包 (v0.2.1) 于 2018 年 12 月 19 日创建}

^{Created on 2018-12-19 by the reprex package (v0.2.1)}

相反，我想评估引用的输入.

Instead I would like to evaluate the quoted input.

#> # A tibble: 5 x 4
#>   color1 color2 value color3
#>   <chr>  <chr>  <int> <chr> 
#> 1 blue   black      1 black 
#> 2 blue   black      2 black 
#> 3 blue   black      3 black 
#> 4 blue   black      4 black 
#> 5 blue   black      5 black

推荐答案

我们可以使用 rlang 中的 sym 将字符串转换为符号，然后计算 (!!)

We can use sym from rlang to convert the string to symbol and then evaluate (!!)

library(dplyr)
df %>%
   mutate(color3 = !!(rlang::sym(paste0("color", num))))
# A tibble: 5 x 4
#  color1 color2 value color3
#  <chr>  <chr>  <int> <chr> 
#1 blue   black      1 black 
#2 blue   black      2 black 
#3 blue   black      3 black 
#4 blue   black      4 black 
#5 blue   black      5 black 

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