本文介绍了如何使用 dplyr 编程语法创建和评估变量名称的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
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
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
这篇关于如何使用 dplyr 编程语法创建和评估变量名称的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!