[英]understanding rlang: mutate with variable col name and variable column
I'd like to define a function that takes a data.frame and a column name and returns the data.frame in with that column transformed (eg to lowercase). 我想定义一个带有data.frame和列名的函数,并返回data.frame,并将该列转换为(例如小写)。 When the column name is known in advance, this is simple:
当列名提前知道时,这很简单:
diamonds %>% mutate(cut = tolower(cut))
How do I define a function foo
, such that: 如何定义函数
foo
,例如:
col <- "cut"
foo(diamonds, col)
does this same behavior? 做同样的行为吗? (Not looking for a base R or
data.table
answer since I want to preserve dplyr
's ability to translate this into a lazily-evaluated SQL call). (不寻找基本R或
data.table
答案,因为我想保留dplyr
将其转换为懒惰评估的SQL调用的能力)。
If I just wanted my function to work using: foo(diamonds, cut)
, I just need enquo
and !!
如果我只是想让我的功能使用:
foo(diamonds, cut)
,我只需要enquo
和!!
foo <- function(df, col){
x <- enquo(col)
mutate(df, !!x := tolower(!!x))
}
If I want to take the column name in quotes, foo(diamonds, "cut")
, adding ensym
is sufficient: 如果我想用引号中的列名,
foo(diamonds, "cut")
,添加ensym
就足够了:
foo <- function(df, col){
col <- ensym(col)
x <- enquo(col)
mutate(df, !!x := tolower(!!x))
}
but this fails when given a variable for an argument: 但是当给出参数的变量时,这会失败:
col <- "cut"
foo(diamonds, col)
Error in ~col : object 'col' not found
What am I missing that can evaluate the variable? 我错过了什么可以评估变量?
You can also avoid tidy eval entirely here by using mutate_at()
. 您还可以使用
mutate_at()
完全避免整理eval。
library(tidyverse)
(x <- tibble(
num = 1:3,
month = month.abb[num]
))
#> # A tibble: 3 x 2
#> num month
#> <int> <chr>
#> 1 1 Jan
#> 2 2 Feb
#> 3 3 Mar
x %>%
mutate(month = tolower(month))
#> # A tibble: 3 x 2
#> num month
#> <int> <chr>
#> 1 1 jan
#> 2 2 feb
#> 3 3 mar
foo <- function(df, col) {
mutate_at(df, .vars = col, .funs = tolower)
}
foo(x, "month")
#> # A tibble: 3 x 2
#> num month
#> <int> <chr>
#> 1 1 jan
#> 2 2 feb
#> 3 3 mar
this <- "month"
foo(x, this)
#> # A tibble: 3 x 2
#> num month
#> <int> <chr>
#> 1 1 jan
#> 2 2 feb
#> 3 3 mar
Created on 2019-03-09 by the reprex package (v0.2.1.9000) 由reprex包创建于2019-03-09(v0.2.1.9000)
library(tidyverse)
col <- "cut"
foo <- function(df, col) {
df %>%
mutate(!!sym(col) := tolower(!!sym(col)))
}
foo(diamonds, col)
Check out Pass a string as variable name in dplyr::filter . 检查在dplyr :: filter中将字符串作为变量名传递 。
Going back to your original example, just use ensym()
to convert text arguments to symbols, there is no need for a quosure in this case. 回到原始示例,只需使用
ensym()
将文本参数转换为符号,在这种情况下不需要quosure。
library(ggplot2)
col <- "cut"
foo <- function(df, col){
col <- rlang::sym(col)
dplyr::mutate(df, !!col := tolower(!!col))
}
foo(diamonds, col)
#> # A tibble: 53,940 x 10
#> carat cut color clarity depth table price x y z
#> <dbl> <chr> <ord> <ord> <dbl> <dbl> <int> <dbl> <dbl> <dbl>
#> 1 0.23 ideal E SI2 61.5 55 326 3.95 3.98 2.43
#> 2 0.21 premium E SI1 59.8 61 326 3.89 3.84 2.31
#> 3 0.23 good E VS1 56.9 65 327 4.05 4.07 2.31
#> 4 0.290 premium I VS2 62.4 58 334 4.2 4.23 2.63
#> 5 0.31 good J SI2 63.3 58 335 4.34 4.35 2.75
#> 6 0.24 very good J VVS2 62.8 57 336 3.94 3.96 2.48
#> 7 0.24 very good I VVS1 62.3 57 336 3.95 3.98 2.47
#> 8 0.26 very good H SI1 61.9 55 337 4.07 4.11 2.53
#> 9 0.22 fair E VS2 65.1 61 337 3.87 3.78 2.49
#> 10 0.23 very good H VS1 59.4 61 338 4 4.05 2.39
#> # … with 53,930 more rows
Created on 2019-03-11 by the reprex package (v0.2.1) 由reprex包创建于2019-03-11(v0.2.1)
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.