如何将参数传递给函数内的 dplyr 连接函数? [英] How to pass by argument to dplyr join function within a function?

问题描述

我想将一个不带引号的变量名 x 传递给一个 left_join 函数.我期望的输出与我运行时相同:

left_join(mtcars, mtcars, by = c('mpg' = 'mpg'))

我正在尝试:

 ff <- function(x) {x <- enquo(x)left_join(mtcars, mtcars, by = c(x = x))}ff(mpg)

<块引用>

匹配错误(x，表，nomatch = 0L):匹配"需要向量论据

解决方案

您需要字符串作为 by 的输入，因此您需要使用 quo_name 打破 quosure 并返回一个字符串.

库(rlang)图书馆(tidyverse)ff <- 函数(x) {x <- enquo(x)left_join(mtcars，mtcars，by = quo_name(x))}头(FF(mpg))#>mpg cyl.x disp.x hp.x drat.x wt.x qsec.x vs.x am.x gear.x carb.x cyl.y#>1 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4 6#>2 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4 6#>3 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4 6#>4 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4 6#>5 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1 4#>6 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1 4#>disp.y hp.y drat.y wt.y qsec.y vs.y am.y gear.y carb.y#>1 160.0 110 3.90 2.620 16.46 0 1 4 4#>2 160.0 110 3.90 2.875 17.02 0 1 4 4#>3 160.0 110 3.90 2.620 16.46 0 1 4 4#>4 160.0 110 3.90 2.875 17.02 0 1 4 4#>5 108.0 93 3.85 2.320 18.61 1 1 4 1#>6 140.8 95 3.92 3.150 22.90 1 0 4 2

对 LHS & 使用 xby的RHS，我们需要使用set_names
归功于此答案

ff2 <- function(x) {x <- enquo(x)by = set_names(quo_name(x), quo_name(x))left_join(mtcars, mtcars, by = by)}头(ff2(mpg))#>mpg cyl.x disp.x hp.x drat.x wt.x qsec.x vs.x am.x gear.x carb.x cyl.y#>1 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4 6#>2 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4 6#>3 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4 6#>4 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4 6#>5 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1 4#>6 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1 4#>disp.y hp.y drat.y wt.y qsec.y vs.y am.y gear.y carb.y#>1 160.0 110 3.90 2.620 16.46 0 1 4 4#>2 160.0 110 3.90 2.875 17.02 0 1 4 4#>3 160.0 110 3.90 2.620 16.46 0 1 4 4#>4 160.0 110 3.90 2.875 17.02 0 1 4 4#>5 108.0 93 3.85 2.320 18.61 1 1 4 1#>6 140.8 95 3.92 3.150 22.90 1 0 4 2

I would like to pass an unquoted variable name x to a left_join function. The output I expect is the same as if I ran:

left_join(mtcars, mtcars, by = c('mpg' = 'mpg'))

I'm trying this:

 ff <- function(x) {
      x <- enquo(x)
      left_join(mtcars, mtcars, by = c(x = x))
    }
    ff(mpg)

Error in match(x, table, nomatch = 0L) : 'match' requires vector arguments

解决方案

You need strings as input for by therefore you need to use quo_name break a quosure and return a string.

library(rlang)
library(tidyverse)

ff <- function(x) {
  x <- enquo(x)
  left_join(mtcars, mtcars, by = quo_name(x))
}

head(ff(mpg))
#>    mpg cyl.x disp.x hp.x drat.x  wt.x qsec.x vs.x am.x gear.x carb.x cyl.y
#> 1 21.0     6    160  110   3.90 2.620  16.46    0    1      4      4     6
#> 2 21.0     6    160  110   3.90 2.620  16.46    0    1      4      4     6
#> 3 21.0     6    160  110   3.90 2.875  17.02    0    1      4      4     6
#> 4 21.0     6    160  110   3.90 2.875  17.02    0    1      4      4     6
#> 5 22.8     4    108   93   3.85 2.320  18.61    1    1      4      1     4
#> 6 22.8     4    108   93   3.85 2.320  18.61    1    1      4      1     4
#>   disp.y hp.y drat.y  wt.y qsec.y vs.y am.y gear.y carb.y
#> 1  160.0  110   3.90 2.620  16.46    0    1      4      4
#> 2  160.0  110   3.90 2.875  17.02    0    1      4      4
#> 3  160.0  110   3.90 2.620  16.46    0    1      4      4
#> 4  160.0  110   3.90 2.875  17.02    0    1      4      4
#> 5  108.0   93   3.85 2.320  18.61    1    1      4      1
#> 6  140.8   95   3.92 3.150  22.90    1    0      4      2

To use x for both LHS & RHS of by, we need to use set_names
Credit to this answer

ff2 <- function(x) {
  x <- enquo(x)
  by = set_names(quo_name(x), quo_name(x))
  left_join(mtcars, mtcars, by = by)
}

head(ff2(mpg))
#>    mpg cyl.x disp.x hp.x drat.x  wt.x qsec.x vs.x am.x gear.x carb.x cyl.y
#> 1 21.0     6    160  110   3.90 2.620  16.46    0    1      4      4     6
#> 2 21.0     6    160  110   3.90 2.620  16.46    0    1      4      4     6
#> 3 21.0     6    160  110   3.90 2.875  17.02    0    1      4      4     6
#> 4 21.0     6    160  110   3.90 2.875  17.02    0    1      4      4     6
#> 5 22.8     4    108   93   3.85 2.320  18.61    1    1      4      1     4
#> 6 22.8     4    108   93   3.85 2.320  18.61    1    1      4      1     4
#>   disp.y hp.y drat.y  wt.y qsec.y vs.y am.y gear.y carb.y
#> 1  160.0  110   3.90 2.620  16.46    0    1      4      4
#> 2  160.0  110   3.90 2.875  17.02    0    1      4      4
#> 3  160.0  110   3.90 2.620  16.46    0    1      4      4
#> 4  160.0  110   3.90 2.875  17.02    0    1      4      4
#> 5  108.0   93   3.85 2.320  18.61    1    1      4      1
#> 6  140.8   95   3.92 3.150  22.90    1    0      4      2

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