使用 mutate 和 case_when 的用户定义函数

Question

我有学校级别的数据显示每个种族群体中的学生百分比（前黑人学生/总学生数）。

我的样本数据如下：

School  Race    perc_race
1   EnrollBlack 3
2   EnrollBlack 67
3   EnrollWhite 4
4   EnrollWhite 8
5   EnrollHis   55
6   EnrollHis   88
7   EnrollAsian 43
8   EnrollAsian 34

我试图为每个种族创建一个虚拟变量，显示一所学校属于哪个三分位数。 例如，如果一所学校有 20% 的黑人学生，则黑人的值为 1，因为该学校属于第一个三分位数。 如果一所学校有 67% 的黑人，那么他们就属于第三个三分位数，并且在黑色栏中会有“3”。

School  Race    Percent_race    black   white   hisp    asian
1   EnrollBlack       3         1           
2   EnrollBlack       67        3           
3   EnrollWhite       4                    1        
4   EnrollWhite       8                    1        
5   EnrollHis         55                          2 
6   EnrollHis         88                          3 
7   EnrollAsian       43                                  2
8   EnrollAsian 3     4                                   2

我可以为数据集中的每个种族重复此代码块，但通过相应地替换种族（即“EnrollWhite”、“EnrollHis”...）

  mutate(black = case_when(race=='EnrollBlack' & perc_race>66.66 ~"3",
                           race=='EnrollBlack' & perc_race>33.33 ~"2",
                           race=='EnrollBlack' & perc_race<=33.33 ~"1"))

我没有复制粘贴这 5 次，而是试图想出一个用户定义的函数，例如这样。

  def_tercile <- function(x,y){
  mutate(y = case_when(race=='x' & perc_race>66.66 ~"3",
                           race=='x' & perc_race>33.33 ~"2",
                           race=='x' & perc_race<=33.33 ~"1"))
  }

其中 data %>% def_tercile(EnrollWhite, White) 将返回一个新列，该列定义了学校所属的“白色”terciles。

我不确定 dplyr 是否可以以这种方式在函数中使用（当我运行该函数时它不断抛出错误）。 关于我应该如何解决这个问题的任何想法？

Answer 1

library("tidyverse")

df <- read_table2("School  Race    perc_race
1   EnrollBlack 3
2   EnrollBlack 67
3   EnrollWhite 4
4   EnrollWhite 8
5   EnrollHis   55
6   EnrollHis   88
7   EnrollAsian 43
8   EnrollAsian 34")

为了得到三分位数，我们可以除以33.33并加上1 。

df %>%
  group_by(Race) %>%
  mutate(
    tercile = 1 + perc_race %/% (100/3)
  )
#> # A tibble: 8 x 4
#> # Groups:   Race [4]
#>   School Race        perc_race tercile
#>    <dbl> <chr>           <dbl>   <dbl>
#> 1      1 EnrollBlack         3       1
#> 2      2 EnrollBlack        67       3
#> 3      3 EnrollWhite         4       1
#> 4      4 EnrollWhite         8       1
#> 5      5 EnrollHis          55       2
#> 6      6 EnrollHis          88       3
#> 7      7 EnrollAsian        43       2
#> 8      8 EnrollAsian        34       2

然后我们可以使用pivot_wider为它们提供自己的列。

df %>%
  group_by(Race) %>%
  mutate(
    tercile = 1 + perc_race %/% (100/3),
    simple_race = Race %>% str_replace("Enroll", "") %>% str_to_lower()
  ) %>%
  pivot_wider(names_from = simple_race, values_from = tercile)
#> # A tibble: 8 x 7
#> # Groups:   Race [4]
#>   School Race        perc_race black white   his asian
#>    <dbl> <chr>           <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1      1 EnrollBlack         3     1    NA    NA    NA
#> 2      2 EnrollBlack        67     3    NA    NA    NA
#> 3      3 EnrollWhite         4    NA     1    NA    NA
#> 4      4 EnrollWhite         8    NA     1    NA    NA
#> 5      5 EnrollHis          55    NA    NA     2    NA
#> 6      6 EnrollHis          88    NA    NA     3    NA
#> 7      7 EnrollAsian        43    NA    NA    NA     2
#> 8      8 EnrollAsian        34    NA    NA    NA     2

---

回答你关于dplyr函数的问题，你想定义的函数可以这样写。 对于将race_name作为列名处理的函数，我们需要使用!! 和:=语法。

def_tercile <- function(data, race_value, race_name) {
  mutate(data,
    !!race_name := case_when(
      Race == race_value & perc_race > 66.66 ~ "3",
      Race == race_value & perc_race > 33.33 ~"2",
      Race == race_value & perc_race <= 33.33 ~"1")
  )
}

df %>%
  def_tercile("EnrollBlack", "black") %>%
  def_tercile("EnrollWhite", "white") %>%
  def_tercile("EnrollHis", "his") %>%
  def_tercile("EnrollAsian", "asian")
#> # A tibble: 8 x 7
#>   School Race        perc_race black white his   asian
#>    <dbl> <chr>           <dbl> <chr> <chr> <chr> <chr>
#> 1      1 EnrollBlack         3 1     NA    NA    NA   
#> 2      2 EnrollBlack        67 3     NA    NA    NA   
#> 3      3 EnrollWhite         4 NA    1     NA    NA   
#> 4      4 EnrollWhite         8 NA    1     NA    NA   
#> 5      5 EnrollHis          55 NA    NA    2     NA   
#> 6      6 EnrollHis          88 NA    NA    3     NA   
#> 7      7 EnrollAsian        43 NA    NA    NA    2    
#> 8      8 EnrollAsian        34 NA    NA    NA    2