Question

我有学校数据显示每个种族组中的学生百分比（黑人学生/总学生）。

我的示例数据如下：

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

我正在尝试为每个种族创建一个虚拟变量，以显示一所学校所处的位置。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等项。如果学校有67％的黑色，则它们将落入第三等分线，并且在黑色栏中会显示“ 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比赛重复此代码块，但是可以通过相应地替换比赛（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5次，而是想出一个用户定义的函数，例如this。

  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"))
  }

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

我不确定dplyr是否可以通过这种方式在函数中使用（运行函数时，它会不断抛出错误）。关于如何处理此问题有任何想法吗？

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")

要获得韧性，我们只需将33.33除以并加上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为他们提供自己的列。

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

要回答有关dplyr函数的问题，可以这样编写要定义的函数。对于将race_name作为列名进行处理的函数，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

用户定义的函数使用mutate和case_when

1 个答案: