使用 mutate 和 case_when 的用戶定義函數

Question

我有學校級別的數據顯示每個種族群體中的學生百分比（前黑人學生/總學生數）。

我的樣本數據如下：

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

我試圖為每個種族創建一個虛擬變量，顯示一所學校屬於哪個三分位數。 例如，如果一所學校有 20% 的黑人學生，則黑人的值為 1，因為該學校屬於第一個三分位數。 如果一所學校有 67% 的黑人，那么他們就屬於第三個三分位數，並且在黑色欄中會有“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

我可以為數據集中的每個種族重復此代碼塊，但通過相應地替換種族（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"))

我沒有復制粘貼這 5 次，而是試圖想出一個用戶定義的函數，例如這樣。

  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) 將返回一個新列，該列定義了學校所屬的“白色”terciles。

我不確定 dplyr 是否可以以這種方式在函數中使用（當我運行該函數時它不斷拋出錯誤）。 關於我應該如何解決這個問題的任何想法？

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")

為了得到三分位數，我們可以除以33.33並加上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

然后我們可以使用pivot_wider為它們提供自己的列。

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

---

回答你關於dplyr函數的問題，你想定義的函數可以這樣寫。 對於將race_name作為列名處理的函數，我們需要使用!! 和:=語法。

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