R:对所有值使用 group_by

Question

我正在使用 R 编程语言。

我有以下数据集:

library(dplyr)

df = structure(list(ethnicity = c("c", "c", "c", "b", "c", "b", "b", 
"b", "c", "a", "b", "b", "a", "b", "c", "a", "c", "c", "a", "a", 
"a", "a", "c", "b", "c", "b", "a", "b", "c", "b", "a", "c", "c", 
"a", "c", "b", "a", "c", "a", "a", "b", "c", "c", "a", "c", "a", 
"c", "b", "a", "b", "a", "a", "c", "a", "b", "a", "a", "c", "a", 
"b", "a", "c", "a", "c", "b", "c", "b", "b", "c", "b", "b", "c", 
"c", "a", "b", "b", "a", "b", "a", "a", "b", "c", "c", "a", "b", 
"a", "b", "a", "c", "c", "b", "c", "a", "b", "b", "c", "b", "a", 
"c", "c"), number_of_degrees = c(3L, 2L, 2L, 3L, 1L, 1L, 3L, 
2L, 2L, 2L, 2L, 2L, 2L, 3L, 2L, 1L, 2L, 2L, 2L, 3L, 2L, 3L, 2L, 
3L, 1L, 3L, 3L, 3L, 1L, 3L, 3L, 2L, 2L, 2L, 3L, 3L, 3L, 2L, 1L, 
2L, 1L, 3L, 3L, 2L, 1L, 3L, 1L, 3L, 2L, 2L, 1L, 3L, 2L, 1L, 3L, 
3L, 3L, 1L, 2L, 2L, 1L, 2L, 3L, 3L, 1L, 2L, 1L, 2L, 3L, 3L, 1L, 
3L, 2L, 1L, 1L, 2L, 3L, 1L, 2L, 2L, 1L, 2L, 1L, 1L, 2L, 1L, 3L, 
1L, 1L, 2L, 1L, 2L, 1L, 1L, 1L, 3L, 3L, 2L, 1L, 2L)), class = "data.frame", row.names = c(NA, 
-100L))


df %>%
    # Group the data by number_of_degrees
    group_by(number_of_degrees) %>%
    # Calculate the percentage of each ethnicity within each group
    summarize(
        percent_a = mean(ethnicity == "a") * 100,
        percent_b = mean(ethnicity == "b") * 100,
        percent_c = mean(ethnicity == "c") * 100
    )

这会产生以下输出:

# A tibble: 3 x 4
  number_of_degrees percent_a percent_b percent_c
              <int>     <dbl>     <dbl>     <dbl>
1                 1      33.3      36.7      30  
2                 2      31.6      21.1      47.4
3                 3      34.4      40.6      25  

我的问题:是否有更“紧凑”的方式来编写此代码，这样我就不必手动编写“percent_a”、“percent_b”等？这样，它会更快并自动针对所有种族值(value)观执行此操作。

Answer 1

也许您可以尝试这个基本 R 选项(列名称可能与所需的输出有点不同)

> aggregate(. ~ number_of_degrees, df, \(x) proportions(table(x)))
  number_of_degrees ethnicity.a ethnicity.b ethnicity.c
1                 1   0.3333333   0.3666667   0.3000000
2                 2   0.3157895   0.2105263   0.4736842
3                 3   0.3437500   0.4062500   0.2500000

或

reshape(
    as.data.frame(proportions(table(df), 2)),
    direction = "wide",
    idvar = "number_of_degrees",
    timevar = "ethnicity"
)

这给出了

  number_of_degrees    Freq.a    Freq.b    Freq.c
1                 1 0.3333333 0.3666667 0.3000000
4                 2 0.3157895 0.2105263 0.4736842
7                 3 0.3437500 0.4062500 0.2500000

---

或者，使用 dplyr 提供一个不太紧凑的选项(抱歉我的 tidyverse 知识有限)

table(rev(df)) %>%
    proportions(1) %>%
    as.data.frame.matrix() %>%
    rownames_to_column(var = "number_of_degrees") %>%
    mutate(number_of_degrees = as.integer(number_of_degrees))

这给出了

  number_of_degrees         a         b         c
1                 1 0.3333333 0.3666667 0.3000000
2                 2 0.3157895 0.2105263 0.4736842
3                 3 0.3437500 0.4062500 0.2500000