如何计算和计算 R data.frame 中两列的百分比？答案

【问题标题】：How to count and calculate percentages for two columns in an R data.frame?如何计算和计算 R data.frame 中两列的百分比？
【发布时间】：2022-01-14 20:40:32
【问题描述】：

在 R 中，我有一个这样的 data.frame：

df1 <- data.frame(
  grade = rep(LETTERS[1:5], 4),
  sex = c(rep("male", 5), rep("female", 5), rep("male", 4), rep("female", 6)),
  class = c(rep(1, 10), rep(2, 10))
)

df1

   grade    sex class
1      A   male     1
2      B   male     1
3      C   male     1
4      D   male     1
5      E   male     1
6      A female     1
7      B female     1
8      C female     1
9      D female     1
10     E female     1
11     A   male     2
12     B   male     2
13     C   male     2
14     D   male     2
15     E female     2
16     A female     2
17     B female     2
18     C female     2
19     D female     2
20     E female     2

我想计算每个班级的性别百分比并制作另一个 data.frame，例如：

Class Male_percent Female_percentage 
1     50%          50% 
2     40%          60%

有人可以教我怎么做吗？这个问题以前可能有人问过，但我不知道这个问题的关键字是什么。如果我再次问同样的问题，我很抱歉。

【问题讨论】：

标签： r dataframe

【解决方案1】：

试试看门人包中的tabyl：

library(janitor)
df1 %>%
  tabyl(class, sex) %>%
  adorn_percentages()

 class female male
     1    0.5  0.5
     2    0.6  0.4

如果您想格式化为百分比，请添加adorn_pct_formatting():

df1 %>%
  tabyl(class, sex) %>%
  adorn_percentages() %>%
  adorn_pct_formatting()

 class female  male
     1  50.0% 50.0%
     2  60.0% 40.0%

免责声明：我是这些函数的作者。

【讨论】：

【解决方案2】：

你可以试试

 prop.table(table(df1[3:2]),1)*100
 #    sex
 #class female male
 #  1     50   50
 #  2     60   40

或者data.table

 library(data.table)
 setDT(df1)[, .N, by = .(class, sex)
          ][, .(Male_percent = paste0(100 * N[sex == 'male'] / sum(N), '%'), 
              Female_percent = paste0(100 * N[sex == 'female'] / sum(N), '%')), 
           by = class] 
 #   class Male_percent   Female_percent
 #1:     1          50%              50%
 #2:     2          40%              60%

或使用dplyr

 library(dplyr)
 df1 %>%
     group_by(class) %>% 
     summarise(Male_Percent= sprintf('%d%%', 100*sum(sex=='male')/n()), 
             Female_Percent = sprintf('%d%%', 100*sum(sex=='female')/n()))
 #    class Male_Percent Female_Percent
 #1     1          50%            50%
 #2     2          40%            60%

或者

  library(sqldf)
  res1 <- sqldf('select class, 
            100*sum(sex=="male")/count(sex) as m, 
            100*sum(sex=="female")/count(sex) as f,
            "%" as p
             from df1
             group by class')
   sqldf("select class,
           m||p as Male_Percent, 
           f||p as Female_Percent 
           from res1")
   #  class Male_Percent Female_Percent
   #1     1          50%            50%
   #2     2          40%            60%

更新

基于@G.Grothendieck 的 cmets，sqldf cmets 可以简化为

   sqldf("select class,
        (100 * avg(sex = 'male')) || '%' as Male_Percent,
        (100 * avg(sex = 'female')) || '%' as Female_Percent
        from df1 group
         by class")
   #     class Male_Percent Female_Percent
   #1     1        50.0%          50.0%
   #2     2        40.0%          60.0%

【讨论】：

@G.Grothendieck 谢谢，好多了。

【解决方案3】：

使用data.table 包，您可以执行以下操作

setDT(df)[ , .(
                Male_Percent = paste0(( nrow(.SD[sex == "male"]) / .N ) * 100 , "%")   , 
                Female_Percent = paste0(( nrow(.SD[sex == "female"]) / .N ) * 100 , "%")
              )   , 
           by = class
         ]

结果

#     class      Male_Percent  Female_Percent
# 1:     1          50%            50%
# 2:     2          40%            60%

另一个dplyr 解决方案将是

df %>%
  group_by(sex , class) %>%
  summarise(n = n()) %>%
  group_by(class) %>%
  summarise(
    Male_Percent = paste0((n[sex == "male"] / sum(n)) * 100 , "%")    , 
    Female_Percent = paste0((n[sex == "female"] / sum(n) * 100) , "%")   
  )

#  class   Male_Percent     Female_Percent
#   1          50%            50%
#   2          40%            60%

【讨论】：

我想你打算评论另一个答案:)