r - 如何使用 R 计算选项的响应百分比？

Question

这是我的数据框的示例:

    Question Student Option
1          1       1      4
2          1       2      1
3          1       3      1
4          1       4      5
5          2       1      3
6          2       2      2
7          2       3      2
8          2       4      1

我正在尝试计算每个问题的选项 0-5 的响应百分比。我设法通过这样的问题计算每个选项的响应百分比:

Global_Response <- group_by(ExamData,Question) %>% #group data by question
  summarise(Global_0 = length(which(Option=="0"))/length(Option)*100, #Percentage of response option "0" by question
        Global_1 = length(which(Option=="1"))/length(Option)*100,
        Global_2 = length(which(Option=="2"))/length(Option)*100,
        Global_3 = length(which(Option=="3"))/length(Option)*100,
        Global_4 = length(which(Option=="4"))/length(Option)*100,
        Global_5 = length(which(Option=="5"))/length(Option)*100)

最后，我想要一个新的数据框，其中包含每个选项的响应百分比。与此示例类似:

Question Global_0 Global_1 Global_2 Global_3 Global_4 Global_5
  <int>    <dbl>    <dbl>    <dbl>    <dbl>    <dbl>    <dbl>
   1        0       50.0      0         0       25.0     25.0
   2        0       25.0      50.0     25.0       0       0  

现在，我想做同样的事情，但只需几行代码。我尝试过 for 循环，但没有成功。我不知道什么更适合做我想做的事。

Answer 1

也许这可以帮助......

df <- dplyr::tibble(
  Question=c(1,1,1,1,1,1,2,3),
  Student=c(1,2,3,4,5,6,1,1),
  Option=c(4,1,1,3,4,2,5,1)
) 

global_response <- df %>%
  dplyr::count(Question,Option) %>%
  dplyr::group_by(Question) %>%
  dplyr::mutate(pct=n/sum(n)*100) %>%
  dplyr::ungroup()

    # A tibble: 6 x 4
  Question Option     n   pct
     <dbl>  <dbl> <int> <dbl>
1        1      1     2  33.3
2        1      2     1  16.7
3        1      3     1  16.7
4        1      4     2  33.3
5        2      5     1 100  
6        3      1     1 100  

如果您想要更宽的格式，就这样做

global_response %>%
  tidyr::pivot_wider(id_cols=Question,
                     names_from=Option,
                     values_from=pct,
                     values_fill=list(pct=0))

# A tibble: 3 x 6
  Question   `1`   `2`   `3`   `4`   `5`
     <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1        1  33.3  16.7  16.7  33.3     0
2        2   0     0     0     0     100
3        3 100     0     0     0       0