dplyr总结了不扁平数据的组

Question

我有一个数据集：

df <- structure(list(ID = c(101188, 101192, 101193, 101196, 101198, 
101202, 101203, 101206, 101211, 101212, 101216, 101219, 101220, 
101222, 101223, 101224, 101226, 101227, 101228, 101229), LA = c("Barking and Dagenham", 
"Barking and Dagenham", "Barking and Dagenham", "Barking and Dagenham", 
"Barking and Dagenham", "Barking and Dagenham", "Barking and Dagenham", 
"Barking and Dagenham", "Barking and Dagenham", "Barking and Dagenham", 
"Barking and Dagenham", "Barking and Dagenham", "Barking and Dagenham", 
"Barking and Dagenham", "Barking and Dagenham", "Barking and Dagenham", 
"Barking and Dagenham", "Barking and Dagenham", "Barking and Dagenham", 
"Barking and Dagenham"), EstablishmentGroup = c("Local authority maintained schools", 
"Local authority maintained schools", "Local authority maintained schools", 
"Local authority maintained schools", "Local authority maintained schools", 
"Local authority maintained schools", "Local authority maintained schools", 
"Local authority maintained schools", "Local authority maintained schools", 
"Local authority maintained schools", "Local authority maintained schools", 
"Local authority maintained schools", "Local authority maintained schools", 
"Local authority maintained schools", "Local authority maintained schools", 
"Local authority maintained schools", "Local authority maintained schools", 
"Local authority maintained schools", "Local authority maintained schools", 
"Local authority maintained schools")), row.names = c(NA, -20L
), class = c("tbl_df", "tbl", "data.frame"))

如果运行以下代码，我希望最终的总结将数据弄平并告诉我

df %>%
  group_by(LA) %>%
  mutate(All_schools = n()) %>%
  ungroup() %>%
  group_by(LA, EstablishmentGroup, All_schools) %>%
  summarise(total = n(),
            per = total/All_schools)

Barking和Dagenham地方当局维护学校20 20 1

但它给了我20行。我可以使用独特的，但不确定我做错了什么。

Answer 1

您可以先总结计数，然后突变以计算百分比。

df %>%
  group_by(LA) %>%
  mutate(All_schools = n()) %>%
  ungroup() %>% 
  group_by(LA, EstablishmentGroup, All_schools) %>% 
  summarise(total = n()) %>% 
  mutate(per = total/All_schools)

输出：

# A tibble: 1 x 5
# Groups:   LA, EstablishmentGroup [1]
  LA                   EstablishmentGroup                 All_schools total   per
  <chr>                <chr>                                    <int> <int> <dbl>
1 Barking and Dagenham Local authority maintained schools          20    20     1