在dplyr中获得具有独特功能的唯一值

Question

我的数据如下

library(tibble)
df <- tibble(ID = c(100000L, 100000L, 100000L, 100000L, 100001L, 100001L, 100001L, 100001L, 100002L, 100002L, 100002L, 100002L, 100003L, 100003L, 100003L), subject_result2 = c("OTHERPassedTerm1", "OTHERPassedTerm1", "OTHERPassedTerm1", "MATHPassedTerm1", "OTHERPassedTerm1", "OTHERPassedTerm1", "OTHERPassedTerm1", "OTHERFailedTerm1", "OTHERPassedTerm1", "OTHERPassedTerm1", "MATHPassedTerm1", "MATHFailedTerm1", "OTHERPassedTerm1", "MATHPassedTerm1", "OTHERPassedTerm1"))

# A tibble: 15 x 2
       ID subject_result2 
    <int> <chr>           
 1 100000 OTHERPassedTerm1
 2 100000 OTHERPassedTerm1
 3 100000 OTHERPassedTerm1
 4 100000 MATHPassedTerm1 
 5 100001 OTHERPassedTerm1
 6 100001 OTHERPassedTerm1
 7 100001 OTHERPassedTerm1
 8 100001 OTHERFailedTerm1
 9 100002 OTHERPassedTerm1
10 100002 OTHERPassedTerm1
11 100002 MATHPassedTerm1 
12 100002 MATHFailedTerm1 
13 100003 OTHERPassedTerm1
14 100003 MATHPassedTerm1 
15 100003 OTHERPassedTerm1

我想基于每个subject_result2获得一个唯一的ID。如下所示，但此代码不起作用

library(dplyr)
df %>%
 group_by(ID) %>%
 distinct(subject_result2)

您能解决我的问题吗？谢谢

所需结果：

#   <int> <chr>           
#1 100000 OTHERPassedTerm1
#2 100000 MATHPassedTerm1 
#3 100001 OTHERPassedTerm1
#4 100001 OTHERFailedTerm1
#5 100002 OTHERPassedTerm1
#6 100002 MATHPassedTerm1 
#7 100002 MATHFailedTerm1 
#8 100003 OTHERPassedTerm1
#9 100003 MATHPassedTerm1 

Answer 1

您只需要做

distinct(df)
# A tibble: 9 x 2
#      ID subject_result2 
#   <int> <chr>           
#1 100000 OTHERPassedTerm1
#2 100000 MATHPassedTerm1 
#3 100001 OTHERPassedTerm1
#4 100001 OTHERFailedTerm1
#5 100002 OTHERPassedTerm1
#6 100002 MATHPassedTerm1 
#7 100002 MATHFailedTerm1 
#8 100003 OTHERPassedTerm1
#9 100003 MATHPassedTerm1 

Answer 2

您可以做的一件事是计算ID和subject_result2组合的实例。

new_df <- df %>%
          group_by(ID, subject_result2) %>%
          summarise(id = n()) %>%distinct() %>%
          select(-id)


new_df