按应用程序和用户 ID 分组后检索特定文本的所有行

Question

当用户以数字方式完成一个步骤时，is_digitally_signed 将变为 YES。 我正在尝试做的事情：如果任何步骤以数字方式完成，我想检索相同application_id 和user_id 的所有行。请检查我想要的输出下方。

复制我的数据集的 R 代码

df <- data.table(application_id = c(1,1,1,2,2,2,3,3,3), 
                 user_id = c(123,123,123,456,456,456,789,789,789), 
                 application_status = c("incomplete", "details_verified", "complete"),
                 date = c("01/01/2018", "02/01/2018", "03/01/2018"),
                 is_digitally_signed = c("NULL", "NULL", "YES", "NULL", "NULL", "NULL", "NULL", "YES", "NULL")) %>%
  mutate(date = as.Date(date, "%d/%m/%Y"))

带输出

df
  application_id user_id application_status       date is_digitally_signed
              1     123         incomplete  2018-01-01                NULL
              1     123   details_verified  2018-01-02                NULL
              1     123           complete  2018-01-03                 YES
              2     456         incomplete  2018-01-01                NULL
              2     456   details_verified  2018-01-02                NULL
              2     456           complete  2018-01-03                NULL
              3     789         incomplete  2018-01-01                NULL
              3     789   details_verified  2018-01-02                 YES
              3     789           complete  2018-01-03                NULL

我的（不成功的）努力

df %>% group_by(application_id,user_id) %>% filter_all(all.vars(. == "YES"))

期望的结果

application_id user_id application_status       date is_digitally_signed
              1     123         incomplete 2018-01-01                NULL
              1     123   details_verified 2018-01-02                NULL
              1     123           complete 2018-01-03                 YES
              3     789         incomplete 2018-01-01                NULL
              3     789   details_verified 2018-01-02                 YES
              3     789           complete 2018-01-03                NULL

Answer 1

dplyr

我们可以将filter 与any 一起使用，这将检查给定组中是否至少有一条带有is_digitally_signed == 'YES' 的记录：

library(dplyr)

df %>% 
  group_by(application_id, user_id) %>%
  filter(any(is_digitally_signed == "YES"))

或使用all 函数对不是所有is_digitally_signed == "NULL" 的组进行子集化：

df %>% 
  group_by(application_id, user_id) %>%
  filter(!all(is_digitally_signed == "NULL"))

数据表

我们也可以使用data.table，因为您已经将数据加载为 DT：

library(data.table)
dt = setDT(df)
dt[dt[,.I[any(is_digitally_signed == "YES")], by=.(application_id, user_id)]$V1,]

或.SD:

dt[,.SD[any(is_digitally_signed == "YES")], by=.(application_id, user_id)]

输出：

# A tibble: 6 x 5
# Groups:   application_id, user_id [2]
  application_id user_id application_status date       is_digitally_signed
           <dbl>   <dbl> <fct>              <date>     <fct>              
1              1     123 incomplete         2018-01-01 NULL               
2              1     123 details_verified   2018-01-02 NULL               
3              1     123 complete           2018-01-03 YES                
4              3     789 incomplete         2018-01-01 NULL               
5              3     789 details_verified   2018-01-02 YES                
6              3     789 complete           2018-01-03 NULL

Answer 2

由于只有一列要测试，我们可以简单地使用filter 和any

library(dplyr)
df %>% 
   group_by(application_id,user_id) %>% 
    filter(any(is_digitally_signed  == "YES"))
# A tibble: 6 x 5
# Groups:   application_id, user_id [2]
#  application_id user_id application_status date       is_digitally_signed
#           <dbl>   <dbl> <chr>              <date>     <chr>              
#1              1     123 incomplete         2018-01-01 NULL               
#2              1     123 details_verified   2018-01-02 NULL               
#3              1     123 complete           2018-01-03 YES                
#4              3     789 incomplete         2018-01-01 NULL               
#5              3     789 details_verified   2018-01-02 YES                
#6              3     789 complete           2018-01-03 NULL               

---

或者另一种选择是使用%in% 来返回一个被回收的TRUE/FALSE 输出

df %>% 
   group_by(application_id,user_id) %>% 
   filter("YES" %in% is_digitally_signed)

---

或者我们可以使用base R

df[with(df, ave(is_digitally_signed == "YES", application_id,user_id, FUN = any)),]