使用 dplyr 根据 dplyr 中每组其他列中的条件添加列，数据为长格式答案

【问题标题】：Add column based on condition in other column per group in dplyr with data in a long format, using dplyr使用 dplyr 根据 dplyr 中每组其他列中的条件添加列，数据为长格式
【发布时间】：2021-07-14 11:10:54
【问题描述】：

我有 3 名患者重复测量（4 或 5 次）的长格式数据：

library(dplyr)
library(magrittr)

questiondata <- structure(list(ID = c(2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 
4), time = c("time1", "time2", "time3", "time4", "time1", "time2", 
"time3", "time4", "time5", "time1", "time2", "time3", "time4", 
"time5"), drug_use = structure(c(NA, 1L, NA, NA, NA, 2L, NA, 
NA, NA, NA, 1L, NA, NA, NA), .Label = c("no", "yes"), class = "factor")), row.names = c(NA, 
-14L), class = c("tbl_df", "tbl", "data.frame"))

# Corresponding to the following tibble:
# A tibble: 14 x 3
      ID time  drug_use
   <dbl> <chr> <fct>   
 1     2 time1 NA      
 2     2 time2 no      
 3     2 time3 NA      
 4     2 time4 NA      
 5     3 time1 NA      
 6     3 time2 yes     
 7     3 time3 NA      
 8     3 time4 NA      
 9     3 time5 NA      
10     4 time1 NA      
11     4 time2 no      
12     4 time3 NA      
13     4 time4 NA      
14     4 time5 NA

我正在尝试创建一个关于药物使用的新专栏（名为 'drug_use_2），根据该患者是否在任何 @987654323 使用药物，对每个患者说“是”或“否” @。所以想要的结果应该是这样的：

 A tibble: 14 x 4
      ID time  drug_use drug_use2
   <dbl> <chr> <fct>    <fct>    
 1     2 time1 NA       no       
 2     2 time2 no       no       
 3     2 time3 NA       no       
 4     2 time4 NA       no       
 5     3 time1 NA       yes       
 6     3 time2 yes      yes      
 7     3 time3 NA       yes      
 8     3 time4 NA       yes      
 9     3 time5 NA       yes      
10     4 time1 NA       no      
11     4 time2 no       no       
12     4 time3 NA       no       
13     4 time4 NA       no       
14     4 time5 NA       no

我已尝试根据 PXE 编号和/或 drug_use 进行分组，然后将 mutate 与 case when 一起使用，但这会卡在 drug_use 的缺失值上：

questiondata <- questiondata %>% 
  group_by(ID) %>% 
  mutate(drug_use2=
           case_when(
             drug_use=="yes" ~ "yes", 
             drug_use=="no" ~ "no", 
             TRUE ~ "missing"))

# A tibble: 14 x 4
# Groups:   ID [3]
      ID time  drug_use drug_use2
   <dbl> <chr> <fct>    <chr>    
 1     2 time1 NA       missing  
 2     2 time2 no       no       
 3     2 time3 NA       missing  
 4     2 time4 NA       missing  
 5     3 time1 NA       missing  
 6     3 time2 yes      yes      
 7     3 time3 NA       missing  
 8     3 time4 NA       missing  
 9     3 time5 NA       missing  
10     4 time1 NA       missing  
11     4 time2 no       no       
12     4 time3 NA       missing  
13     4 time4 NA       missing  
14     4 time5 NA       missing

我是否使用了 group_by 错误？提前致谢

【问题讨论】：

为什么是 row10，是吗？
您好，感谢您的回复。那是一个错字。我编辑了它。

标签： r dplyr case-when

【解决方案1】：

使用any

questiondata %>%
  group_by(ID) %>%
  mutate(druguse2 = case_when(any(drug_use == 'yes') ~ 'yes',
                              TRUE ~ 'no'))

# A tibble: 14 x 4
# Groups:   ID [3]
      ID time  drug_use druguse2
   <dbl> <chr> <fct>    <chr>   
 1     2 time1 NA       no      
 2     2 time2 no       no      
 3     2 time3 NA       no      
 4     2 time4 NA       no      
 5     3 time1 NA       yes     
 6     3 time2 yes      yes     
 7     3 time3 NA       yes     
 8     3 time4 NA       yes     
 9     3 time5 NA       yes     
10     4 time1 NA       no      
11     4 time2 no       no      
12     4 time3 NA       no      
13     4 time4 NA       no      
14     4 time5 NA       no

【讨论】：

【解决方案2】：

我们也可以在这里使用fill

library(dplyr)
library(tidyr)
questiondata %>%
     mutate(drug_use2 = drug_use) %>%
     group_by(ID) %>% 
     fill(drug_use2, .direction = 'updown') %>% 
     ungroup
# A tibble: 14 x 4
      ID time  drug_use drug_use2
   <dbl> <chr> <fct>    <fct>    
 1     2 time1 <NA>     no       
 2     2 time2 no       no       
 3     2 time3 <NA>     no       
 4     2 time4 <NA>     no       
 5     3 time1 <NA>     yes      
 6     3 time2 yes      yes      
 7     3 time3 <NA>     yes      
 8     3 time4 <NA>     yes      
 9     3 time5 <NA>     yes      
10     4 time1 <NA>     no       
11     4 time2 no       no       
12     4 time3 <NA>     no       
13     4 time4 <NA>     no       
14     4 time5 <NA>     no

或者coalesce的另一个选项

questiondata %>% 
   group_by(ID) %>%
    mutate(drug_use2 = coalesce(drug_use[which(drug_use %in% 
        'yes')[1]], "no")) %>% 
   ungroup
# A tibble: 14 x 4
      ID time  drug_use drug_use2
   <dbl> <chr> <fct>    <chr>    
 1     2 time1 <NA>     no       
 2     2 time2 no       no       
 3     2 time3 <NA>     no       
 4     2 time4 <NA>     no       
 5     3 time1 <NA>     yes      
 6     3 time2 yes      yes      
 7     3 time3 <NA>     yes      
 8     3 time4 <NA>     yes      
 9     3 time5 <NA>     yes      
10     4 time1 <NA>     no       
11     4 time2 no       no       
12     4 time3 <NA>     no       
13     4 time4 <NA>     no       
14     4 time5 <NA>     no

或者也可以在没有分组的情况下这样做

questiondata %>% 
  mutate(drug_use2 =  c("no", "yes")[1 + (ID %in% ID[drug_use == 'yes'])])
# A tibble: 14 x 4
      ID time  drug_use drug_use2
   <dbl> <chr> <fct>    <chr>    
 1     2 time1 <NA>     no       
 2     2 time2 no       no       
 3     2 time3 <NA>     no       
 4     2 time4 <NA>     no       
 5     3 time1 <NA>     yes      
 6     3 time2 yes      yes      
 7     3 time3 <NA>     yes      
 8     3 time4 <NA>     yes      
 9     3 time5 <NA>     yes      
10     4 time1 <NA>     no       
11     4 time2 no       no       
12     4 time3 <NA>     no       
13     4 time4 <NA>     no       
14     4 time5 <NA>     no

类似于base R

transform(questiondata, drug_use2 =  c("no", "yes")[1 +
        (ID %in% ID[drug_use == 'yes'])])

【讨论】：