使用R根据时间戳拆分数据帧中的行答案

【问题标题】：Split the rows in data frame based on timestamp using R使用R根据时间戳拆分数据帧中的行
【发布时间】：2021-01-08 16:21:41
【问题描述】：

我有以下非结构化票务数据集，其中包含工作说明更新。每张工单都有多个基于时间戳的工作笔记。我需要将 Work notes 列拆分为具有时间戳的每一行及其对应的更新，类似于 Expected output

中显示的更新

I.NO    Ticket No:               Worknotes                  
0         198822       2015-06-19 01:57:11 -Account Service
1         198822       Event closed
2         198822     Acknowledged 
3         198822     2015-06-19 01:58:33- Lawrence David 
4         198822     Data unavialable and hence ticket closed     
5         198824     2015-06-19 02:07:01- Account Service
6         198824     User requested for database information   
7         198824     2015-06-19 02:07:34- Cecilia Trandau 
8         198824     Backup in progress. Under discusion 
9         198824     2015-06-20 02:07:01- Account Service
10        198824     Auto closed 

########## Edited    **Output of dput**

structure(list(I.NO = c(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10), `Ticket No:` = c(198822, 
198822, 198822, 198822, 198822, 198824, 198824, 198824, 198824, 
198824, 198824), Worknotes = c("2015-06-19 01:57:11 -Account Service", 
"Event closed", "Acknowledged", "2015-06-19 01:58:33- Lawrence David", 
"Data unavialable and hence ticket closed", "2015-06-19 02:07:01- Account Service", 
"User requested for database information", "2015-06-19 02:07:34- Cecilia Trandau", 
"Backup in progress. Under discusion", "2015-06-20 02:07:01- Account Service", 
"Auto closed")), row.names = c(NA, -11L), class = c("tbl_df", 
"tbl", "data.frame"))
# A tibble: 6 x 3
   I.NO `Ticket No:` Worknotes                               
  <dbl>        <dbl> <chr>                                   
1     0       198822 2015-06-19 01:57:11 -Account Service    
2     1       198822 Event closed                            
3     2       198822 Acknowledged                            
4     3       198822 2015-06-19 01:58:33- Lawrence David     
5     4       198822 Data unavialable and hence ticket closed
6     5       198824 2015-06-19 02:07:01- Account Service  

###########################

**Expected Output**

   **Ticket No:**       **Worknotes**                  
    198822     2015-06-19 01:57:11 -Account Service
                      Event closed
                      Acknowledge
    198822     2015-06-19 01:58:33- Lawrence David 
               Data unavailable and hence ticket closed 
    198824     2015-06-19 02:07:01- Account Service
               User requested for database information
    198824     2015-06-19 02:07:34- Cecilia Trandau 
               Backup in progress. Under discusion 

    198824     2015-06-20 02:07:01- Account Service
               Auto closed

【问题讨论】：

您能解释一下您预期输出的数据结构吗？对于每个可以包含多个值的票号，工作笔记列中是否应该有一个向量？一个列表？还是在一张 19822 票的表中有三行，其中两行在“票号”列中根本没有条目？
嗨，Akshi，请不要基本上发帖 the same question 两次。正如我上次评论的那样，尚不清楚您的数据在 R 中是如何格式化的，因此我们很难提供帮助。请用head(dput(data)) 的输出edit 你的问题将data 替换为你的数据对象的名称。
我已经添加了 dput 输出。谢谢！

标签： r dplyr strsplit

【解决方案1】：

这是一种对cumsum 和str_detect 进行分组的方法：

library(tidyverse)
data %>%
  mutate(grouper = cumsum(str_detect(Worknotes,"^[0-9\\-]{10}"))) 
# A tibble: 11 x 4
    I.NO `Ticket No:` Worknotes                                grouper
   <dbl>        <dbl> <chr>                                      <int>
 1     0       198822 2015-06-19 01:57:11 -Account Service           1
 2     1       198822 Event closed                                   1
 3     2       198822 Acknowledged                                   1
 4     3       198822 2015-06-19 01:58:33- Lawrence David            2
 5     4       198822 Data unavialable and hence ticket closed       2
 6     5       198824 2015-06-19 02:07:01- Account Service           3
 7     6       198824 User requested for database information        3
 8     7       198824 2015-06-19 02:07:34- Cecilia Trandau           4
 9     8       198824 Backup in progress. Under discusion            4
10     9       198824 2015-06-20 02:07:01- Account Service           5
11    10       198824 Auto closed                                    5

从这里，我们可以group_by、summarise和paste：

data %>%
    mutate(grouper = cumsum(str_detect(Worknotes,"^[0-9\\-]{10}"))) %>%
    group_by(`Ticket No:`, grouper) %>%
    summarise(Worknotes = paste(Worknotes, collapse = "\n")) %>%
    select(-grouper) -> result
result
  `Ticket No:` Worknotes                                                                      
         <dbl> <chr>                                                                          
1       198822 "2015-06-19 01:57:11 -Account Service\nEvent closed\nAcknowledged"             
2       198822 "2015-06-19 01:58:33- Lawrence David\nData unavialable and hence ticket closed"
3       198824 "2015-06-19 02:07:01- Account Service\nUser requested for database information"
4       198824 "2015-06-19 02:07:34- Cecilia Trandau\nBackup in progress. Under discusion"    
5       198824 "2015-06-20 02:07:01- Account Service\nAuto closed"

请注意，\n 在 R 中不会使用 print() 进行解析，但会使用 cat() 进行解析：

cat(as.matrix(result[1,2]))
2015-06-19 01:57:11 -Account Service
Event closed
Acknowledged

【讨论】：