如何使组之间的时间差异答案

【问题标题】：How to make timing differences between groups如何使组之间的时间差异
【发布时间】：2018-07-07 05:53:54
【问题描述】：

我遇到了与时间差异有关的问题，我正在尝试通过dplyr 解决。我的初始数据框如下所示：

Paper <- data.frame(
  Student = c("A", "A", "A", "A", "A", "A", "B", "B", "B", "B"), 
  Dates = c("2014-04-17", "2014-04-17", "2014-04-17", "2014-04-17", "2014-04-18", "2014-04-18", "2014-04-18", "2014-04-18", "2014-04-18","2014-04-18"),
  Time = c("10:35:00", "11:25:00", "19:15:00", "21:00:00", "22:00:00", "22:21:26", "10:25:00", "11:15:00", "16:05:00", "17:25:00"),
  Connection = c("Initial", "Final", "Initial", "Final", "Initial", "Final", "Initial", "Final", "Initial", "Final")
)

或

   Student      Dates     Time  Connection
       A    2014-04-17  10:35:00    Initial
       A    2014-04-17  11:25:00      Final
       A    2014-04-17  19:15:00    Initial
       A    2014-04-17  21:00:00      Final
       A    2014-04-18  22:00:00    Initial
       A    2014-04-18  22:21:26      Final
       B    2014-04-18  10:25:00    Initial
       B    2014-04-18  11:15:00      Final
       B    2014-04-18  16:05:00    Initial
       B    2014-04-18  17:25:00      Final

考虑到计算的实时时间在"Initial" 和"Final" Connection 之间，我想知道每个Date 专用的时间。

所以我预期的数据框应该是这样的：

  Student    Dates    Time (Minutes)
     A    14-04-17     155
     A    14-04-18   21.43
     B    14-04-18     130

我已经尝试过了，我几乎得到了解决方案，但我不知道如何考虑计算连接之间的时间差（"Initial"/"Final"）所以我得到了这个：

Paper$Dates <- as.Date(Paper$Dates, "%Y-%m-%d")

Paper$Time <- as.numeric(as.POSIXct(as.character(Paper$Time),
                         format = "%H:%M:%S"))

FinalPaper <- 
  Paper %>% 
  group_by(Student, Dates) %>% 
  summarise(TimeSpent = sum(diff(Time))) %>% 
  mutate(TimeSpent = TimeSpent/60) %>% 
  mutate(TimeSpent = round(TimeSpent, digits = 2))

结果

  Student      Dates   TimeSpent
1       A   2014-04-17    625.00
2       A   2014-04-18     21.43
3       B   2014-04-18    420.00

从TimeSpent 中可以看出，时间更高，这是因为我没有考虑连接，所以它计算了错误的时间。例如对于学生 A，它正在计算 10:35:00 和 21:00:00 之间的时间，这是错误的。

非常感谢！！

【问题讨论】：

好问题，很好解释的问题和可重现的数据和代码。谢谢你；）但有一件事；在您预期的data.frame 中，A 14-04-18 60 行错了？
好的，谢谢！

标签： r datetime dataframe dplyr

【解决方案1】：

您可以使用cumsum(Connection == "Initial") 为每个“会话”添加一个ID。这样做的先决条件是数据按照您在此处显示的方式进行排序。然后我们可以计算每个会话的时间差，并再次聚合以获得每个学生每个日期花费的总时间：

Paper <- data.frame(
  Student = c("A", "A", "A", "A", "A", "A", "B", "B", "B", "B"), 
  Dates= c("2014-04-17", "2014-04-17", "2014-04-17", "2014-04-17", "2014-04-18", "2014-04-18", "2014-04-18", "2014-04-18", "2014-04-18","2014-04-18"),
  Time = c("10:35:00", "11:25:00", "19:15:00", "21:00:00", "22:00:00", "22:21:26", "10:25:00", "11:15:00", "16:05:00", "17:25:00"), 
  Connection = c("Initial", "Final", "Initial", "Final", "Initial", "Final", "Initial", "Final", "Initial", "Final")
)

Paper$Dates <- as.Date(Paper$Dates, "%Y-%m-%d")
Paper$Time <- as.numeric(as.POSIXct(as.character(Paper$Time),
                                    format = "%H:%M:%S"))

FinalPaper <- Paper %>% 
  mutate(seqid = cumsum(Connection == "Initial")) %>% 
  group_by(Student, Dates, seqid) %>% 
  summarise(TimeSpent = sum(diff(Time))) %>% 
  group_by(Student, Dates) %>% 
  summarise(TimeSpent = round(sum(TimeSpent)/60,2))

输出：

# A tibble: 3 x 3
# Groups:   Student [2]
  Student      Dates TimeSpent
   <fctr>     <date>     <dbl>
1       A 2014-04-17    155.00
2       A 2014-04-18     21.43
3       B 2014-04-18    130.00

希望这会有所帮助！

【讨论】：

【解决方案2】：

这是一个基于data.table 的解决方案：

library(data.table)
setDT(Paper)
Paper[order(Student, Time), .(
    TimeSpend = sum(c(0,diff(Time))[Connection == "Final"])/60
  ), by = .(Student, Dates)]

   Student      Dates TimeSpend
1:       A 2014-04-17 155.00000
2:       A 2014-04-18  21.43333
3:       B 2014-04-18 130.00000

【讨论】：