检查日期时间中的前一行，如果时间大于某个值，则放在一个组中并以秒为单位获取其持续时间（R，dplyr，lubridate）

Question

我有一个数据集，df：（该数据集包含超过 4000 行）

  DATEB

  9/9/2019 7:51:58 PM
  9/9/2019 7:51:59 PM
  9/9/2019 7:51:59 PM
  9/9/2019 7:52:00 PM
  9/9/2019 7:52:01 PM
  9/9/2019 7:52:01 PM
  9/9/2019 7:52:02 PM
  9/9/2019 7:52:03 PM
  9/9/2019 7:54:00 PM
  9/9/2019 7:54:02 PM
  9/10/2019 8:00:00PM

我想分组（如果时间不在上一行的10秒以内），然后取新组的持续时间。

期望的输出：

Group   Duration

 a       5 sec
 b       2 sec
 c       0 sec




 dput:


  structure(list(DATEB = structure(c(2L, 3L, 3L, 4L, 5L, 5L, 6L, 
  7L, 8L, 9L, 1L), .Label = c("      9/10/2019 8:00:00 PM", "      9/9/2019 7:51:58 PM", 
  "      9/9/2019 7:51:59 PM", "      9/9/2019 7:52:00 PM", "      9/9/2019 7:52:01 PM", 
  "      9/9/2019 7:52:02 PM", "      9/9/2019 7:52:03 PM", "      9/9/2019 7:54:00 PM", 
  "      9/9/2019 7:54:02 PM"), class = "factor")), class = "data.frame", row.names = c(NA, 
  -11L))

我已经尝试了下面的代码，效果很好，除了，我只想要以秒为单位的单位。下面的代码给出了分钟和秒的单位。

  library(dplyr)
  library(lubridate)


  df2 <- mutate(df,
          DATEB = lubridate::mdy_hms(DATEB))

 df2$time_since_last_row <- df2$DATEB - lag(df2$DATEB)
 df2$time_since_last_row[[1]] <- 0 # replace the first NA
 df2$group_10s <- 0

 for ( i in 2:nrow(df2))
   {
 if(df2$time_since_last_row[[i]]>seconds(10))
 df2$group_10s[[i]] <-  df2$group_10s[[i-1]] +1 
 else 
 df2$group_10s[[i]] <-  df2$group_10s[[i-1]]
      }





 df3 <- group_by(df2,
            group_10s) %>%
   summarise(volume_in_group=n(),
   min_DATEB=min(DATEB),
   max_DATEB=max(DATEB),
   group_duration = max_DATEB - min_DATEB)



   #nirgrahamuk-R community

欢迎任何建议。

Answer 1

其实我之前也做过类似的事情。您可以修改最后一个块：

df3 <- group_by(df2, group_10s) %>%
  summarise(
    volume_in_group=n(),
    min_DATEB=min(DATEB),
    max_DATEB=max(DATEB),
    group_duration = as.numeric(max_DATEB - min_DATEB, units = "secs")
  )

Answer 2

这就是我要做的：

gap_threshold <- 10
df %>%
  mutate(DATEB = lubridate::mdy_hms(DATEB), 
         gap = c(0, diff(DATEB))) %>% 
  group_by(grp = cumsum(gap > gap_threshold)) %>% 
  summarise(begin = min(DATEB), end = max(DATEB), 
            duration = difftime(end, begin, units = "secs"))

# A tibble: 3 x 4
    grp begin               end                 duration
  <int> <dttm>              <dttm>              <drtn>  
1     0 2019-09-09 19:51:58 2019-09-09 19:52:03 5 secs  
2     1 2019-09-09 19:54:00 2019-09-09 19:54:02 2 secs  
3     2 2019-09-10 20:00:00 2019-09-10 20:00:00 0 secs

请注意，输出中的列数比要求的多，只是为了演示。

只要两个后续行之间的间隙大于给定的 gap_threshold，组计数 grp 就会提前一个。最后，对每个组取 min() 和 max()，并根据这些计算持续时间。

检查日期时间中的前一行，如果时间大于某个值，则放在一个组中并以秒为单位获取其持续时间（R，dplyr，lubridate）

Check previous row in datetime, if time is greater than a certain value, place in a group and take its duration in seconds (R, dplyr, lubridate)

loops

r

lubridate

dplyr