如果时间超过一定量（R，Dplyr），则创建新部分并取时差

Question

我有一个数据集 df，它有 10,000 行：

DateA

9/9/2019 7:52:16 PM
9/9/2019 7:52:16 PM
9/9/2019 7:52:17 PM
9/9/2019 7:52:18 PM
9/9/2019 7:52:18 PM
9/9/2019 7:52:19 PM
9/10/2019 1:02:23 AM
9/10/2019 1:02:25 AM
9/10/2019 1:02:26 AM
9/10/2019 1:02:27 AM
9/10/2019 1:02:27 AM
9/10/2019 1:02:29 AM
9/10/2019 1:02:29 AM
9/10/2019 1:03:29 AM    
9/10/2019 1:03:29 AM    
9/10/2019 1:03:31 AM    
9/10/2019 1:03:32 AM    
9/10/2019 4:18:48 AM    
9/10/2019 4:18:50 AM    
9/10/2019 4:18:51 AM

我想要这个输出：

Group   Duration

 a       3 sec
 b       6 sec
 c       3 sec
 d       3 sec

我想将阈值设置为 1 分钟或 60 秒。如果检测到超过 60 秒的流逝，将创建一个新组及其持续时间。

输出：

 structure(list(DateA = structure(c(12L, 12L, 13L, 14L, 14L,   15L, 
 1L, 2L, 3L, 4L, 4L, 5L, 5L, 6L, 6L, 7L, 8L, 9L, 10L, 11L),      .Label = c("9/10/2019 1:02:23 AM", 
"9/10/2019 1:02:25 AM", "9/10/2019 1:02:26 AM", "9/10/2019   1:02:27 AM", 
"9/10/2019 1:02:29 AM", "9/10/2019 1:03:29 AM", "9/10/2019 1:03:31 AM", 
"9/10/2019 1:03:32 AM", "9/10/2019 4:18:48 AM", "9/10/2019 4:18:50 AM", 
"9/10/2019 4:18:51 AM", "9/9/2019 7:52:16 PM", "9/9/2019 7:52:17 PM", 
"9/9/2019 7:52:18 PM", "9/9/2019 7:52:19 PM"), class =  "factor")), class =   "data.frame", row.names = c(NA, 
 -20L))

我试过：

 thresh1 <-60

 library(data.table)
 setDT(df)[, DateA := as.ITime(as.character(DateA))][, 
  .(Duration = difftime(max(as.POSIXct(DateA)),         min(as.POSIXct(DateA)), 
   unit = 'sec')),.(group = letters[cumsum(c(TRUE, diff(DateA) >     thresh1))])]

但是，我做错了什么，因为我只得到 1 行的输出。

  group  Duration

   a       0

不确定我做错了什么？任何建议表示赞赏。

Answer 1

我们可以将 DateA 转换为 POSIXct class，format 它只包含精确到分钟的信息，并找出 max 和 min 每组持续时间。

library(dplyr)

df %>%
  mutate(DateA = lubridate::dmy_hms(DateA), 
         temp = format(DateA, "%Y-%m-%d %H:%M")) %>%
  group_by(temp) %>%
  summarise(duration = difftime(max(DateA), min(DateA), units = "secs"))

# A tibble: 4 x 2
#  temp             duration
#  <chr>            <drtn>  
#1 2019-09-09 19:52 3 secs  
#2 2019-10-09 01:02 6 secs  
#3 2019-10-09 01:03 3 secs  
#4 2019-10-09 04:18 3 secs

如果时间超过一定量（R，Dplyr），则创建新部分并取时差

Create new section and take time difference if time exceeds a certain amount (R, Dplyr)

r

dplyr

difftime