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我在计算两个变量之间的时间差时遇到问题。正好是入睡和醒来之间的时间。我考虑过使用 chron 这样做，这是我的例子。

DF <- data.frame(time_start = c("20:00", "21:30", "22:00", "23:00", "00:30", "02:00", "04:00"),
                 time_end = c("03:00", "06:30", "07:00", "09:00", "5:30", "09:00", "10:00"))

library(chron)
DF$time_start <- paste0(DF$time_start, ":00")
DF$time_start <- chron(times. =DF$time_start)
DF$time_end <- paste0(DF$time_end, ":00")
DF$time_end <- chron(times. =DF$time_end)
DF$time_duration <- times(DF$time_end - DF$time_start)

使用 chron 格式化变量部分按预期工作。time_end 的摘要看起来不错。另一方面， time_start 相当混乱。time_duration 是十进制的。

> summary(DF$time_start)
    Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 
00:30:00 03:00:00 20:00:00 13:17:09 21:45:00 23:00:00 
> summary(DF$time_end)
    Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 
03:00:00 06:00:00 07:00:00 07:08:34 09:00:00 10:00:00 
> summary(DF$time_duration)
      Min.    1st Qu.     Median       Mean    3rd Qu.       Max. 
-0.7083333 -0.6250000 -0.5833333 -0.2559524  0.2291667  0.2916667

我的问题：

如何获得 time_start 和

如何以更易读的方式（如小时）获得 time_start 和 time_end 之间的差异？提前感谢您的帮助！

4

1 回答 1

1

end_time如果与 相同或第二天，这将给出正确的持续时间start_time：

library(tidyverse)

DF <- data.frame(time_start = c("20:00", "21:30", "22:00", "23:00", "00:30", "02:00", "04:00"),
                 time_end = c("03:00", "06:30", "07:00", "09:00", "5:30", "09:00", "10:00")) %>% 
  mutate_at(vars(starts_with("time")), hm) %>% 
  mutate(time_duration = if_else(
    time_end > time_start,
    time_end - time_start,
    hm("24:00") - time_start + time_end)
  )

DF
  time_start  time_end time_duration
1  20H 0M 0S  3H 0M 0S      7H 0M 0S
2 21H 30M 0S 6H 30M 0S      9H 0M 0S
3  22H 0M 0S  7H 0M 0S      9H 0M 0S
4  23H 0M 0S  9H 0M 0S     10H 0M 0S
5     30M 0S 5H 30M 0S      5H 0M 0S
6   2H 0M 0S  9H 0M 0S      7H 0M 0S
7   4H 0M 0S 10H 0M 0S      6H 0M 0S
于 2020-04-27T11:24:28.753 回答