如何将十进制时间转换为时间格式

Question

我想计算两个时间集之间的差异。我可以做到这一点，但我只能得到小数位数的差异，我想知道如何将它们转换为 "Minutes:Second".

中的格式

所以，我将分和秒作为字符：

video_begin <- c("8:14", "4:47", "8:27", "4:59", "4:57", "7:51", "6:11", "5:30")
video_end <- c("39:08", "47:10", "49:51", "44:31", "39:41", "47:12", "40:13", "46:52")

我用 as.POSIXct 将它们转换成时间值，制作一个 df 并将差值添加为第三列，简单易行...

video_begin <- as.POSIXct(video_begin, format = "%M:%S")
video_end <- as.POSIXct(video_end, format = "%M:%S")
video <- data.frame(video_begin, video_end)
video$video_duration <- video_end - video_begin

这就是我得到的 video:

  video_begin         video_end            video_duration
1 2017-09-12 00:08:14 2017-09-12 00:39:08  30.90000 mins
2 2017-09-12 00:04:47 2017-09-12 00:47:10  42.38333 mins
3 2017-09-12 00:08:27 2017-09-12 00:49:51  41.40000 mins
4 2017-09-12 00:04:59 2017-09-12 00:44:31  39.53333 mins
5 2017-09-12 00:04:57 2017-09-12 00:39:41  34.73333 mins
6 2017-09-12 00:07:51 2017-09-12 00:47:12  39.35000 mins
7 2017-09-12 00:06:11 2017-09-12 00:40:13  34.03333 mins
8 2017-09-12 00:05:30 2017-09-12 00:46:52  41.36667 mins

如何将 video$video_duration 的格式从十进制更改为与 video$video_begin 和 video$video_end 相同的格式："Minutes:Seconds"（我不关心日期、月、年和时）？

我试过了：

video$video_duration <- as.POSIXct(video$video_duration, format = "%M:%S")

和

strptime(video$video_duration, format="%M:%S")

但是不...

我找到了一些答案，但我对它们不是很满意：

How convert decimal to POSIX time

难道没有更...更方便、更简单的方法吗？

谢谢！

Answer 1

这将创建 MM:SS 格式：

df$video_duration <- gsub('\s+[[:alpha:]]{4}','',df$video_duration)
x <- as.numeric(df$video_duration)
paste(floor(x), round((x-floor(x))*60), sep=":")

输出：

[1] "30:54" "42:23" "41:24" "39:32" "34:44" "39:21" "34:2"  "41:22"

Answer 2

使用 difftime 包装在函数中。可能有更优雅的解决方案，但嘿它有效且灵活。

time_diff <- function(time1, time2){
  hours <- difftime(time1, time2, units="hours")
  minutes <- round(60 * (hours - floor(hours)), 0)

  paste(floor(hours), minutes, sep=":")
}
time_diff(video_end, video_begin)

给予

[1] "0:31" "0:42" "0:41" "0:40" "0:35" "0:39" "0:34" "0:41"

Answer 3

另一个选项：

library(lubridate)
video$video_duration <- as.numeric(video_end - video_begin, units = "secs")
video$video_duration  <- seconds_to_period(video$video_duration)

          video_begin           video_end video_duration
1 2017-09-12 00:08:14 2017-09-12 00:39:08        30M 54S
2 2017-09-12 00:04:47 2017-09-12 00:47:10        42M 23S
3 2017-09-12 00:08:27 2017-09-12 00:49:51        41M 24S
4 2017-09-12 00:04:59 2017-09-12 00:44:31        39M 32S
5 2017-09-12 00:04:57 2017-09-12 00:39:41        34M 44S
6 2017-09-12 00:07:51 2017-09-12 00:47:12        39M 21S
7 2017-09-12 00:06:11 2017-09-12 00:40:13         34M 2S
8 2017-09-12 00:05:30 2017-09-12 00:46:52        41M 22S

如何将十进制时间转换为时间格式

How to convert decimal time to time format

r

time

decimal

lubridate

strptime