Question

我想计算两个时间集之间的差异。我可以做到这一点，但我只得到小数的差异，我想知道如何将它们转换为格式，如“分钟：秒”。

所以，我将分钟和秒数作为字符：

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将它们转换为时间值，制作一个df并添加差异作为第三列，轻松腻......

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

这就是我对video的看法：

  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中相同的格式：“分钟：秒”（我不关心日，月），年和小时）？

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

Algorithm to convert Text time to Decimal Time

是不是有更方便，更容易的方法呢？

谢谢！

Answer 1

这会创建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

使用函数中包含的difftime。可能有一个更优雅的解决方案，但它有效且灵活。

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)

给出

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

Answer 3

另一种选择：

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

3 个答案: