以日历天为单位分割一个间隔的持续时间 [英] Split a duration of an interval in calendar days

问题描述

我有以下数据集显示一集的开始和结束(日期和时间)

I have the following data set which shows the start and the end of an episode (date and time)

ep <- data.frame(start=c("2009-07-13 23:45:00", "2009-08-14 08:30:00", 
                         "2009-09-16 15:30:00"), 
                 end=c("2009-07-14 00:03:00", "2009-08-15 08:35:00", 
                       "2009-09-19 07:30:00"))

我需要将它转换成一个数据框，它会在每个日历日显示有多少分钟的剧集.对于上面的示例，它将是:

I need to convert it into a data frame which would show in each calendar day how many minutes of episodes there were. For the above example it would be:

2009-07-13  15
2009-07-14  3
2009-08-14  930
2009-08-15  515
2009-09-16  510
2009-09-17  1440
2009-09-18  1440
2009-09-19  450

感谢您的帮助

推荐答案

这有效，但似乎有点不雅.首先，创建一个向量，它是每个开始和结束时间之间的时间序列，以分钟为单位:

This works, but seems slightly inelegant. First, create a vector that is a sequence of times between each start and end time by minutes:

tmp <- do.call(c, apply(ep, 1, 
                        function(x) head(seq(from = as.POSIXct(x[1]), 
                                             to = as.POSIXct(x[2]),by = "mins"), 
                                         -1)))

我们使用 head(...., -1) 从每个序列中删除最后一分钟，以便分钟匹配您想要的.

We use head(...., -1) to remove the last minute from each sequence so as the minutes match what you wanted.

接下来，将此向量拆分为发生在个别日期的分钟，并计算每天有多少分钟:

Next, split this vector into minutes occurring on individual days, and count how many minuts there are per day:

tmp <- sapply(split(tmp, format(tmp, format = "%Y-%m-%d")), length)

请注意，出于某种原因(可能与时区相关)，我们不能仅使用 as.Date(tmp) 来获取日期向量，我们需要将时间显式格式化为仅显示日期部分.

Note that for some reason (probably time-zone related) that we can't just use as.Date(tmp) to get a vector of dates, we need to explicitly format the times to show only the date parts.

最后一步是将包含我们需要的所有内容的 tmp 对象排列成您要求的格式:

The final step is to arrange the tmp object that contains everything we need into the format you requested:

mins <- data.frame(Date = names(tmp), Minutes = tmp, row.names = NULL)

这给出:

> mins
        Date Minutes
1 2009-07-13      15
2 2009-07-14       3
3 2009-08-14     930
4 2009-08-15     515
5 2009-09-16     510
6 2009-09-17    1440
7 2009-09-18    1440
8 2009-09-19     450

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