根据开始和停止时间计算每天的分钟数

时间：2013-06-29 22:43:13

标签： r time

如何从一个时间间隔中提取分钟数，以便返回每个日期的分钟数？

我有一个带有案例ID的数据框，start＆amp;停止时间，以及以分钟为单位的总耗用时间间隔。我需要弄清楚每个案件每个日期的分钟数。

这有点类似于this question，但我无法将这种方法转移到这种情况。

原始数据集的结构如下，但最终解决方案需要扩展以允许更多情况（请参阅下面的替代数据集）。

原始数据 

>so
  Case               Start                Stop   Minutes
1    A 2013-01-10 18:00:00 2013-01-10 20:30:00  150 mins
2    B 2013-01-15 22:45:00 2013-01-16 00:15:00   90 mins
3    C 2013-01-20 22:00:00 2013-01-22 02:00:00 1680 mins
4    D 2013-01-27 12:00:00 2013-01-28 06:00:00 1080 mins
5    E 2013-01-01 00:00:00 2013-01-01 02:00:00  120 mins
6    F 2013-01-02 08:00:00 2013-01-03 07:00:00 1380 mins

所需的输出 

  Case       Date Minutes
1    A 2013-01-10     150
2    B 2013-01-15      75
3    B 2013-01-16      15
4    C 2013-01-20     120
5    C 2013-01-21    1440
6    C 2013-01-22     120
7    D 2013-01-27     720
8    D 2013-01-28     360
9    E 2013-01-01     120
10   F 2013-01-02     960 
11   F 2013-01-03     420

原始数据的DPUT 

> dput(so)  
structure(list(  
Case = structure(1:5, .Label = c("A", "B", "C", "D", "E"), class = "factor"), 
Start = structure(c(2L, 3L, 4L, 5L, 1L), .Label = c("2013-01-01 00:00:00", "2013-01-10 18:00:00", "2013-01-15 22:45:00", "2013-01-20 22:00:00", "2013-01-27 12:00:00"), class = "factor"),  
Stop = structure(c(2L, 3L, 4L, 5L, 1L), .Label = c("2013-01-01 02:00:00", "2013-01-10 20:30:00", "2013-01-16 00:15:00", "2013-01-22 02:00:00", "2013-01-28 06:00:00"), class = "factor"), 
Minutes = structure(c(150, 90, 1680, 1080, 120), tzone = "", units = "mins", class = "difftime")
), 
.Names = c("Case","Start", "Stop", "Minutes"), 
row.names = c(NA, -5L), 
class = "data.frame")

其他较大数据集的DPUT 

> dput(so_alt)
structure(list(Case = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 1L, 2L), .Label = c("A", "B", "C", "D", "E", "F", "G", "H"), class = "factor"),     Start = structure(c(12L, 13L, 4L, 5L, 19L, 20L, 21L, 30L, 22L, 23L, 24L, 25L, 26L, 27L, 28L, 29L, 9L, 6L, 7L, 8L, 3L, 10L, 11L, 14L, 15L, 16L), .Label = c("2012-10-25 18:10:00", "2012-11-01 20:18:00", "2012-11-16 17:45:00", "2012-12-06 01:00:00", "2012-12-06 08:00:00", "2012-12-26 13:15:00", "2012-12-29 19:50:00", "2012-12-30 00:00:00", "2013-01-01 01:46:00", "2013-01-10 20:15:00", "2013-01-11 09:00:00", "2013-01-29 17:00:00", "2013-02-05 21:30:00", "2013-02-21 01:50:00", "2013-02-21 09:25:00", "2013-02-21 12:20:00", "2013-02-22 21:45:00", "2013-02-24 15:30:00", "2013-03-01 10:10:00", "2013-03-06 16:15:00", "2013-03-07 20:00:00", "2013-03-12 21:00:00", "2013-03-13 05:15:00", "2013-03-14 00:45:00", "2013-03-14 11:30:00", "2013-03-15 21:00:00", "2013-03-16 08:15:00", "2013-03-17 06:45:00", "2013-03-18 18:04:00", "2013-03-21 21:40:00"), class = "factor"), Stop = structure(c(12L, 13L, 4L, 5L, 19L, 20L, 21L, 30L, 22L, 23L, 24L, 25L, 26L, 27L, 28L, 29L, 9L, 6L, 7L, 8L, 3L, 10L, 11L, 14L, 15L, 16L), .Label = c("2012-10-25 21:30:00", "2012-11-03 08:00:00", "2012-11-17 08:45:00", "2012-12-06 05:00:00", "2012-12-06 09:30:00", "2012-12-29 19:50:00", "2012-12-30 00:00:00", "2012-12-31 20:00:00", "2013-01-02 06:00:00", "2013-01-10 22:13:00", "2013-01-12 03:30:00", "2013-01-30 07:45:00", "2013-02-06 07:30:00", "2013-02-21 09:25:00", "2013-02-21 12:20:00", "2013-02-22 11:30:00", "2013-02-24 10:15:00", "2013-02-24 16:45:00", "2013-03-06 07:45:00", "2013-03-07 07:30:00", "2013-03-08 08:00:00", "2013-03-13 05:00:00", "2013-03-14 00:45:00", "2013-03-14 10:15:00", "2013-03-14 19:00:00", "2013-03-15 22:36:00", "2013-03-16 19:00:00", "2013-03-18 15:15:00", "2013-03-18 22:00:00", "2013-03-22 08:30:00"), class = "factor"), Minutes = structure(c(885, 600, 240, 90, 7055, 915, 720, 650, 480, 1170, 570, 450, 96, 645, 1950, 236, 1694, 4715, 250, 2640, 900, 118, 1110, 455, 175, 1390), class = "difftime", units = "mins")), .Names = c("Case", "Start", "Stop", "Minutes"), row.names = c(NA, 26L), class = "data.frame")  
>

2 个答案:

答案 0 :(得分：2)

这是一种方式。

library(plyr)

ddply(ddply(so, 1, function(x) {
    d1 <- as.Date(x$Start)
    d2 <- as.Date(x$Stop)
    if (d1 == d2) data.frame(date = d1, min = x$Minutes)
    else {
        ret <- rbind(data.frame(date=d1, min = difftime(as.POSIXct(paste(d1+1, "00:00:00")), as.POSIXct(x$Start), units = "mins")), 
        data.frame(date=d2, min=difftime(as.POSIXct(x$Stop), as.POSIXct(paste(d2, "00:00:00")), units = "mins")))
        if (d2-d1>1) {
            ret <- rbind(ret, data.frame(date = seq(d1+1, d2-1, by = "day"), min = 60*24))
        }
        ret
    }
}), .(Case, date), summarise, min = sum(min))

输出是：

  Case       date       min
1    A 2013-01-10  150 mins
2    B 2013-01-15   75 mins
3    B 2013-01-16   15 mins
4    C 2013-01-20  120 mins
5    C 2013-01-21 1440 mins
6    C 2013-01-22  120 mins

The final version here:

ddply(ddply(data.frame(.id = 1:nrow(so_alt), so_alt), .(.id, Case), function(x) {
    d1 <- as.Date(x$Start)
    d2 <- as.Date(x$Stop)
    if (d1 == d2) data.frame(date = d1, min = x$Minutes)
    else {
        ret <- rbind(data.frame(date=d1, min = difftime(as.POSIXct(paste(d1+1, "00:00:00")), as.POSIXct(x$Start), units = "mins")), 
        data.frame(date=d2, min=difftime(as.POSIXct(x$Stop), as.POSIXct(paste(d2, "00:00:00")), units = "mins")))
        if (d2-d1>1) {
            ret <- rbind(ret, data.frame(date = seq(d1+1, d2-1, by = "day"), min = 60*24))
        }
        ret
    }
}), .(Case, date), summarise, min = sum(min))

并输出：

   Case       date       min
1     A 2013-01-01 1334 mins
2     A 2013-01-02  360 mins
3     A 2013-01-29  420 mins
4     A 2013-01-30  465 mins
5     A 2013-02-21  175 mins
6     A 2013-03-12  180 mins
7     A 2013-03-13  300 mins
8     B 2012-12-26  645 mins
9     B 2012-12-27 1440 mins
10    B 2012-12-28 1440 mins
11    B 2012-12-29 1190 mins
12    B 2013-02-05  150 mins
13    B 2013-02-06  450 mins
14    B 2013-02-21  700 mins
15    B 2013-02-22  690 mins
16    B 2013-03-13 1125 mins
17    B 2013-03-14   45 mins
18    C 2012-12-06  240 mins
19    C 2012-12-29  250 mins
20    C 2012-12-30    0 mins
21    C 2013-03-14  570 mins
22    D 2012-12-06   90 mins
23    D 2012-12-30 1440 mins
24    D 2012-12-31 1200 mins
25    D 2013-03-14  450 mins
26    E 2012-11-16  375 mins
27    E 2012-11-17  525 mins
28    E 2013-03-01  830 mins
29    E 2013-03-02 1440 mins
30    E 2013-03-03 1440 mins
31    E 2013-03-04 1440 mins
32    E 2013-03-05 1440 mins
33    E 2013-03-06  465 mins
34    E 2013-03-15   96 mins
35    F 2013-01-10  118 mins
36    F 2013-03-06  465 mins
37    F 2013-03-07  450 mins
38    F 2013-03-16  645 mins
39    G 2013-01-11  900 mins
40    G 2013-01-12  210 mins
41    G 2013-03-07  240 mins
42    G 2013-03-08  480 mins
43    G 2013-03-17 1035 mins
44    G 2013-03-18  915 mins
45    H 2013-02-21  455 mins
46    H 2013-03-18  236 mins
47    H 2013-03-21  140 mins
48    H 2013-03-22  510 mins

答案 1 :(得分：1)

这是一个data.table解决方案。这个想法非常简单（可能效率不高）： 为每个Case创建一个分钟向量，将其转换为天矢量并使用表计算频率。

DT <- as.data.table(dat)
DT[,list(seqD= seq(as.Date(Start),as.Date(Stop),'day'),
         duration= { 
           seq.minus <- seq(as.POSIXct(Start),
                                   as.POSIXct(Stop),'min')[-1]
           as.data.frame(table(format(seq.minus,'%Y-%m-%d')))$Freq
           }),
   by = list(Case)]

   Case       seqD duration
1:    A 2013-01-10      150
2:    B 2013-01-15       74
3:    B 2013-01-16       16
4:    C 2013-01-20      119
5:    C 2013-01-21     1440
6:    C 2013-01-22      121
相关问题
最新问题