使用 r 根据时间戳检查制造工厂的班次答案

【问题标题】：using r to check the shift in manufacturing plant based on the timestamp使用 r 根据时间戳检查制造工厂的班次
【发布时间】：2015-12-05 12:29:50
【问题描述】：

通过使用生产单元的时间戳，我想检查它是在哪个班次生产的。基本上每天两班制进行生产。班次时间为 06:00 至 18:00 和 18:00 至 06:00。下面的班次数据框显示了 12 月班次的计划。

让我说得更清楚

2015-12-01         A shift(2015-12-01 06:00:00 to 2015-12-01 17:59:59)                       
2015-12-01         D shift(2015-12-01 18:00:00 to 2015-12-02 05:59:59)
2015-12-02         A shift(2015-12-02 06:00:00 to 2015-12-02 17:59:59)
2015-12-02         D shift(2015-12-02 18:00:00 to 2015-12-03 05:59:59)
and so on..

head(shifts)
        date day_shift night_shift
1 2015-12-01         A           D
2 2015-12-02         A           D
3 2015-12-03         B           A
4 2015-12-04         B           A
5 2015-12-05         C           B
6 2015-12-06         C           B 


shifts <- structure(list(date = structure(1:31, .Label = c("2015-12-01", 
    "2015-12-02", "2015-12-03", "2015-12-04", "2015-12-05", "2015-12-06", 
    "2015-12-07", "2015-12-08", "2015-12-09", "2015-12-10", "2015-12-11", 
    "2015-12-12", "2015-12-13", "2015-12-14", "2015-12-15", "2015-12-16", 
    "2015-12-17", "2015-12-18", "2015-12-19", "2015-12-20", "2015-12-21", 
    "2015-12-22", "2015-12-23", "2015-12-24", "2015-12-25", "2015-12-26", 
    "2015-12-27", "2015-12-28", "2015-12-29", "2015-12-30", "2015-12-31"
    ), class = "factor"), day_shift = structure(c(1L, 1L, 2L, 2L, 
    3L, 3L, 4L, 4L, 1L, 1L, 2L, 2L, 3L, 3L, 4L, 4L, 1L, 1L, 2L, 2L, 
    3L, 3L, 4L, 4L, 1L, 1L, 2L, 2L, 3L, 3L, 4L), .Label = c("A", 
    "B", "C", "D"), class = "factor"), night_shift = structure(c(4L, 
    4L, 1L, 1L, 2L, 2L, 3L, 3L, 4L, 4L, 1L, 1L, 2L, 2L, 3L, 3L, 4L, 
    4L, 1L, 1L, 2L, 2L, 3L, 3L, 4L, 4L, 1L, 1L, 2L, 2L, 3L), .Label = c("A", 
    "B", "C", "D"), class = "factor")), .Names = c("date", "day_shift", 
    "night_shift"), class = "data.frame", row.names = c(NA, -31L))

在检查数据框中，我有每个生产单元的时间戳。通过使用这些时间戳，我想检查该单元是在哪个班次生产的。

head(check)
            eventtime
1 2015-12-01 06:10:08
2 2015-12-01 10:10:24
3 2015-12-01 19:01:15
4 2015-12-02 01:54:54
5 2015-12-02 06:24:14
6 2015-12-02 08:15:47

check <- structure(list(eventtime = structure(c(1448946608, 1448961024, 
1448992875, 1449017694, 1449033854, 1449040547, 1449076903, 1449085710, 
1449100168, 1449119720), class = c("POSIXct", "POSIXt"), tzone = "")), .Names = "eventtime", row.names = c(NA, 
-10L), class = "data.frame")

期望的结果：

ds
             eventtime shift
1  2015-12-01 06:10:08     A
2  2015-12-01 10:10:24     A
3  2015-12-01 19:01:15     D
4  2015-12-02 01:54:54     D
5  2015-12-02 06:24:14     A
6  2015-12-02 08:15:47     A
7  2015-12-02 18:21:43     D
8  2015-12-02 20:48:30     D
9  2015-12-03 00:49:28     D
10 2015-12-03 06:15:20     B

为了简单起见，我只展示了 12 月份的轮班计划。实际上，我需要检查完整的年份。

【问题讨论】：

标签： r data.table time-series iranges

【解决方案1】：

这是使用 lubridate 及其%within% 函数检查日期是否在时间间隔内的答案。根据您的原始数据是否实际存储为因子，您可以通过删除一些转换来简化代码。

library(lubridate)

day_shift_start <- as.POSIXct(shifts$date) + hms("06:00:00")
day_shift_end <- as.POSIXct(shifts$date) + hms("17:59:59")
night_shift_start <- as.POSIXct(shifts$date) + hms("18:00:00")
night_shift_end <- as.POSIXct(shifts$date) + days(1) + hms("05:59:59")

shift_intervals <- data.frame(intervals = c(interval(day_shift_start, day_shift_end),
                                            interval(night_shift_start, night_shift_end)),
                              shift = c(as.character(shifts$day_shift),
                                        as.character(shifts$night_shift)))
check$shift <- unlist(lapply(check$eventtime, function(x) {
                  shift_intervals$shift[x %within% shift_intervals$intervals]
               }))

check

#              eventtime shift
# 1  2015-12-01 06:10:08     A
# 2  2015-12-01 10:10:24     A
# 3  2015-12-01 19:01:15     D
# 4  2015-12-02 01:54:54     D
# 5  2015-12-02 06:24:14     A
# 6  2015-12-02 08:15:47     A
# 7  2015-12-02 18:21:43     D
# 8  2015-12-02 20:48:30     D
# 9  2015-12-03 00:49:28     D
# 10 2015-12-03 06:15:20     B

【讨论】：