如何在 MATLAB 中计算除周末之外的时间戳之间的营业时间？

Question

我想知道是否有人知道如何计算MATLAB 中时间戳之间的营业时间或分钟？假设我有两个时间戳：

started: 22/06/2017 18:00
ended: 26/06/2017 09:00

我想计算08:00 到17:00 之间的分钟数，不包括周末。应该是600 分钟。

有谁知道如何编写MATLAB 函数来做到这一点？ 我也可以使用python 和/或R。

新请求：
如果每个工作日的工作时间更改为9:30-11:30 和13:00-15:00。不知道有没有人知道怎么弄？

提前致谢！

更新：
@gnovice 我对您的原始代码进行了一些更改。它适用于使用两个单独的范围！想知道你对此有何感想？

function workMins = work_time(startTime, endTime)
  dateBlock = repmat((dateshift(startTime, 'start', 'day'):...
                      dateshift(endTime, 'start', 'day')).', 1, 4); %'
  dateBlock(:, 1) = dateBlock(:, 1)+hours(9)+minutes(30);
  dateBlock(:, 2) = dateBlock(:, 2)+hours(11)+minutes(30);
  dateBlock(:, 3) = dateBlock(:, 3)+hours(13)+minutes(00);
  dateBlock(:, 4) = dateBlock(:, 4)+hours(15);  
  dateBlock(1, 1) = max(dateBlock(1, 1), startTime);
  dateBlock(1, 3) = max(dateBlock(1, 3), startTime);  
  dateBlock(end, 2) = min(dateBlock(end, 2), endTime);
  dateBlock(end, 4) = min(dateBlock(end, 4), endTime);  
  dateBlock((datestr(dateBlock(:, 1), 'd') == 'S'), :) = [];
  workMins = max(diff(dateBlock, 1, 2), 0);
  workMins(:,2) = [];
  workMins = minutes(sum(workMins(:)));
end

Answer 1

MATLAB 解决方案并不像大家想象的那么难看（尽管使用Financial Toolbox 可能会对其进行简化）。切入正题，这里是一个接受两个 datetime 值的函数的解决方案：

function workMins = work_time(startTime, endTime)
  dateBlock = repmat((dateshift(startTime, 'start', 'day'):...
                      dateshift(endTime, 'start', 'day')).', 1, 2); %'
  dateBlock(:, 1) = dateBlock(:, 1)+hours(8);
  dateBlock(:, 2) = dateBlock(:, 2)+hours(17);
  dateBlock(1, 1) = max(dateBlock(1, 1), startTime);
  dateBlock(end, 2) = min(dateBlock(end, 2), endTime);
  dateBlock((datestr(dateBlock(:, 1), 'd') == 'S'), :) = [];
  workMins = minutes(sum(max(diff(dateBlock, 1, 2), 0)));
end

这就是它的工作原理......

首先，我们将开始和结束时间设置为 datetime 值：

startTime = datetime('22/06/2017 18:00');
endTime = datetime('26/06/2017 09:00');

现在我们可以创建一个N-by-2 日期时间矩阵。每行将涵盖从startTime 到endTime 范围内的一天，时间现在设置为0:00:00：

dateBlock = repmat((dateshift(startTime, 'start', 'day'):...
                    dateshift(endTime, 'start', 'day')).', 1, 2);

现在我们将第一列时间设置为8:00:00，将第二列时间设置为17:00:00：

dateBlock(:, 1) = dateBlock(:, 1)+hours(8);
dateBlock(:, 2) = dateBlock(:, 2)+hours(17);

现在我们分别添加startTime 和endTime 作为第一个和最后一个元素，将它们剪辑到8:00:00 到17:00:00 时间范围：

dateBlock(1, 1) = max(dateBlock(1, 1), startTime);
dateBlock(end, 2) = min(dateBlock(end, 2), endTime);

接下来，我们删除 datestr 日值为 'S'（即周末）的行：

dateBlock((datestr(dateBlock(:, 1), 'd') == 'S'), :) = [];

最后，我们对列的差异求和（忽略负值），然后转换为分钟：

workMins = minutes(sum(max(diff(dateBlock, 1, 2), 0)));

我们得到了想要的结果：

workMins =

   600

---

编辑：

关于新请求，您可以稍微更改函数以允许传递工作日的开始和结束时间，如下所示：

function workMins = work_time(startTime, endTime, workDayStart, workDayEnd)
  dateBlock = repmat((dateshift(startTime, 'start', 'day'):...
                      dateshift(endTime, 'start', 'day')).', 1, 2); %'
  dateBlock(:, 1) = dateBlock(:, 1)+hours(workDayStart);
  dateBlock(:, 2) = dateBlock(:, 2)+hours(workDayEnd);
  ...

现在您可以使用单独的工作时间范围调用它并添加结果：

startTime = datetime('22/06/2017 18:00');
endTime = datetime('26/06/2017 09:00');
workTotal = work_time(startTime, endTime, 9.5, 11.5) ...
            + work_time(startTime, endTime, 13, 15);

或者从一个较大的范围中减去一个子范围：

workTotal = work_time(startTime, endTime, 9.5, 15) ...
            - work_time(startTime, endTime, 11.5, 13);

如果您希望将工作日时间指定为字符数组，您可以编写如下函数：

function workMins = work_time(startTime, endTime, workDayStart, workDayEnd)
  dateBlock = repmat((dateshift(startTime, 'start', 'day'):...
                      dateshift(endTime, 'start', 'day')).', 1, 2); %'
  workDayStart = datevec(workDayStart);
  workDayEnd = datevec(workDayEnd);
  dateBlock(:, 1) = dateBlock(:, 1)+duration(workDayStart(4:6));
  dateBlock(:, 2) = dateBlock(:, 2)+duration(workDayEnd(4:6));
  ...

并像这样使用它：

workTotal = work_time(startTime, endTime, '9:30', '11:30') ...
            + work_time(startTime, endTime, '13:00', '15:00');

Answer 2

这不漂亮，必须有一种更经典的方式，但这应该在基础 R 中解决问题：

x <- seq(from = as.POSIXct("22/06/2017-18:00", format = '%d/%m/%Y-%H:%M'), to = as.POSIXct("26/06/2017-09:00", format = '%d/%m/%Y-%H:%M'), by= (60*60))
x <- x[!weekdays(x) %in% c('Saturday','Sunday')]
x <- x[x %in% as.POSIXct(unlist(lapply(unique(as.Date(x)), function(x){paste0(x, ' ', ifelse(nchar(8:16) == 1, as.character(paste0('0', 8:16)), 8:16))})), format = "%Y-%m-%d %H")]
(length(x)-1)*60

考虑一下，生成具有所需时间跨度的工作日并将每天的 difftime 的结果相加可能会更聪明。

Answer 3

这是R中的解决方案

数据

start = "22/06/2017 18:00"
end = "26/06/2017 09:00"

功能

foo = function(start, end, filter_weekend = TRUE,
        daystart = "08:00:00", dayend = "17:00:00", units = "mins"){
#start and end should be POSIXct

    require(lubridate)

    #Get a sequence of all dates between start and end
    df = data.frame(col1 = seq.Date(from = as.Date(start),
                                    to = as.Date(end),
                                    by = "days"))

    #Set start of each day at daystart
    df$start = ymd_hms(paste(df$col1, daystart))

    #Set end of each day at dayend
    df$end = ymd_hms(paste(df$col1, dayend))
    df$days = weekdays(df$start)

    #Set the first day$start to be correct date and time
    df$start[1] = start
    #Set the last day$end to be correct date and time
    df$end[NROW(df)] = end

    if (filter_weekend == TRUE){
        #Remove weekends
        df = df[!df$days %in% c("Saturday", "Sunday"), ]
    }
    #Remove rows if end is smaller than start (relevant in first and last row)
    df = df[df$end > df$start, ]

    #Compute time difference for each row
    df$time = difftime(time1 = df$end, time2 = df$start, units = units)

    #Output
    return(sum(df$time))
}

用法

library(lubridate)
foo(start = dmy_hm(start), end = dmy_hm(end))
#Time difference of 600 mins

foo(start = dmy_hms("22/06/2017 14:21:19"),
        end = dmy_hms("26/06/2017 06:20:53"),
        units = "secs", filter_weekend = TRUE)
#Time difference of 41921 secs

Answer 4

这是一个自定义函数，用于计算工作分钟数。它有点冗长，但其中有一些检查有助于确保正确计算值。它还将单位作为参数，因此您可以将总数计算为秒、分钟、小时、天或周。

work.time <- function(Date1, Date2, work.start, work.end, unit){

  # If dates are equal return 0
  if(Date1 == Date2){
    return(0)
  }

  # Check to make sure Date1 is always before Date2
  if(Date1 > Date2){
    D <- Date1
    Date1 <- Date2
    Date2 <- D
    rm(D)
  }

  # Get workday start and end timestamps for both days
  D1.start <- as.POSIXct(format(Date1, format = paste("%d/%m/%Y", work.start)), format = "%d/%m/%Y %H:%M")
  D1.end <- as.POSIXct(format(Date1, format = paste("%d/%m/%Y", work.end)), format = "%d/%m/%Y %H:%M")

  D2.start <- as.POSIXct(format(Date2, format = paste("%d/%m/%Y", work.start)), format = "%d/%m/%Y %H:%M")
  D2.end <- as.POSIXct(format(Date2, format = paste("%d/%m/%Y", work.end)), format = "%d/%m/%Y %H:%M")

  # Calculate value for a full workday
  full.day <- as.numeric(difftime(D1.end, D1.start, units = unit))

  # Calculate value if dates fall on the same day
  if(as.Date(Date1) == as.Date(Date2)){
    if(weekdays(as.Date(Date1)) %in% c("Saturday", "Sunday")){
      val <- 0
    } else if(Date1 >= D1.start & Date2 <= D1.end){
      val <- as.numeric(difftime(Date2, Date1, units = unit))
    } else if(Date1 >= D1.start & Date2 > D1.end){
      val <- as.numeric(difftime(D1.end, Date1, units = unit))
    } else if(Date1 < D1.start & Date2 <= D1.end){
      val <- as.numeric(difftime(Date2, D1.start, units = unit))
    } else if(Date1 < D1.start & Date2 > D1.end){
      val <- full.day
    } else{
      val <- 0
    }
    return(val)
  }

  # Calculate value for first workday
  if(weekdays(as.Date(Date1)) %in% c("Saturday", "Sunday")){
    d1.val <- 0
  } else 
    if(Date1 >= D1.start & Date1 <= D1.end){
    d1.val <- as.numeric(difftime(D1.end, Date1, units = unit))
  } else if(Date1 > D1.end){
    d1.val <- 0
  } else if(Date1 < D1.start){
    d1.val <- full.day
  }

  # Calculate value for last workday
  if(weekdays(as.Date(Date2)) %in% c("Saturday", "Sunday")){
    d2.val <- 0
  } else if(Date2 >= D2.start & Date2 <= D2.end){
    d2.val <- as.numeric(difftime(Date2, D2.start, units = unit))
  } else if(Date2 > D2.end){
    d2.val <- full.day
  } else if(Date2 < D2.start){
    d2.val <- 0
  }

  # Calculate work value for all workdays between Date1 and Date2
  work.days <- sum(!weekdays(seq(as.Date(Date1) + 1, as.Date(Date2) - 1, "days")) %in% c("Saturday", "Sunday"))

  # Add up final work value total
  work.val <- (work.days * full.day) + d1.val + d2.val

  return(work.val)
}


Date1 <- as.POSIXct("24/06/2017 18:00", format = "%d/%m/%Y %H:%M")
Date2 <- as.POSIXct("26/06/2017 09:00", format = "%d/%m/%Y %H:%M") 

work.time(Date1, Date2, work.start = "08:00", work.end = "17:00", unit = "mins")