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r - 如果开始和结束时间可用，R 每天汇总数据

问题描述

我有以下问题。我有以下结构的数据框：

        startdatetime         enddatetime  type amount
1 2019-02-01 03:35:00 2019-02-03 06:35:00 prod1  1e+03
2 2019-02-03 06:35:00 2019-02-05 09:35:00 prod1  5e+03
3 2019-02-05 09:35:00 2019-02-06 01:35:00 prod2  3e+07
4 2019-02-06 01:35:00 2019-02-06 03:35:00 prod1  1e+02

表示在特定时间跨度（开始日期时间和结束日期时间）内生产的数量。现在我想每天汇总这些数据。让我们忽略不完整的一天 2019-02-01，从 2019-02-02 开始。第一个产品 1 在 2019-02-01 03:35:00 和 2019-02-03 06:35:00 之间生产，总共生产了 1000 公斤。因此，例如，在 2019 年 2 月 2 日：24/51*1000= 生产了 prod 1 的 470.58 个，因为24h + 21h + 6h = 51h. 到目前为止，我的解决方案是基于一个 for 和一个 while 循环，但我想有一个基于包“lubridate”的更快的解决方案，或者我没有找到。有什么建议吗？在我的代码下面

#create test data set
mydata <- data.frame(startdatetime=c(as.POSIXct("2019-02-01 03:35:00"), as.POSIXct("2019-02-03 06:35:00"),as.POSIXct("2019-02-05 09:35:00"),as.POSIXct("2019-02-06 01:35:00")),
                     enddatetime  =c(as.POSIXct("2019-02-03 06:35:00"), as.POSIXct("2019-02-05 09:35:00"),as.POSIXct("2019-02-06 01:35:00"),as.POSIXct("2019-02-06 03:35:00")),
                     type=c("prod1","prod1","prod2","prod1"),
                     amount=c(1000,5000,30000000,100)) 

# take only full days into account and ignore the first and the last day
minstartday = min(mydata$startdatetime)+24*60*60
maxendday   = max(mydata$enddatetime)-24*60*60

#create a day index
timesindex <- seq(from = as.Date(format(minstartday, format = "%Y/%m/%d")), 
                  to   = as.Date(format(maxendday, format = "%Y/%m/%d")), by = "day")

# create an empty dataframe which will be filled with the production data for each day
prodperday <- data.frame(Date=as.Date(timesindex),
                         prod1=replicate(length(timesindex),0), 
                         prod2=replicate(length(timesindex),0), 
                         stringsAsFactors=FALSE) 

# loop over all entries and separate them into produced fractions per day
for (irow in 1:dim(mydata)[1]){
  timestart = mydata[irow,"startdatetime"]
  datestart = as.Date(format(timestart, format = "%Y/%m/%d"))
  timeend = timestart
  tota_run_time_in_h = (as.numeric((mydata[irow,"enddatetime"]-mydata[irow,"startdatetime"])))*24.
  while (timeend < mydata[irow,"enddatetime"]){
    timeend = min (as.POSIXct(datestart, format = "%Y/%m/%d %H:%M:%S")+23*60*60-1,
                   mydata[irow,"enddatetime"])
    tdiff = as.numeric(timeend-timestart)
    fraction_prod = (tdiff/tota_run_time_in_h)*mydata[irow,"amount"]
    if (datestart %in% prodperday$Date){
      prodperday[prodperday$Date == datestart,as.character(mydata[irow,"type"])] = 
        prodperday[prodperday$Date == datestart,as.character(mydata[irow,"type"])] + fraction_prod
    }

    timestart = timeend+1
    datestart = as.Date(format(timestart, format = "%Y/%m/%d"))
    timeend = timestart
  }
}

结果：

        Date     prod1   prod2
1 2019-02-02  470.5828       0
2 2019-02-03 1836.5741       0
3 2019-02-04 2352.9139       0
4 2019-02-05  939.5425 1126280

标签： rtimetime-seriesaggregatetransformation

解决方案

我提出的解决方案并不完美，因为边界存在问题，但按小时转换生产数据并按天汇总后的想法可能是个好主意。

我随意使用这两个库：

library(lubridate)
library(dplyr)

参考时间：

ref.times <- seq(from = min(mydata$startdatetime),
           to = max(mydata$enddatetime),
           by = "hour")

构建按小时生产的数据库：

newdata <- data.frame(hour = floor_date(ref.times, unit = "hour"),
                      prod1 = 0,
                      prod2 = 0,
                      day = floor_date(newdata$hour, unit= "day"))
for(i in 1:nrow(mydata)){
  ref.times <- seq(from = mydata$startdatetime[i],
                   to = mydata$enddatetime[i],
                   by = "hour")
  n <- length(floor_date(ref.times, "hour"))
  if(mydata[i, 3] == "prod1"){
    newdata[newdata$hour %in%  floor_date(ref.times, unit = "hour"), 2] <-
      rep(mydata[i, 4] / n, n)
  }else{
    newdata[newdata$hour %in%  floor_date(ref.times, unit = "hour"), 3] <-
      rep(mydata[i, 4] / n, n)
  }
}

按天汇总：

newdata %>% group_by(day) %>% summarise(prod1 = sum(prod1),
                                        prod2 = sum(prod2))

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