如何计算大型数据集每分钟出现的次数答案

【问题标题】：How to calculate number of occurrences per minute for a large dataset如何计算大型数据集每分钟出现的次数
【发布时间】：2013-06-14 15:38:44
【问题描述】：

我有一个数据集，其中包含 50 万个约会，持续时间在 5 到 60 分钟之间。

tdata <- structure(list(Start = structure(c(1325493000, 1325493600, 1325494200, 1325494800, 1325494800, 1325495400, 1325495400, 1325496000, 1325496000, 1325496600, 1325496600, 1325497500, 1325497500, 1325498100, 1325498100, 1325498400, 1325498700, 1325498700, 1325499000, 1325499300), class = c("POSIXct", "POSIXt"), tzone = "GMT"), End = structure(c(1325493600, 1325494200, 1325494500, 1325495400, 1325495400, 1325496000, 1325496000, 1325496600, 1325496600, 1325496900, 1325496900, 1325498100, 1325498100, 1325498400, 1325498700, 1325498700, 1325499000, 1325499300, 1325499600, 1325499600), class = c("POSIXct", "POSIXt"), tzone = "GMT"), Location = c("LocationA", "LocationA", "LocationA", "LocationA", "LocationA", "LocationA", "LocationA", "LocationA", "LocationA", "LocationB", "LocationB", "LocationB", "LocationB", "LocationB", "LocationB", "LocationB", "LocationB", "LocationB", "LocationB", "LocationB"), Room = c("RoomA", "RoomA", "RoomA", "RoomA", "RoomB", "RoomB", "RoomB", "RoomB", "RoomB", "RoomB", "RoomA", "RoomA", "RoomA", "RoomA", "RoomA", "RoomA", "RoomA", "RoomA", "RoomA", "RoomA")), .Names = c("Start", "End", "Location", "Room"), row.names = c(NA, 20L), class = "data.frame")

> head(tdata)
                Start                 End  Location  Room
1 2012-01-02 08:30:00 2012-01-02 08:40:00 LocationA RoomA
2 2012-01-02 08:40:00 2012-01-02 08:50:00 LocationA RoomA
3 2012-01-02 08:50:00 2012-01-02 08:55:00 LocationA RoomA
4 2012-01-02 09:00:00 2012-01-02 09:10:00 LocationA RoomA
5 2012-01-02 09:00:00 2012-01-02 09:10:00 LocationA RoomB
6 2012-01-02 09:10:00 2012-01-02 09:20:00 LocationA RoomB

我想计算每个位置和每个房间（以及原始数据集中的其他几个因素）的 并发约会数。

我曾尝试使用mysql 包来执行左连接，它适用于小数据集，但对于整个数据集需要很长时间：

# SQL Join.
start.min <- min(tdata$Start, na.rm=T)
end.max <- max(tdata$End, na.rm=T)
tinterval <- seq.POSIXt(start.min, end.max, by = "mins")
tinterval <- as.data.frame(tinterval)

library(sqldf)
system.time(
  output <- sqldf("SELECT *
              FROM tinterval 
              LEFT JOIN tdata 
              ON tinterval.tinterval >= tdata.Start
              AND tinterval.tinterval < tdata.End "))

head(output)
            tinterval               Start                 End  Location  Room
1 2012-01-02 09:30:00 2012-01-02 09:30:00 2012-01-02 09:40:00 LocationA RoomA
2 2012-01-02 09:31:00 2012-01-02 09:30:00 2012-01-02 09:40:00 LocationA RoomA
3 2012-01-02 09:32:00 2012-01-02 09:30:00 2012-01-02 09:40:00 LocationA RoomA
4 2012-01-02 09:33:00 2012-01-02 09:30:00 2012-01-02 09:40:00 LocationA RoomA
5 2012-01-02 09:34:00 2012-01-02 09:30:00 2012-01-02 09:40:00 LocationA RoomA
6 2012-01-02 09:35:00 2012-01-02 09:30:00 2012-01-02 09:40:00 LocationA RoomA

它创建一个数据框，其中列出了每分钟的所有“活动”约会。大型数据集涵盖一整年（约 525600 分钟）。平均预约持续时间为 18 分钟，我希望 sql 连接创建一个包含约 500 万行的数据集，我可以用它来为不同的因素（位置/房间等）创建占用图。

基于How to count number of concurrent users 中建议的 sapply 解决方案，我尝试使用data.table 和snowfall，如下所示：

require(snowfall) 
require(data.table)
sfInit(par=T, cpu=4)
sfLibrary(data.table)

tdata <- data.table(tdata)
tinterval <- seq.POSIXt(start.min, end.max, by = "mins")
setkey(tdata, Start, End)
sfExport("tdata") # "Transport" data to cores

system.time( output <- data.frame(tinterval,sfSapply(tinterval, function(i) length(tdata[Start <= i & i < End,Start]) ) ) )

> head(output)
            tinterval sfSapply.tinterval..function.i..length.tdata.Start....i...i...
1 2012-01-02 08:30:00                                                              1
2 2012-01-02 08:31:00                                                              1
3 2012-01-02 08:32:00                                                              1
4 2012-01-02 08:33:00                                                              1
5 2012-01-02 08:34:00                                                              1
6 2012-01-02 08:35:00                                                              1

此解决方案速度很快，大约需要 18 秒来计算 1 天（全年大约需要 2 小时）。缺点是我无法为某些因素（位置、房间等）创建并发约会数量的子集。我觉得必须有更好的方法来做到这一点.. 有什么建议吗？

更新：根据 Geoffrey 的回答，最终解决方案看起来像这样。该示例显示了如何确定每个位置的占用率。

setkey(tdata, Location, Start, End)
vecTime <- seq(from=tdata$Start[1],to=tdata$End[nrow(tdata)],by=60)
res <- data.frame(time=vecTime)

for(i in 1:length(unique(tdata$Location)) ) { 
  addz <- array(0,length(vecTime))
  remz <- array(0,length(vecTime))

  tdata2 <- tdata[J(unique(tdata$Location)[i]),] # Subset a certain location.

  startAgg <- aggregate(tdata2$Start,by=list(tdata2$Start),length)
  endAgg <- aggregate(tdata2$End,by=list(tdata2$End),length)
  addz[which(vecTime %in% startAgg$Group.1 )] <- startAgg$x
  remz[which(vecTime %in% endAgg$Group.1)] <- -endAgg$x

  res[,c( unique(tdata$Location)[i] )] <- cumsum(addz + remz)
}

> head(res)
                 time LocationA LocationB
1 2012-01-01 03:30:00         1         0
2 2012-01-01 03:31:00         1         0
3 2012-01-01 03:32:00         1         0
4 2012-01-01 03:33:00         1         0
5 2012-01-01 03:34:00         1         0
6 2012-01-01 03:35:00         1         0

【问题讨论】：

很高兴为有用的答案投票。只是一个指针。

标签： r time data.table sqldf snowfall

【解决方案1】：

这样更好吗。

创建一个空白时间向量和一个空白计数向量。

 vecTime <- seq(from=tdata$Start[1],to=tdata$End[nrow(tdata)],by=60)
 addz <- array(0,length(vecTime))
 remz <- array(0,length(vecTime))


 startAgg <- aggregate(tdata$Start,by=list(tdata$Start),length)
 endAgg <- aggregate(tdata$End,by=list(tdata$End),length)
 addz[which(vecTime %in% startAgg$Group.1 )] <- startAgg$x
 remz[which(vecTime %in% endAgg$Group.1)] <- -endAgg$x
 res <- data.frame(time=vecTime,occupancy=cumsum(addz + remz))

【讨论】：

感谢 Geoffrey，但这不包括在特定时间段内活动的约会数量。这告诉我有两个约会从 9:00 开始，但不考虑活动约会（已经开始但未结束）。我需要每分钟的占用图才能研究真正繁忙时期的峰值。
Hee Goeffrey，您的解决方案对我的整个数据集耗时 9 秒。我已经为此苦苦挣扎了好几个小时。非常感谢您的意见。我一直在寻找错误的方向：汇总所有约会的开始和结束时间并据此确定入住率真的很聪明。考虑到计算速度，我可以使用一些 for 循环为每个位置或每个房间构建占用图，所以我认为我的问题得到了回答。

【解决方案2】：

如果我了解您的目标，我不确定。不过，这可能有用：

#I changed the example to actually have concurrent appointments
DF <- read.table(text="                Start,                 End,  Location,  Room
1, 2012-01-02 08:30:00, 2012-01-02 08:40:00, LocationA, RoomA
2, 2012-01-02 08:40:00, 2012-01-02 08:50:00, LocationA, RoomA
3, 2012-01-02 08:50:00, 2012-01-02 09:55:00, LocationA, RoomA
4, 2012-01-02 09:00:00, 2012-01-02 09:10:00, LocationA, RoomA
5, 2012-01-02 09:00:00, 2012-01-02 09:10:00, LocationA, RoomB
6, 2012-01-02 09:10:00, 2012-01-02 09:20:00, LocationA, RoomB",header=TRUE,sep=",",stringsAsFactors=FALSE)

DF$Start <- as.POSIXct(DF$Start,format="%Y-%d-%m %H:%M:%S",tz="GMT")
DF$End <- as.POSIXct(DF$End,format="%Y-%d-%m %H:%M:%S",tz="GMT")

library(data.table)
DT <- data.table(DF)
DT[,c("Start_num","End_num"):=lapply(.SD,as.numeric),.SDcols=1:2]

fun <- function(s,e) {
  require(intervals)
  mat <- cbind(s,e)
  inter <- Intervals(mat,closed=c(FALSE,FALSE),type="R")
  io <- interval_overlap( inter, inter )
  tablengths <- table(sapply(io,length))[-1]
  sum(c(0,as.vector(tablengths/as.integer(names(tablengths)))))
}

#number of overlapping events per room and location
DT[,fun(Start_num,End_num),by=list(Location,Room)]
#     Location   Room V1
#1:  LocationA  RoomA  1
#2:  LocationA  RoomB  0

我没有对此进行测试，尤其是速度方面。

【讨论】：

谢谢罗兰。有趣的方法，但我正在寻找每分钟的总入住人数，并能够对位置和房间的入住人数进行子集。

【解决方案3】：

这是一个策略 - 按开始时间排序，然后通过开始、结束、开始、结束...取消列出数据，并查看该向量是否需要重新排序。如果没有，则没有冲突，如果有，您可以查看有多少约会（如果您愿意，还有哪些约会）相互冲突。

# Using Roland's example:
DF <- read.table(text="                Start,                 End,  Location,  Room
1,2012-01-02 08:30:00,2012-01-02 08:40:00,LocationA,RoomA
2,2012-01-02 08:40:00,2012-01-02 08:50:00,LocationA,RoomA
3,2012-01-02 08:50:00,2012-01-02 09:55:00,LocationA,RoomA
4,2012-01-02 09:00:00,2012-01-02 09:10:00,LocationA,RoomA
5,2012-01-02 09:00:00,2012-01-02 09:10:00,LocationA,RoomB
6,2012-01-02 09:10:00,2012-01-02 09:20:00,LocationA,RoomB",header=TRUE,sep=",",stringsAsFactors=FALSE)

dt = data.table(DF)

# the conflicting appointments
dt[order(Start),
   .SD[unique((which(order(c(rbind(Start, End))) != 1:(2*.N)) - 1) %/% 2 + 1)],
   by = list(Location, Room)]
#    Location  Room               Start                 End
#1: LocationA RoomA 2012-01-02 08:50:00 2012-01-02 09:55:00
#2: LocationA RoomA 2012-01-02 09:00:00 2012-01-02 09:10:00

# and a speedier version of the above, that avoids constructing the full .SD:
dt[dt[order(Start),
      .I[unique((which(order(c(rbind(Start, End))) != 1:(2*.N)) - 1) %/% 2 + 1)],
      by = list(Location, Room)]$V1]

也许上面从不匹配顺序到正确索引的公式可以简化，我没有花太多时间思考它，只是使用了完成工作的第一件事。

【讨论】：