R 如何计算大型数据集每分钟出现的次数

我有一个数据集，有500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")

我想计算每个地点和每个房间（以及原始数据集中的几个其他因素）的并发预约总数

我曾尝试使用

mysql

包执行左连接，该连接适用于小数据集，但对整个数据集来说需要花费很长时间：

# 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

此解决方案速度快，计算1天需要约18秒（全年约2小时）。缺点是我无法为某些因素（地点、房间等）创建并发预约数的子集。我觉得一定有更好的办法。。有什么建议吗

更新： 根据杰弗里的回答，最终的解决方案是这样的。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

setkey（数据、位置、开始、结束）
vecTime这样更好吗

创建空白时间向量和空白计数向量

 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))

vecTime我不太确定，我是否理解你的目标。尽管如此，这可能还是有用的：

#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

#我将示例更改为实际有并发约会
DF这里有一个策略-按开始时间排序，然后按开始、结束、开始、结束……取消列出数据，。。。看看这个向量是否需要重新排序。如果没有，那么就没有冲突，如果有，你可以看到有多少约会（如果你愿意，还有哪些约会）彼此冲突

# 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]

#以罗兰为例：
DF感谢Geoffrey，但这不包括在某一时期内活跃的约会数量。这告诉我有两个约会在9:00开始，但是不考虑活动约会（已经开始但没有结束）。我需要每分钟的占用率曲线图，以便能够研究真正繁忙时期的峰值。Hee Goeffrey，你的解决方案花了9秒来处理我的整个数据集。我已经为此挣扎了好几个小时了。非常感谢你的投入。我一直在往错误的方向看：把所有约会的开始和结束时间加起来，并以此为基础确定入住率，这真是聪明。考虑到计算速度，我可以在每个位置或每一个房间中建立占用地块，其中一些是循环的，所以我考虑我的问题。谢谢罗兰。有趣的方法，但我在寻找每分钟的总占用率，并能够根据位置和房间对占用率进行子集。很高兴能投票选出有用的答案。只是一个指针。
# 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]