r - 将函数应用于data.table的每一行

Question

我希望用它data.table来提高给定函数的速度,但我不确定我是以正确的方式实现它:

数据

鉴于两个data.tables(dt和dt_lookup)

library(data.table)
set.seed(1234)
t <- seq(1,100); l <- letters; la <- letters[1:13]; lb <- letters[14:26]
n <- 10000
dt <- data.table(id=seq(1:n), 
                 thisTime=sample(t, n, replace=TRUE), 
                 thisLocation=sample(la,n,replace=TRUE),
                 finalLocation=sample(lb,n,replace=TRUE))
setkey(dt, thisLocation)

set.seed(4321)
dt_lookup <- data.table(lkpId = paste0("l-",seq(1,1000)),
                        lkpTime=sample(t, 10000, replace=TRUE),
                        lkpLocation=sample(l, 10000, replace=TRUE))
## NOTE: lkpId is purposly recycled
setkey(dt_lookup, lkpLocation)

我有找到的函数lkpId同时包含thisLocation和finalLocation,并具有"最近" lkpTime(即最小的非负的值thisTime - lkpTime)

功能

## function to get the 'next' lkpId (i.e. the lkpId with both thisLocation and finalLocation,
## with the minimum non-negative time between thisTime and dt_lookup$lkpTime)
getId <- function(thisTime, thisLocation, finalLocation){

  ## filter lookup based on thisLocation and finalLocation,
  ## and only return values where the lkpId has both 'this' and 'final' locations
  tempThis <- unique(dt_lookup[lkpLocation == thisLocation,lkpId])
  tempFinal <- unique(dt_lookup[lkpLocation == finalLocation,lkpId])
  availServices <- tempThis[tempThis %in% tempFinal]

  tempThisFinal <- dt_lookup[lkpId %in% availServices & lkpLocation==thisLocation, .(lkpId, lkpTime)]

  ## calcualte time difference between 'thisTime' and 'lkpTime' (from thisLocation)
  temp2 <- thisTime - tempThisFinal$lkpTime

  ## take the lkpId with the minimum non-negative difference
  selectedId <- tempThisFinal[min(which(temp2==min(temp2[temp2>0]))),lkpId]
  selectedId
}

尝试解决方案

我需要得到lkpId每一行dt.因此,我最初的本能是使用一个*apply函数,但是时间太长(对我来说)n/nrow > 1,000,000.所以我试图实现一个data.table解决方案,看看它是否更快:

selectedId <- dt[,.(lkpId = getId(thisTime, thisLocation, finalLocation)),by=id]

但是,我相当新data.table,并且这种方法似乎没有提供任何性能提升*apply解决方案:

lkpIds <- apply(dt, 1, function(x){
  thisLocation <- as.character(x[["thisLocation"]])
  finalLocation <- as.character(x[["finalLocation"]])
  thisTime <- as.numeric(x[["thisTime"]])
  myId <- getId(thisTime, thisLocation, finalLocation)
})

对于n = 10,000,两者都需要约30秒.

题

有没有更好的方法data.table用于getId在每一行上应用函数dt？

2015年8月12日更新

感谢来自@eddi的指针,我重新设计了我的整个算法,并且正在使用滚动连接(一个很好的介绍),从而正确使用data.table.我稍后会写一个答案.

Answer 1

自从提出这个问题以来，我花了时间研究必须提供什么data.table，data.table通过@eddi的指针研究连接（例如data.table 上的滚动连接和不等式的内部连接），我想出了一个解决方案。

其中一个棘手的部分是摆脱“对每一行应用函数”的想法，并重新设计解决方案以使用联接。

而且，毫无疑问会有更好的编程方法，但这是我的尝试。

## want to find a lkpId for each id, that has the minimum difference between 'thisTime' and 'lkpTime'
## and where the lkpId contains both 'thisLocation' and 'finalLocation'

## find all lookup id's where 'thisLocation' matches 'lookupLocation'
## and where thisTime - lkpTime > 0
setkey(dt, thisLocation)
setkey(dt_lookup, lkpLocation)

dt_this <- dt[dt_lookup, {
  idx = thisTime - i.lkpTime > 0
  .(id = id[idx],
    lkpId = i.lkpId,
    thisTime = thisTime[idx],
    lkpTime = i.lkpTime)
},
by=.EACHI]

## remove NAs
dt_this <- dt_this[complete.cases(dt_this)]

## find all matching 'finalLocation' and 'lookupLocaiton'
setkey(dt, finalLocation)
## inner join (and only return the id columns)
dt_final <- dt[dt_lookup, nomatch=0, allow.cartesian=TRUE][,.(id, lkpId)]

## join dt_this to dt_final (as lkpId must have both 'thisLocation' and 'finalLocation')
setkey(dt_this, id, lkpId)
setkey(dt_final, id, lkpId)

dt_join <- dt_this[dt_final, nomatch=0]

## take the combination with the minimum difference between 'thisTime' and 'lkpTime'
dt_join[,timeDiff := thisTime - lkpTime]

dt_join <- dt_join[ dt_join[order(timeDiff), .I[1], by=id]$V1]  

## equivalent dplyr code
# library(dplyr)
# dt_this <- dt_this %>%
#   group_by(id) %>%
#   arrange(timeDiff) %>%
#   slice(1) %>%
#   ungroup