Question

我正在处理名为 orderFlow 的data.table并计算 potentialWelfare.tmp 作为输出。到目前为止，以下基于plyr的方法一直是我的解决方案，但由于输入 orderFlow 具有数百万行，我更倾向于利用R中的data.table性能的解决方案。

    # solution so far, poor performance on huge orderFlow input data.table
    require(plyr)
    potentialWelfare.tmp = ddply(orderFlow, 
                       .variables = c("simulationrun_id", "db"), 
                       .fun = calcPotentialWelfare, 
                       .progress = "text", 
                       .parallel=TRUE)

Edit1：简而言之，自定义功能会检查df中是否有更多出价或要求，并对 NbAsks 排序（按估值）出价的估值求和。这样做是为了选择最有价值的出价并总结其估值。代码是遗留的，可能效率不高，但它与plyr和普通的data.frames结合使用。

    calcPotentialWelfare <- function(df){
       NbAsks = dim(df[df$type=="ask",])[1]
    #   print(NbAsks)
      Bids = df[df$type == "bid",]
    #         dd[with(dd, order(-z, b)), ]
      Bids = Bids[with(Bids,order(valuation,decreasing = TRUE)),]
      NbBids = dim(df[df$type == "bid",])[1]
    #   print(Bids)
      if (NbAsks > 0){
        Bids = Bids[1:min(NbAsks,NbBids),]
        potentialWelfare = sum(Bids$valuation)
        return(potentialWelfare)
      }
      else{
        potentialWelfare = 0
        return(potentialWelfare)
      }
    }

不幸的是，我找不到使用data.table实现此方法的工作方式。到目前为止我使用？data.table和相应的常见问题解答得到的是：

    #   trying to use data.table, but it doesn't work so far.
    potentialWelfare.tmp = orderFlow[, lapply(.SD, calcPotentialWelfare), by = list(simulationrun_id, db),.SDcols=c("simulationrun_id", "db")]

我得到的是

    Error in `[.data.frame`(orderFlow, , lapply(.SD, calcPotentialWelfare),  : unused arguments (by = list(simulationrun_id, db), .SDcols = c("simulationrun_id", "db"))

这是输入：

    > head(orderFlow)
      type  valuation price               dateCreation                    dateDue                dateMatched id
    1  ask 0.30000000   0.3 2012-01-01 00:00:00.000000 2012-01-01 00:30:00.000000 2012-01-01 00:01:01.098307  1
    2  bid 0.39687633   0.0 2012-01-01 00:01:01.098307 2012-01-01 00:10:40.024807 2012-01-01 00:01:01.098307  2
    3  bid 0.96803384    NA 2012-01-01 00:03:05.660811 2012-01-01 00:06:26.368941                       <NA>  3
    4  bid 0.06163186    NA 2012-01-01 00:05:25.413959 2012-01-01 00:09:06.189893                       <NA>  4
    5  bid 0.57017143    NA 2012-01-01 00:10:10.344876 2012-01-01 00:57:58.998516                       <NA>  5
    6  bid 0.37188442    NA 2012-01-01 00:11:25.761372 2012-01-01 00:43:24.274176                       <NA>  6
              created_at updated_at simulationrun_id db
    1 2013-12-10 14:37:29.065634         NA             7004  1
    2 2013-12-10 14:37:29.065674         NA             7004  1
    3 2013-12-10 14:37:29.065701         NA             7004  1
    4 2013-12-10 14:37:29.065726         NA             7004  1
    5 2013-12-10 14:37:29.065750         NA             7004  1
    6 2013-12-10 14:37:29.065775         NA             7004  1

我期待像这样的输出，即函数calcPotentialWelfare以某种特殊的方式从data.table orderFlow的列'valu'聚合数据。

    > head(potentialWelfare.tmp)
      simulationrun_id db potentialWelfare
    1                1  1         16.86684
    2                2  1         18.44314
    3                4  1         16.86684
    4                5  1         18.44314
    5                7  1         16.86684
    6                8  1         18.44314

看到这个问题得到解决真的很兴奋。谢谢你的阅读！

EDIT2：

    > dput(head(orderFlow))
    structure(list(type = c("ask", "bid", "bid", "bid", "bid", "bid"
    ), valuation = c(0.3, 0.39687632952068, 0.968033835246625, 0.0616318564942726, 
    0.570171430446081, 0.371884415116724), price = c(0.3, 0, NA, 
    NA, NA, NA), dateCreation = c("2012-01-01 00:00:00.000000", "2012-01-01 00:01:01.098307", 
    "2012-01-01 00:03:05.660811", "2012-01-01 00:05:25.413959", "2012-01-01 00:10:10.344876", 
    "2012-01-01 00:11:25.761372"), dateDue = c("2012-01-01 00:30:00.000000", 
    "2012-01-01 00:10:40.024807", "2012-01-01 00:06:26.368941", "2012-01-01 00:09:06.189893", 
    "2012-01-01 00:57:58.998516", "2012-01-01 00:43:24.274176"), 
        dateMatched = c("2012-01-01 00:01:01.098307", "2012-01-01 00:01:01.098307", 
        NA, NA, NA, NA), id = 1:6, created_at = c("2013-12-10 14:37:29.065634", 
        "2013-12-10 14:37:29.065674", "2013-12-10 14:37:29.065701", 
        "2013-12-10 14:37:29.065726", "2013-12-10 14:37:29.065750", 
        "2013-12-10 14:37:29.065775"), updated_at = c(NA_real_, NA_real_, 
        NA_real_, NA_real_, NA_real_, NA_real_), simulationrun_id = c(7004L, 
        7004L, 7004L, 7004L, 7004L, 7004L), db = c(1L, 1L, 1L, 1L, 
        1L, 1L)), .Names = c("type", "valuation", "price", "dateCreation", 
    "dateDue", "dateMatched", "id", "created_at", "updated_at", "simulationrun_id", 
    "db"), row.names = c(NA, 6L), class = "data.frame")

Answer 1

我认为这应该更快。您使用data.table的方式存在一些错误。我建议你仔细阅读介绍，通过实例，阅读常见问题。

calcPotentialWelfare <- function(dt){
  NbAsks = nrow(dt["ask", nomatch=0L]) # binary search based subset/join - very fast
  Bids   = dt["bid", nomatch=0L] # binary search based subset/join - very fast
  NbBids = nrow(Bids)
  # for each 'type', the 'valuation' will always be sorted, 
  # but in ascending order - but you need descending order
  # so you can just use the function 'tail' to fetch the last 'n' items... as follows.
  if (NbAsks > 0) return(sum(tail(Bids, min(NbAsks, NbBids))$valuation))
  else return(0)
}

# setkey on 'type' column to use binary search based subset/join in the function
# also on valuation so that we don't have to 'order' for every group 
# inside the function - we can use 'tail'
setkey(orderFlow, type, valuation) 
potentialWelfare.tmp =
  orderFlow[, calcPotentialWelfare(.SD), 
            by=.(simulationrun_id, db),
            .SDcols=c("type", "valuation")]

.SD是一个特殊变量，它为每个分组创建一个data.table，其中包含by=中未提及的所有列（如果未指定.SDcols）。如果指定了.SDcols，那么为每个groupw创建.SD，只指定那些列，并且数据对应于该组。

使用lapply(.SD, ...)为每个列提供了该功能，这不是您需要的。您需要将整个数据发送到该函数。但是，由于您只需要函数中的“类型”和“评估”列，因此可以通过提供.SDcols=c('type', 'valuation')来加快速度。通过忽略其他列，这将节省大量时间。

在data.table上应用自定义函数而不是使用plyr和ddply

1 个答案: