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我遇到了一段需要很长时间才能执行的代码的挑战，我想知道优化此代码执行时间的关键技巧是什么。我不得不承认输入 data.frame 很重要（140,000 行），而输出 data.frame 大约有 220,000 行。

输入 data.frame 的示例：

head(extremes)
X_BusinessIDDescription     min         max         month
ID105                       2007-12-01  2008-06-01  2007-12-01
ID206                       2007-12-01  2009-07-01  2007-12-01
ID204                       2007-12-01  2008-02-01  2007-12-01
ID785                       2008-07-01  2010-08-01  2008-07-01
ID125                       2007-11-01  2008-07-01  2007-11-01
ID107                       2007-11-01  2011-06-01  2007-11-01

将使用循环扩展的 data.frame。启动 data.frame 以使结构到位。

output <- extremes[1,]
output
X_BusinessIDDescription     min         max         month
ID105                       2007-12-01  2008-06-01  2007-12-01

其他值

IDcounter <- 1
IDmax <- nrow(extremes)
linecounter <- 1

我想优化的while循环：

while (IDcounter <= IDmax){
    start <- extremes$min[IDcounter]
    end <- extremes$max[IDcounter] # add three months
    while(start <= end){
        output[linecounter,] <- extremes[IDcounter,]
        output$month[linecounter] <- start
        linecounter <- linecounter+1
        start <- seq(start, by ="month", length=2)[2]
    }
    IDcounter <- IDcounter + 1
}

对于少数行，此代码执行得非常快，但随着输出的扩展，它似乎正在减慢。

输出看起来像这样：

head(output)
X_BusinessIDDescription     min         max         month
ID105                       2007-12-01  2008-06-01  2007-12-01
ID105                       2007-12-01  2008-06-01  2008-01-01
ID105                       2007-12-01  2008-06-01  2008-02-01
ID105                       2007-12-01  2008-06-01  2008-03-01
ID105                       2007-12-01  2008-06-01  2008-04-01
ID105                       2007-12-01  2008-06-01  2008-05-01

对于极端文件中 min 和 max 之间的间隔中的每个月，都会创建一行。

我也有兴趣了解如何让这段代码为可用的计算资源的多个核心做好准备。好的，我承认这并不是真正的优化，但它会减少执行时间，这也很重要。

约赫姆

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1 回答 1

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正如@CarlWitthoft 已经提到的，由于许多重复数据，您必须重新考虑您的数据结构。

在这里，您可以找到一种简单的矢量化方法：

  ## create all possible ranges of months
  ranges <- mapply(function(mi, ma) {seq(from=mi, to=ma, by="month")}, mi=extremes$min, ma=extremes$max)

  ## how many months per ID?
  n <- unlist(lapply(ranges, length))

  ## create new data.frame
  output <- data.frame(X_BusinessIDDescription=rep(extremes$X_BusinessIDDescription, n),
                      min=rep(extremes$min, n),
                      max=rep(extremes$max, n),
                      month=as.Date(unlist(ranges), origin="1970-01-01"), stringsAsFactors=FALSE)

与您的方法比较：

extremes <- data.frame(X_BusinessIDDescription=c("ID105", "ID206", "ID204", "ID785", "ID125", "ID107"),
                      min=as.Date(c("2007-12-01", "2007-12-01", "2007-12-01", "2008-07-01", "2007-11-01", "2007-11-01")),
                      max=as.Date(c("2008-06-01", "2009-07-01", "2008-02-01", "2010-08-01", "2008-07-01", "2011-06-01")),
                      month=as.Date(c("2007-12-01", "2007-12-01", "2007-12-01", "2008-07-01", "2007-11-01", "2007-11-01")),
                      stringsAsFactors=FALSE)

approachWhile <- function(extremes) {
  output <- data.frame(X_BusinessIDDescription=NA, min=as.Date("1970-01-01"), max=as.Date("1970-01-01"), month=as.Date("1970-01-01"), stringsAsFactors=FALSE)
  IDcounter <- 1
  IDmax <- nrow(extremes)
  linecounter <- 1
  while (IDcounter <= IDmax){
    start <- extremes$min[IDcounter]
    end <- extremes$max[IDcounter] # add three months
    while(start <= end){
        output[linecounter,] <- extremes[IDcounter,]
        output$month[linecounter] <- start
        linecounter <- linecounter+1
        start <- seq(start, by ="month", length=2)[2]
    }
    IDcounter <- IDcounter + 1
  }
  return(output)
}

approachMapply <- function(extremes) {                       
  ranges <- mapply(function(mi, ma) {seq(from=mi, to=ma, by="month")}, mi=extremes$min, ma=extremes$max)

  n <- unlist(lapply(ranges, length))

  output <- data.frame(X_BusinessIDDescription=rep(extremes$X_BusinessIDDescription, n),
                      min=rep(extremes$min, n),
                      max=rep(extremes$max, n),
                      month=as.Date(unlist(ranges), origin="1970-01-01"), stringsAsFactors=FALSE)
  return(output)
}

identical(approachWhile(extremes), approachMapply(extremes)) ## TRUE

library("rbenchmark")

benchmark(approachWhile(extremes), approachMapply(extremes), order="relative")
#                      test replications elapsed relative user.self sys.self
#2 approachMapply(extremes)          100   0.176     1.00     0.172    0.000
#1  approachWhile(extremes)          100   6.102    34.67     6.077    0.008
于 2012-11-20T15:22:34.117 回答