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我有一个xts包含多个股票代码时间序列的对象。我需要将xts对象拆分为特定于符号的子组并处理每个符号的数据，然后重新组合xts包含完整行集的原始矩阵中的所有子组。每个符号是一个介于 1 到 4 个字符之间的字段，用作将矩阵拆分为子组的因子索引。

这些是报告在调用时拆分我的矩阵的时间by()，lapply()并且ddply()：

> dim(ets)
[1] 442750     24
> head(ets)
                    Symbol DaySec  ExchTm             LclTm              Open      High      Low       Close     CloseRet       
2011-07-22 09:35:00 "AA"   "34500" "09:34:54.697.094" "09:34:54.697.052" " 158100" " 158400" " 157900" " 158200" " 6.325111e-04"
2011-07-22 09:35:00 "AAPL" "34500" "09:34:59.681.827" "09:34:59.681.797" "3899200" "3899200" "3892200" "3894400" "-1.231022e-03"
2011-07-22 09:35:00 "ABC"  "34500" "09:34:49.805.994" "09:34:49.806.008" " 400100" " 401800" " 400100" " 401600" " 3.749063e-03"
2011-07-22 09:35:00 "ALL"  "34500" "09:34:59.009.001" "09:34:59.008.810" " 285500" " 285500" " 285300" " 285300" "-7.005254e-04"
2011-07-22 09:35:00 "AMAT" "34500" "09:34:59.982.447" "09:34:59.982.423" " 130200" " 130500" " 130200" " 130500" " 2.304147e-03"
2011-07-22 09:35:00 "AMZN" "34500" "09:34:48.012.576" "09:34:48.012.565" "2137400" "2139100" "2137400" "2139100" " 7.953588e-04"
... (15 more columns)
> system.time(by(ets, ets$Symbol, function(x) { return(x) }))
   user  system elapsed 
 78.725   0.932  79.735
> system.time(ddply(as.data.frame(ets), "Symbol", function(x) { return (x) }))
   user  system elapsed 
100.590   0.416 101.105 
> system.time(lapply(split.default(ets, ets$Symbol), function(x) { return(x) }))
   user  system elapsed 
  1.572   0.280   1.853

这篇出色的博客文章中提供了有关使用数据框和矩阵子组的更多信息。

为什么使用 lapply/split.default 时性能会有如此大的差异？

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1 回答 1

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在数字模式下工作大大减少了处理时间：

> system.time(by(myxts[,c(1,2,3,4,5)], myxts$Symbol, summary))
   user  system elapsed 
 57.768   0.688  58.511 
> system.time(by(myxts[,c(1,2,3,4,5,6,7,8)], myxts$Symbol, summary))
   user  system elapsed 
  62.284   0.620  62.971 
> system.time(by(myxts[,c(1,2,3,4,5,6,7,8, 9, 10, 11, 12)], myxts$Symbol, summary))
    user  system elapsed 
 76.529   0.632  77.232 
> myxts.numeric = myxts
> mode(myxts.numeric) = "numeric"
Warning message:
In as.double.xts(c("AA", "AAPL", "ABC", "ALL", "AMAT", "AMZN", "BAC",  :
  NAs introduced by coercion
> system.time(by(myxts.numeric[,c(1,2,3,4,5,6,7,8, 9, 10, 11, 12)], myxts$Symbol, summary))
   user  system elapsed 
  4.948   0.688   5.642
于 2012-01-17T21:11:26.870 回答