问题描述
我正在 R 中做经典的拆分-应用-重组.随着时间的推移，我的数据集是一堆公司.我正在做的应用是对每个公司进行回归并返回残差，因此，我没有按公司聚合.plyr 对此非常有用，但是当公司数量很大时，它需要很长时间才能运行.有没有办法用 data.table 做到这一点?
I am doing the classic split-apply-recombine thing in R. My data set is a bunch of firms over time. The applying I am doing is running a regression for each firm and returning the residuals, therefore, I am not aggregating by firm. plyr is great for this but it takes a very very long time to run when the number of firms is large. Is there a way to do this with data.table?
样本数据:
dte, id, val1, val2
2001-10-02, 1, 10, 25
2001-10-03, 1, 11, 24
2001-10-04, 1, 12, 23
2001-10-02, 2, 13, 22
2001-10-03, 2, 14, 21
我需要按每个 id(即 1 和 2)进行拆分.运行回归，返回残差并将其作为一列附加到我的数据中.有没有办法使用 data.table 来做到这一点?
I need to split by each id (namely 1 and 2). Run a regression, return the residuals and append it as a column to my data. Is there a way to do this using data.table?
推荐答案
我猜这需要按id"排序才能正确排列.幸运的是，当您设置密钥时，这会自动发生:
I'm guessing this needs to be sorted by "id" to line up properly. Luckily that happens automatically when you set the key:
dat <-read.table(text="dte, id, val1, val2
2001-10-02, 1, 10, 25
2001-10-03, 1, 11, 24
2001-10-04, 1, 12, 23
2001-10-02, 2, 13, 22
2001-10-03, 2, 14, 21
", header=TRUE, sep=",")
dtb <- data.table(dat)
setkey(dtb, "id")
dtb[, residuals(lm(val1 ~ val2)), by="id"]
#---------------
cbind(dtb, dtb[, residuals(lm(val1 ~ val2)), by="id"])
#---------------
dte id val1 val2 id.1 V1
[1,] 2001-10-02 1 10 25 1 1.631688e-15
[2,] 2001-10-03 1 11 24 1 -3.263376e-15
[3,] 2001-10-04 1 12 23 1 1.631688e-15
[4,] 2001-10-02 2 13 22 2 0.000000e+00
[5,] 2001-10-03 2 14 21 2 0.000000e+00
> dat <- data.frame(dte=Sys.Date()+1:1000000,
id=sample(1:2, 1000000, repl=TRUE),
val1=runif(1000000), val2=runif(1000000) )
> dtb <- data.table(dat)
> setkey(dtb, "id")
> system.time( cbind(dtb, dtb[, residuals(lm(val1 ~ val2)), by="id"]) )
user system elapsed
1.696 0.798 2.466
> system.time( dtb[,transform(.SD,r = residuals(lm(val1~val2))),by = "id"] )
user system elapsed
1.757 0.908 2.690
来自马修的编辑:这对于 CRAN 上的 v1.8.0 来说都是正确的.j 中的 transform 是 data.table wiki 点 2:为了速度，不要按组transform()，cbind()> 之后".但是，:= 现在在 v1.8.1 中按组工作，既简单又快速.请参阅我的回答以进行说明(但无需投票).
EDIT from Matthew :This is all correct for v1.8.0 on CRAN. With the small addition that transform in j is the subject of data.table wiki point 2: "For speed don't transform() by group, cbind() afterwards". But, := now works by group in v1.8.1 and is both simple and fast. See my answer for illustration (but no need to vote for it).
好吧，我投了赞成票.这是在 Mac 上安装 v 1.8.1 的控制台命令(如果您有合适的 XCode 工具可用，因为它只在源代码中):
Well, I voted for it. Here is the console command to install v 1.8.1on a Mac (if you have the proper XCode tools avaialble, since it only there in source):
install.packages("data.table", repos= "http://R-Forge.R-project.org", type="source",
lib="/Library/Frameworks/R.framework/Versions/2.14/Resources/lib")
(由于某种原因，我无法让 Mac GUI 软件包安装程序将 r-forge 读取为存储库.)
(For some reason I could not get the Mac GUI Package Installer to read r-forge as a repository.)
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