在我的一个应用程序中,有一段代码可以data.table根据另一个对象的值从对象中检索信息.
# say this table contains customers details
dt <- data.table(id=LETTERS[1:4],
start=seq(as.Date("2010-01-01"), as.Date("2010-04-01"), "month"),
end=seq(as.Date("2010-01-01"), as.Date("2010-04-01"), "month") + c(6,8,10,5),
key="id")
# this one has some historical details
dt1 <- data.table(id=rep(LETTERS[1:4], each=120),
date=seq(as.Date("2010-01-01"), as.Date("2010-04-30"), "day"),
var=rnorm(120),
key="id,date")
# and here I finally retrieve my historical information based one customer detail
#
library(data.table)
myfunc <- function(x) {
# some code
period <- seq(x$start, x$end, "day")
dt1[.(x$id, period)][, mean(var)]
# some code
}
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得到我用的所有结果 adply
library(plyr)
library(microbenchmark)
> adply(dt, 1, myfunc)
id start end V1
1: A 2010-01-01 2010-01-07 0.3143536
2: B 2010-02-01 2010-02-09 -0.5796084
3: C 2010-03-01 2010-03-11 0.1171404
4: D 2010-04-01 2010-04-06 0.2384237
> microbenchmark(adply(dt, 1, myfunc))
Unit: milliseconds
expr min lq median uq max neval
adply(dt, 1, myfunc) 8.812486 8.998338 9.105776 9.223637 88.14057 100
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您是否知道一种避免adply通话的方法并在一个data.table声明中执行上述操作?或者无论如何更快的方法?(标题编辑建议超过欢迎,我想不到更好的,谢谢)
这是使用以下roll参数的好地方data.table:
setkey(dt1, id, date)
setkey(dt, id, start)
dt[dt1, roll = TRUE][end >= start,
list(start = start[1], end = end[1], result = mean(var)), by = id]
# benchmark
microbenchmark(OP = adply(dt, 1, myfunc),
Frank = dt[dt1[as.list(dt[,seq.Date(start,end,"day"),by="id"])][,mean(var),by=id]],
eddi = dt[dt1, roll = TRUE][end >= start,list(start = start[1], end = end[1], result = mean(var)), by = id])
#Unit: milliseconds
# expr min lq median uq max neval
# OP 24.436126 29.184786 30.853094 32.493521 50.898664 100
# Frank 9.115676 11.303691 12.081000 13.122753 28.370415 100
# eddi 5.336315 6.323643 6.771898 7.497285 9.531376 100
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随着数据集大小的增长,时差将变得更加显着.