为什么在`j.`中求值比`data.table`中的`$`要快？ [英] Why is it faster to evaluate in `j` than with `$` in a `data.table`?
问题描述
也许这已经回答了，我错过了，但很难搜索。
一个很简单的问题：为什么 dt [ x] 一般比 dt $ x 吗？
/ p>
dt <-data.table（id = 1：1e7，var = rnorm（1e6））
$ b b test< -microbenchmark（times = 100L，
dt [sample（1e7，size = 200000），var]，
dt [sample（1e7，size = 200000），] $ var）$ b b
test [，expr]< -c（in j，$）
单位：毫秒
expr min lq平均中位数uq max neval
$ 14.28863 15.88779 18.84229 17.23109 18.41577 53.63473 100
in j 14.35916 15.97063 18.87265 17.99266 18.37939 54.19944 100
无论如何，在 j
microbenchmark 可能会吐出一些直方图）。 为什么是这种情况？
使用 j <
$ b
（和你的调用），你是 [。data.table 中的子集，然后选择 $
你实际上是调用2个函数而不是1，因此时间上有一个可以忽略的差异。
比较返回相同结果
dt <-data.table（id = 1：1e7，var = rnorm（1e6））
setkey（dt，id）
ii < = 200000）
microbenchmark（in j= dt [。（ii），var]，$= dt [。（ii）] $ var，'[[' = dt [。（ii）] [['var']]，.subset2（dt [。（ii）]，'var'），dt [。（ii）] [[2]]，dt [['var ']] [ii]，dt $ var [ii]，.subset2（dt，'var'）[ii]）
单位：毫秒
expr min lq平均中位数uq max neval cld
in j 39.491156 40.358669 41.570057 40.860342 41.485622 70.202441 100 b
$ 39.957211 40.561965 41.587420 41.136836 41.634584 69.928363 100 b
[[40.046558 40.515480 42.388432 41.244444 41.750946 72.224827 100 b
.subset2（dt [。 ）]，var）39.772781 40.564077 41.561271 41.111630 41.635489 69.252222 100 b
dt [。（ii）] [[2]] 40.004300 40.513669 41.682526 40.927503 41.492866 72.986995 100 b
dt [[var] ] [ii] 4.432346 4.546898 4.946219 4.623416 4.755777 31.761115 100 a
dt $ var [ii] 4.440496 4.539502 4.668361 4.597457 4.729214 5.425125 100 a
.subset2（dt，var）[ii] 4.365939 4.508261 4.660435 4.598815 4.703858 6.072289 100 a
Perhaps this is already answered and I missed it, but it's hard to search.
A very simple question: Why is dt[,x] generally a tiny bit faster than dt$x?
Example:
dt<-data.table(id=1:1e7,var=rnorm(1e6))
test<-microbenchmark(times=100L,
dt[sample(1e7,size=200000),var],
dt[sample(1e7,size=200000),]$var)
test[,"expr"]<-c("in j","$")
Unit: milliseconds
expr min lq mean median uq max neval
$ 14.28863 15.88779 18.84229 17.23109 18.41577 53.63473 100
in j 14.35916 15.97063 18.87265 17.99266 18.37939 54.19944 100
I might not have chosen the best example, so feel free to suggest something perhaps more poignant.
Anyway, evaluating in j is faster at least 75% of the time (though there appears to be a fat upper tail as the mean is higher; side note, it would be nice if microbenchmark could spit me out some histograms).
Why is this the case?
解决方案 With j, you are subsetting and selecting within a call to [.data.table.
With $ (and your call), you are subsetting within [.data.table and then selecting with $
You are in essence calling 2 functions not 1, thus there is a neglible difference in timing.
In your current example you are calling `sampling(1e,200000) each time.
For comparison to return identical results
dt<-data.table(id=1:1e7,var=rnorm(1e6))
setkey(dt, id)
ii <- sample(1e7,size=200000)
microbenchmark("in j" = dt[.(ii),var], "$"=dt[.(ii)]$var, '[[' =dt[.(ii)][['var']], .subset2(dt[.(ii)],'var'), dt[.(ii)][[2]], dt[['var']][ii], dt$var[ii], .subset2(dt,'var')[ii] )
Unit: milliseconds
expr min lq mean median uq max neval cld
in j 39.491156 40.358669 41.570057 40.860342 41.485622 70.202441 100 b
$ 39.957211 40.561965 41.587420 41.136836 41.634584 69.928363 100 b
[[ 40.046558 40.515480 42.388432 41.244444 41.750946 72.224827 100 b
.subset2(dt[.(ii)], "var") 39.772781 40.564077 41.561271 41.111630 41.635489 69.252222 100 b
dt[.(ii)][[2]] 40.004300 40.513669 41.682526 40.927503 41.492866 72.986995 100 b
dt[["var"]][ii] 4.432346 4.546898 4.946219 4.623416 4.755777 31.761115 100 a
dt$var[ii] 4.440496 4.539502 4.668361 4.597457 4.729214 5.425125 100 a
.subset2(dt, "var")[ii] 4.365939 4.508261 4.660435 4.598815 4.703858 6.072289 100 a
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