data.table - 选择组中的前n行

Question

虽然很简单,但我不知道data.table在数据表中选择组中前n行的解决方案.你能帮帮我吗？

Answer 1

作为备选:

dt[, .SD[1:3], cyl]

当您查看示例数据集的速度时,该head方法与.I@eddi的方法相同.与microbenchmark包装相比:

microbenchmark(head = dt[, head(.SD, 3), cyl],
               SD = dt[, .SD[1:3], cyl], 
               I = dt[dt[, .I[1:3], cyl]$V1],
               times = 10, unit = "relative")

结果是:

Unit: relative
 expr      min       lq     mean   median       uq       max neval cld
 head 1.000000 1.000000 1.000000 1.000000 1.000000 1.0000000    10  a 
   SD 2.156562 2.319538 2.306065 2.365190 2.318540 2.1908401    10   b
    I 1.001810 1.029511 1.007371 1.018514 1.016583 0.9442973    10  a 

但是,data.table专门针对大型数据集而设计.所以,再次运行这个比较:

# creating a 30 million dataset
largeDT <- dt[,.SD[sample(.N, 1e7, replace = TRUE)], cyl]
# running the benchmark on the large dataset
microbenchmark(head = largeDT[, head(.SD, 3), cyl],
               SD = largeDT[, .SD[1:3], cyl], 
               I = largeDT[largeDT[, .I[1:3], cyl]$V1],
               times = 10, unit = "relative")

结果是:

Unit: relative
 expr      min       lq     mean   median       uq     max neval cld
 head 2.279753 2.194702 2.221330 2.177774 2.276986 2.33876    10   b
   SD 2.060959 2.187486 2.312009 2.236548 2.568240 2.55462    10   b
    I 1.000000 1.000000 1.000000 1.000000 1.000000 1.00000    10  a 

现在这个.I方法显然是最快的方法.

更新2016-02-12:

使用data.table包的最新开发版本,该.I方法仍然获胜.无论是.SD方法或head()一种方法更快,似乎取决于数据集的大小.现在基准测试给出:

Unit: relative
 expr      min       lq     mean   median       uq      max neval cld
 head 2.093240 3.166974 3.473216 3.771612 4.136458 3.052213    10   b
   SD 1.840916 1.939864 2.658159 2.786055 3.112038 3.411113    10   b
    I 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000    10  a 

但是,如果数据集稍微小一些(但仍然很大),则可能会发生变化:

largeDT2 <- dt[,.SD[sample(.N, 1e6, replace = TRUE)], cyl]

基准测试现在略微支持该head方法的.SD方法:

Unit: relative
 expr      min       lq     mean   median       uq      max neval cld
 head 1.808732 1.917790 2.087754 1.902117 2.340030 2.441812    10   b
   SD 1.923151 1.937828 2.150168 2.040428 2.413649 2.436297    10   b
    I 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000    10  a 

Answer 2

我们可以使用head与.SD

library(data.table)

dt <- data.table(mtcars)

> dt[, head(.SD, 3), by = "cyl"]

   cyl  mpg  disp  hp drat    wt  qsec vs am gear carb
1:   6 21.0 160.0 110 3.90 2.620 16.46  0  1    4    4
2:   6 21.0 160.0 110 3.90 2.875 17.02  0  1    4    4
3:   6 21.4 258.0 110 3.08 3.215 19.44  1  0    3    1
4:   4 22.8 108.0  93 3.85 2.320 18.61  1  1    4    1
5:   4 24.4 146.7  62 3.69 3.190 20.00  1  0    4    2
6:   4 22.8 140.8  95 3.92 3.150 22.90  1  0    4    2
7:   8 18.7 360.0 175 3.15 3.440 17.02  0  0    3    2
8:   8 14.3 360.0 245 3.21 3.570 15.84  0  0    3    4
9:   8 16.4 275.8 180 3.07 4.070 17.40  0  0    3    3