R 為什么 dplyr 按組計算唯一值 (n_distinct) 的速度比 data.table (uniqueN) 快？

Question

據我了解，data.table 比 dplyr 更高效、更快，但我今天在工作中發現了相反的情況。 我創建了一個模擬來解釋這種情況。

library(data.table)
library(dplyr)

library(microbenchmark)

#  data simulated
dt = data.table(A = sample(1:4247,10000, replace = T),
                B = sample(1:119,10000,replace = T),
                C = sample(1:6,10000,replace = T),
                D = sample(1:30,10000,replace = T))

dt[,ID:=paste(A, ":::" , 
              D,":::",
              C)]
# execution time

microbenchmark(
  DATA_TABLE = dt[, .(count=uniqueN(ID)), 
                  by=c("A","B","C")
                  ],
  DPLYR      = dt %>% 
               group_by(A,B,C)  %>% 
               summarise(count = n_distinct(ID)),
  times = 10
              )

結果

Unit: milliseconds
       expr         min          lq        mean      median    uq         max        neval
 DATA_TABLE 14241.57361 14305.67026 15585.80472 14651.16402  16244.22477 21367.56866  10
      DPLYR    35.95123    37.63894    47.62637    48.56598  53.59919    62.63978     10 

你可以看到很大的不同！ 有人知道原因嗎？ 您對何時使用 dplyr 或 data.table 有什么建議嗎？

我現在有 data.table 語法中的完整代碼，我不知道由於這種情況是否需要將一些代碼塊轉換為 dplyr。

提前致謝。

Answer 1

這是另一種選擇：

dt[order(A, B, C), {
        uniqn <- rleidv(c(.SD, .(ID)))
        lastidx <- c(which(diff(rowidv(.SD))<1L), .N)
        c(.SD[lastidx], .(count=c(uniqn[lastidx[1L]], diff(uniqn[lastidx]))))
    }, .SDcols=cols]

計時碼：

cols <- c("A","B","C")
microbenchmark(times=1L,

  DATA_TABLE = a00 <- dt[, .(count=uniqueN(ID)), cols],

  DATA_TABLE1 = a01 <- dt[, .(count=length(unique(ID))), cols],

  DPLYR      = a_dplyr <- dt %>%
    group_by(A,B,C)  %>%
    summarise(count = n_distinct(ID)),

  #https://github.com/Rdatatable/data.table/issues/1120#issuecomment-463584656
  mtd0 = a10 <- unique(dt, by=c(cols, "ID"))[, .(count=.N), cols],

  #https://github.com/Rdatatable/data.table/issues/1120#issuecomment-463597107
  mtd1 = a11 <- dt[, .N, c(cols, "ID")][, .(count=.N), cols],

  mtd2 = a2 <- dt[order(A, B, C), {
    uniqn <- rleidv(c(.SD, .(ID)))
    lastidx <- c(which(diff(rowidv(.SD))<1L), .N)
    c(.SD[lastidx], .(count=c(uniqn[lastidx[1L]], diff(uniqn[lastidx]))))
  }, .SDcols=cols]
)

檢查：

> fsetequal(a00, a01)
[1] TRUE

> fsetequal(a00, setDT(a_dplyr))
[1] TRUE

> fsetequal(a00, a10)
[1] TRUE

> fsetequal(a00, a11)
[1] TRUE

> fsetequal(a00, a2)
[1] TRUE

以下特定數據集的時間安排：

Unit: milliseconds
        expr         min          lq        mean      median          uq         max neval
  DATA_TABLE 147478.1089 147478.1089 147478.1089 147478.1089 147478.1089 147478.1089     1
 DATA_TABLE1   4998.8236   4998.8236   4998.8236   4998.8236   4998.8236   4998.8236     1
       DPLYR 244081.6925 244081.6925 244081.6925 244081.6925 244081.6925 244081.6925     1
        mtd0   4519.4046   4519.4046   4519.4046   4519.4046   4519.4046   4519.4046     1
        mtd1   2866.5808   2866.5808   2866.5808   2866.5808   2866.5808   2866.5808     1
        mtd2    809.7442    809.7442    809.7442    809.7442    809.7442    809.7442     1

具有 1mio 行的數據：

#R-3.6.1 64bit Win10
library(data.table)  #data.table_1.12.8 getDTthreads()==4
library(dplyr)  #dplyr_1.0.0
library(microbenchmark)

#  data simulated
set.seed(0L)
nr <- 1e6
dt = data.table(A = sample(1:424700,nr, replace = T),
  B = sample(1:11900,nr, replace = T),
  C = sample(1:600, nr, replace = T),
  D = sample(1:3000, nr, replace = T))
dt[,ID:=paste(A,":::",D,":::",C)]