Question

我有一个花车向量。我想在各种范围内反复找到该向量的子集。我当前的语法（DT[x > 1.8 & x < 2.9]）似乎是矢量扫描（它相对较慢）。

是否有更快的语法利用二进制搜索基于范围/区间的data.tables子设置？

示例：

set.seed(123L)
x = runif(1E6)
DT = data.table(x, key = "x")

# For foverlaps()
DT[,xtemp:=x]
range = data.table(start = 0.04, end = 0.5, key=c("start", "end"))


microbenchmark::microbenchmark(
    DT[x < 0.5 & x > 0.04], 
    x[x < 0.5 & x > 0.04],
    foverlaps(DT, range, by.x = c("x", "xtemp"))
    )

Unit: milliseconds
                                         expr       min        lq      mean    median        uq      max neval
                       DT[x < 0.5 & x > 0.04]  12.65391  16.10852  18.43412  17.23268  17.76868 104.1112   100
                        x[x < 0.5 & x > 0.04]  16.48126  19.63882  21.65813  20.31534  20.95264 113.7965   100
 foverlaps(DT, range, by.x = c("x", "xtemp")) 116.72732 131.93093 145.56821 140.09218 146.33287 226.6069   100

Answer 1

基于the answer here，这似乎是某种改进。但是，此方案中将包含等于0.5的值：

bs <- function() DT[{ind <- DT[.(c(0.04, 0.5)), which=TRUE, roll=TRUE]; (ind[1]+1):ind[2]}]
vs <- function() x[x < 0.5 & x > 0.04]

x = runif(1E6)
DT = data.table(x, key = "x")

microbenchmark::microbenchmark(
    bs(), 
    vs()
)

#Unit: milliseconds
# expr       min        lq      mean   median        uq        max neval
# bs()  3.594993  4.150932  5.002947  4.44695  4.952283   9.750284   100
# vs() 15.054460 16.217198 18.999877 17.45298 19.554958 113.623699   100

如果我们将vs()修改为：

vs <- function() x[x <= 0.5 & x > 0.04]

两种方法的结果相同：

identical(bs()$x, sort(vs()))
# [1] TRUE

Answer 2

最新版本的data.table添加了封装此行为的%between%和%inrange%运算符。 Psidom的基于滚动的解决方案似乎稍微慢了一些但是按预期处理所有类型（数字/整数）并且更加简洁。见下文。

# data.table version 1.10.4
# R version 3.3.1 (2016-06-21)

set.seed(123L)
library(data.table)
x = runif(1E6)
DT = data.table(x)

#Psidom Answer
psidom <- function() DT[{ind <- DT[.(c(0.04, 0.5)), which=TRUE, roll=TRUE, on=.(x)]; (ind[1]+1):ind[2]}]

# Unkeyed
microbenchmark::microbenchmark(
  DT[x <= 0.5 & x >= 0.04], 
  x[x <= 0.5 & x >= 0.04],
  DT[x %between% c(0.04, 0.5)],
  DT[x %inrange% c(0.04, 0.5)],
  DT[.(0.04, 0.5), on = .(x >= V1, x <= V2), .(x.x)]
)

# Unit: milliseconds
#                                                expr        min         lq      mean    median        uq      max neval  cld
#                            DT[x <= 0.5 & x >= 0.04]  20.346712  23.983928  34.69493  25.21085  26.73657 281.4747   100  b  
#                             x[x <= 0.5 & x >= 0.04]  19.581049  22.935144  31.61551  23.83557  25.99587 145.3632   100  b  
#                        DT[x %between% c(0.04, 0.5)]   8.024091   9.293261  12.19035  11.38171  12.75843 116.5132   100 a   
#                        DT[x %inrange% c(0.04, 0.5)]  77.108485  79.871207  91.05544  81.83722  84.66684 188.8674   100   c 
#  DT[.(0.04, 0.5), on = .(x >= V1, x <= V2), .(x.x)] 189.488658 195.487681 217.55708 198.52696 205.80428 318.1696   100    d

# Keyed
setkey(DT,x)

#Psidom Answer
psidom <- function() DT[{ind <- DT[.(c(0.04, 0.5)), which=TRUE, roll=TRUE, on=.(x)]; (ind[1]+1):ind[2]}]

microbenchmark::microbenchmark(
  DT[x <= 0.5 & x >= 0.04], 
  x[x <= 0.5 & x >= 0.04],
  DT[x %between% c(0.04, 0.5)],
  DT[x %inrange% c(0.04, 0.5)],
  DT[.(0.04, 0.5), on = .(x >= V1, x <= V2), .(x.x)],
  psidom()
)

# Unit: milliseconds
#                                                expr       min        lq      mean    median        uq      max neval cld
#                            DT[x <= 0.5 & x >= 0.04] 14.550788 18.092458 21.012992 18.934781 20.055428 123.1174   100  b 
#                             x[x <= 0.5 & x >= 0.04] 19.403718 22.401709 27.296872 23.707688 24.608270 128.9123   100  b 
#                        DT[x %between% c(0.04, 0.5)]  5.439340  6.819262 10.789330  9.490118 10.561789 111.6523   100 a  
#                        DT[x %inrange% c(0.04, 0.5)] 12.871260 13.894918 21.434823 16.888748 18.128147 123.4275   100  b 
#  DT[.(0.04, 0.5), on = .(x >= V1, x <= V2), .(x.x)] 49.277678 53.516350 61.422212 54.499675 55.869354 158.1861   100   c
#                                            psidom()  4.615421  5.095880  9.482131  5.325707  8.316817 109.9318   100 a

data.table是否基于二进制搜索实现快速范围子集？那是什么语法？

2 个答案: