假设我有一个包含100万个元素的巨大向量x，我想找到最多30个元素的索引。我不特别在意结果是否在这30个元素中排序，只要它们是整个向量中的最大值30。使用order[x][1:30]似乎非常昂贵，因为它必须对整个向量进行排序。我考虑过使用partial中的sort选项，但sort会返回值，并且在指定index.return时不支持partial选项。有没有一种有效的方法来查找索引而不对整个向量进行排序？
答案 0 :(得分：12)
我想使用sort partial参数和which添加混合方法：
whichpart <- function(x, n=30) {
nx <- length(x)
p <- nx-n
xp <- sort(x, partial=p)[p]
which(x > xp)
}
一些基准测试：
library("microbenchmark")
library("data.table")
library("compiler")
set.seed(123)
x <- rnorm(1e6)
y <- sample.int(1e6)
whichpart <- function(x, n=30) {
nx <- length(x)
p <- nx-n
xp <- sort(x, partial=p)[p]
which(x > xp)
}
cpwhichpart <- cmpfun(whichpart)
# using quicksort
quicksort <- function(x, n=30) {
sort(x, method="quick", decreasing=TRUE, index.return=TRUE)$ix[1:n]
}
cpquicksort <- cmpfun(quicksort)
# @Mariam
whichsort <- function(x, n=30) {
which(x >= sort(x, decreasing=TRUE)[30], arr.ind=TRUE)
}
cpwhichsort <- cmpfun(whichsort)
# @Ferdinand.kraft
top <- function(x, n=30) {
result <- numeric()
for(i in 1:n){
j <- which.max(x)
result[i] <- j
x[j] <- -Inf
}
result
}
cptop <- cmpfun(top)
# @Tony Breyal
dtable <- function(x, n=30) {
dt <- data.table(x=x, x.index=seq.int(x))
setkey(dt, "x")
dt$x.index[1:n]
}
cpdtable <- cmpfun(dtable)
# @Roland
roland <- cmpfun(function(x, n=30) {
y <- rep(-Inf, n)
for (i in seq_along(x)) {
if (x[i] > y[1]) {
y[1] <- x[i]
y <- y[order(y)]
}
}
y
})
## rnorm
microbenchmark(whichpart(x), cpwhichpart(x),
quicksort(x), cpquicksort(x),
whichsort(x), cpwhichsort(x),
top(x), cptop(x),
dtable(x), cpdtable(x),
roland(x), times=10)
# Unit: milliseconds
# expr min lq median uq max neval
# whichpart(x) 45.63544 46.05638 47.09077 49.68452 51.42065 10
# cpwhichpart(x) 45.65996 45.77212 47.02808 48.07482 82.20458 10
# quicksort(x) 100.90936 103.00783 105.17506 109.31784 139.83518 10
# cpquicksort(x) 100.53958 102.78017 107.64470 138.96630 142.52882 10
# whichsort(x) 148.86010 151.04350 155.80871 159.47063 184.56697 10
# cpwhichsort(x) 149.05578 150.21183 151.36918 166.58342 173.87567 10
# top(x) 146.10757 182.42089 184.53050 191.37293 193.62272 10
# cptop(x) 155.14354 179.14847 184.52323 196.80644 220.21222 10
# dtable(x) 1041.32457 1042.54904 1049.26096 1065.40606 1080.89969 10
# cpdtable(x) 1042.08247 1043.54915 1051.76366 1084.14360 1310.26485 10
# roland(x) 251.42885 261.47608 273.20838 295.09733 323.96257 10
## integer
microbenchmark(whichpart(y), cpwhichpart(y),
quicksort(y), cpquicksort(y),
whichsort(y), cpwhichsort(y),
top(y), cptop(y),
dtable(y), cpdtable(y),
roland(y), times=10)
# Unit: milliseconds
# expr min lq median uq max neval
# whichpart(y) 11.60703 11.76857 12.03704 12.52871 47.88526 10
# cpwhichpart(y) 11.62885 11.75006 12.53724 13.88563 46.93677 10
# quicksort(y) 88.14924 89.47630 92.42414 103.53439 137.44335 10
# cpquicksort(y) 88.11544 89.15334 92.63420 94.42244 133.78006 10
# whichsort(y) 122.34675 123.13634 124.91990 127.79134 131.43400 10
# cpwhichsort(y) 121.85618 122.91653 125.45211 127.14112 158.61535 10
# top(y) 163.06669 181.19004 211.11557 224.19237 239.63139 10
# cptop(y) 163.37903 173.55113 209.46770 218.59685 226.81545 10
# dtable(y) 499.50807 505.45513 514.55338 537.84129 604.86454 10
# cpdtable(y) 491.70016 498.62664 525.05342 527.14666 580.19429 10
# roland(y) 235.44664 237.52200 242.87925 268.34080 287.71196 10
identical(sort(quicksort(x)), whichpart(x))
# [1] TRUE
编辑：测试@ flodel的建议
# @flodel
whichpartrev <- function(x, n=30) {
which(x >= -sort(-x, partial=n)[n])
}
microbenchmark(whichpart(x), whichpartrev(x), times=100)
# Unit: milliseconds
# expr min lq median uq max neval
# whichpart(x) 45.44940 46.15011 46.51321 48.67986 80.63286 100
# whichpartrev(x) 28.84482 31.30661 32.87695 62.37843 67.84757 100
microbenchmark(whichpart(y), whichpartrev(y), times=100)
# Unit: milliseconds
# expr min lq median uq max neval
# whichpart(y) 11.56135 12.26539 13.05729 13.75199 43.78484 100
# whichpartrev(y) 16.00612 16.73690 17.71687 19.04153 49.02842 100
答案 1 :(得分：1)
vec <- runif(1000000)
index <- which(vec >= sort(vec, decreasing=T)[30], arr.ind=TRUE)
vec[index]
答案 2 :(得分：1)
我不确定你是如何避免排序的。我认为？which.max可能会有所帮助。
无论如何，我会做类似以下的事情：
require(data.table)
set.seed(42)
n <- 1000000
x <- rnorm(n)
dt <- data.table(x = x, x.index = seq.int(1, n))
setkey(dt, "x")
tail(dt, 30)
# x x.index
# 1: 0.9999712 270177
# 2: 0.9999715 521060
# 3: 0.9999723 863876
# 4: 0.9999757 622734
# 5: 0.9999761 48337
# 6: 0.9999764 699984
# 7: 0.9999766 264473
# 8: 0.9999770 212981
# 9: 0.9999782 911943
# 10: 0.9999874 330250
# 11: 0.9999876 695213
# 12: 0.9999879 219101
# 13: 0.9999880 144000
# 14: 0.9999880 459676
# 15: 0.9999887 910525
# 16: 0.9999894 902172
# 17: 0.9999900 474633
# 18: 0.9999905 360481
# 19: 0.9999920 985058
# 20: 0.9999925 17169
# 21: 0.9999926 424703
# 22: 0.9999927 448196
# 23: 0.9999929 254084
# 24: 0.9999932 468090
# 25: 0.9999940 480390
# 26: 0.9999961 765489
# 27: 0.9999966 556407
# 28: 0.9999968 860100
# 29: 0.9999982 879843
# 30: 0.9999989 507889
答案 3 :(得分：0)
使用这个简单的函数，我设法使用长度为10 ^ 7或更长的矢量来保存几秒钟：
top <- function(x, n=30){
result <- numeric()
for(i in 1:n){
j <- which.max(x)
result[i] <- x[j]
x[j] <- -Inf
}
result
}
我的结果：
> x <- runif(1e7)
> system.time(y <- sort(x,decreasing=TRUE)[1:30])
user system elapsed
3.30 0.04 3.39
> system.time(z <- top(x))
user system elapsed
2.49 0.58 3.12
> x <- runif(1e8)
> system.time(y <- sort(x,decreasing=TRUE)[1:30])
user system elapsed
41.74 1.15 43.62
> system.time(z <- top(x))
user system elapsed
25.96 7.61 34.43
答案 4 :(得分：0)
set.seed(42)
x <- rnorm(1e6)
fun1 <- function(x) x[order(x, decreasing = TRUE)][1:30]
library(compiler)
fun2 <- cmpfun(function(x) {
y <- rep(-Inf,30)
for (i in seq_along(x)) {
if (x[i] > y[1]) {
y[1] <- x[i]
y <- y[order(y)]
}
}
y
})
library(microbenchmark)
microbenchmark(
y1 <- fun1(x),
y2 <- fun2(x),
times=5)
#Unit: milliseconds
# expr min lq median uq max neval
#y1 <- fun1(x) 400.1574 411.8172 418.7872 425.8027 426.2981 5
#y2 <- fun2(x) 255.7817 258.2374 258.8088 259.4630 290.6068 5
identical(sort(y1), sort(y2))
#[1] TRUE
答案 5 :(得分：-1)
我会按降序排序，然后先取n：
x = rnorm(1000000, 10, 5)
x = sort(x, decreasing = TRUE)
n = 30
print(head(x, n))