在R中按列排序最快 [英] Fastest by column sort in R

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问题描述

我有一个数据框 full ,我要从中取最后一列,并且一个列 v 。然后我想以最快的方式排序 v 的两列。 code>从csv中读取,但这可以用于测试(包括一些NAs的现实主义):

  n < -  200000 
full < - data.frame(A = runif(n,1,10000),B = floor(runif(n,0,1.9)) )
full [sample(n,10000),'A'] < - NA
v < - 1

我在这里有 v ,但实际上它可以改变,有很多列。






我已经尝试过排序数据框,数据表和矩阵,每个都有订单 sort.list (一些想法来自

 #DATA FRAME 

ord_df< - function(){
a < - full [c(v,length(full))]
a [with(a,order(a [1]),]
}

sl_df < - function(){
a < - full [c(v,length(full))]
a [sort.list(a [[1]]),]
}


#DATA TABLE

require(data.table)

ord_dt< - function(){
a< ; - as.data.table(full [c(v,length(full))])
colnames(a)[1] < - 'values'
a [order $ b}

sl_dt < - function(){
a< - as.data.table(full [c(v,length(full))])
colnames (a)[1]< - 'values'
a [sort.list(values)]
}


#MATRIX
$ b b ord_mat < - function(){
a < - as.matrix(full [c(v,length(full))])
a [order(a [,1]),]
}

sl_mat < - function(){
a < - as.matrix(full [c(v,length(full))])
a [sort。 list(a [,1]),]
}

时间结果:

  ord_df sl_df ord_dt sl_dt ord_mat sl_mat 
0.230 0.1500 0.1300 0.120 0.140 0.1400
中位数0.250 0.1600 0.1400 0.140 0.140 0.1400
平均0.244 0.1610 0.1430 0.136 0.142 0.1450
最大。 0.250 0.1700 0.1600 0.140 0.160 0.1600

或使用 microbenchmark (结果以毫秒为单位):

  min lq median uq max 
1 ord_df()243.0647 248.2768 254.0544 265.2589 352.3984
2 ord_dt()133.8159 140.0111 143.8202 148.4957 181.2647
3 ord_mat()140.5198 146.8131 149.9876 154.6649 191.6897
4 sl_df()152.6985 161.5591 166.5147 171.2891 194.7155
5 sl_dt()132.1414 139.7655 144.1281 149.6844 188.8592
6 sl_mat()139.2420 146.8578 151.6760 156.6174 186.5416

看起来像排序数据表胜利。 order sort.list 之间没有太大区别,除非使用数据框时 sort.list 更快。



在数据表版本中,我也尝试设置 v 作为键(因为它然后根据文档排序),但我不能得到它的工作,因为 v 的内容不是整数。



我希望尽可能提高速度,因为我必须为不同的 v 值。有谁知道我如何能够加快这个过程更进一步?也许值得尝试一个 Rcpp 实现?非常感谢。






这里是我用于计时的代码,如果它对任何人都有用:

  sortMethods<  -  list(ord_df,sl_df,ord_dt,sl_dt,ord_mat,sl_mat)

require(plyr)
定时< - raply(10,sapply(sortMethods,function(x)system.time(x())[[3]]))
colnames(timings)< - c('ord_df','sl_df ','ord_dt','sl_dt','ord_mat','sl_mat')
apply(timing,2,summary)

require(microbenchmark)
mb < microbenchmark(ord_df(),sl_df(),ord_dt(),sl_dt(),ord_mat(),sl_mat())
plot(mb)


解决方案

我不知道如果把这种东西作为一个编辑更好,但它似乎更像是回答所以这里做。更新的测试功能:

  n < -  1e7 
full< - data.frame(A = runif ,1,10000),B = floor(runif(n,0,1.9)))
full [sample(n,100000),'A'] < - NA

fdf < - full
fma< - as.matrix(full)
fdt< - as.data.table(full)
setnames(fdt,colnames(fdt)[1] 'value')

#DATA FRAME
ord_df< - function(){fdf [order(fdf [1]),]}
sl_df < {fdf [sort.list(fdf [[1]]),]}

#DATA TABLE
require(data.table)
ord_dt< - function fdt [order(values)]}

key_dt< - function(){
setkey(fdt,values)
fdt
}

#MATRIX
ord_mat< - function(){fma [order(fma [,1]),]}
sl_mat

结果(使用不同的计算机,R 2.13.1和 data.table 1.8.2):

  ord_df sl_df ord_dt key_dt ord_mat sl_mat 
最小。 37.56 20.86 2.946 2.249 20.22 20.21
1st Qu。 37.73 21.15 2.962 2.255 20.54 20.59
中位数38.43 21.74 3.002 2.280 21.05 20.82
平均值38.76 21.75 3.074 2.395 21.09 20.95
第三。 39.85 22.18 3.151 2.445 21.48 21.42
最大。 40.36 23.08 3.330 2.797 22.41 21.84



所以data.table是明显的赢家。使用一个键比排序更快,并有一个更好的语法,我会争辩。感谢您的帮助。


I have a data frame full from which I want to take the last column and a column v. I then want to sort both columns on v in the fastest way possible. full is read in from a csv but this can be used for testing (included some NAs for realism):

n <- 200000
full <- data.frame(A = runif(n, 1, 10000), B = floor(runif(n, 0, 1.9)))
full[sample(n, 10000), 'A'] <- NA
v <- 1

I have v as one here, but in reality it could change, and full has many columns.


I have tried sorting data frames, data tables and matrices each with order and sort.list (some ideas taken from this thread). The code for all these:

# DATA FRAME

ord_df <- function() {
  a <- full[c(v, length(full))]
  a[with(a, order(a[1])), ]
}

sl_df <- function() {
  a <- full[c(v, length(full))]
  a[sort.list(a[[1]]), ] 
}


# DATA TABLE

require(data.table)

ord_dt <- function() {
  a <- as.data.table(full[c(v, length(full))])
  colnames(a)[1] <- 'values'
  a[order(values)]
}

sl_dt <- function() {
 a <- as.data.table(full[c(v, length(full))])
 colnames(a)[1] <- 'values'
 a[sort.list(values)]
}


# MATRIX

ord_mat <- function() {
  a <- as.matrix(full[c(v, length(full))])
  a[order(a[, 1]), ] 
}

sl_mat <- function() {
  a <- as.matrix(full[c(v, length(full))])
  a[sort.list(a[, 1]), ] 
}

Time results:

         ord_df  sl_df    ord_dt   sl_dt   ord_mat sl_mat
Min.     0.230   0.1500   0.1300   0.120   0.140   0.1400
Median   0.250   0.1600   0.1400   0.140   0.140   0.1400
Mean     0.244   0.1610   0.1430   0.136   0.142   0.1450
Max.     0.250   0.1700   0.1600   0.140   0.160   0.1600

Or using microbenchmark (results are in milliseconds):

             min      lq       median   uq       max
1  ord_df() 243.0647 248.2768 254.0544 265.2589 352.3984
2  ord_dt() 133.8159 140.0111 143.8202 148.4957 181.2647
3 ord_mat() 140.5198 146.8131 149.9876 154.6649 191.6897
4   sl_df() 152.6985 161.5591 166.5147 171.2891 194.7155
5   sl_dt() 132.1414 139.7655 144.1281 149.6844 188.8592
6  sl_mat() 139.2420 146.8578 151.6760 156.6174 186.5416

Seems like ordering the data table wins. There isn't all that much difference between order and sort.list except when using data frames where sort.list is much faster.

In the data table versions I also tried setting v as the key (since it is then sorted according to the documentation) but I couldn't get it work since the contents of v are not integer.

I would ideally like to speed this up as much as possible since I have to do it many times for different v values. Does anyone know how I might be able to speed this process up even further? Also might it be worth trying an Rcpp implementation? Thanks.


Here's the code I used for timing if it's useful to anyone:

sortMethods <- list(ord_df, sl_df, ord_dt, sl_dt, ord_mat, sl_mat)

require(plyr)
timings <- raply(10, sapply(sortMethods, function(x) system.time(x())[[3]]))
colnames(timings) <- c('ord_df', 'sl_df', 'ord_dt', 'sl_dt', 'ord_mat', 'sl_mat')
apply(timings, 2, summary) 

require(microbenchmark)
mb <- microbenchmark(ord_df(), sl_df(), ord_dt(), sl_dt(), ord_mat(), sl_mat())
plot(mb)

解决方案

I don't know if it's better to put this sort of thing in as an edit but it seems more like answer so here will do. Updated test functions:

n <- 1e7
full <- data.frame(A = runif(n, 1, 10000), B = floor(runif(n, 0, 1.9)))
full[sample(n, 100000), 'A'] <- NA

fdf <- full
fma <- as.matrix(full)
fdt <- as.data.table(full)
setnames(fdt, colnames(fdt)[1], 'values')

# DATA FRAME
ord_df <- function() { fdf[order(fdf[1]), ] }
sl_df <- function() { fdf[sort.list(fdf[[1]]), ] }

# DATA TABLE
require(data.table)
ord_dt <- function() { fdt[order(values)] }

key_dt <- function() {
  setkey(fdt, values) 
  fdt
}

# MATRIX
ord_mat <- function() { fma[order(fma[, 1]), ] }
sl_mat <- function() { fma[sort.list(fma[, 1]), ] }

Results (using a different computer, R 2.13.1 and data.table 1.8.2):

         ord_df  sl_df   ord_dt  key_dt  ord_mat sl_mat
Min.     37.56   20.86   2.946   2.249   20.22   20.21
1st Qu.  37.73   21.15   2.962   2.255   20.54   20.59
Median   38.43   21.74   3.002   2.280   21.05   20.82
Mean     38.76   21.75   3.074   2.395   21.09   20.95
3rd Qu.  39.85   22.18   3.151   2.445   21.48   21.42
Max.     40.36   23.08   3.330   2.797   22.41   21.84

So data.table is the clear winner. Using a key is faster than ordering, and has a nicer syntax as well I'd argue. Thanks for the help everyone.

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