对名称以模式开头的所有列进行求和的最有效方法是什么?

时间:2022-06-22 22:24:04

My goal is to sum all values in columns that start with the prefix skill_ in a data.table. I would prefer a solution using data.table but I am not picky.

我的目标是在data.table中对以前缀skill_开头的列中的所有值求和。我更喜欢使用data.table的解决方案,但我不挑剔。

My solution up to now:

我的解决方案到目前为止:

> require(data.table)
> DT <- data.table(x=1:4, skill_a=c(0,1,0,0), skill_b=c(0,1,1,0), skill_c=c(0,1,1,1))
> DT[, row_idx := 1:nrow(DT)]
> DT[, count_skills := 
          sapply(1:nrow(DT), 
                 function(id) sum(DT[row_idx == id, 
                                     grepl("skill_", names(DT)), with=FALSE]))]

> DT
   x skill_a skill_b skill_c row_idx count_skills
1: 1       0       0       0       1            0
2: 2       1       1       1       2            3
3: 3       0       1       1       3            2
4: 4       0       0       1       4            1

But this becomes very slow when DT is very large. Is there a more efficient way to do this?

但是当DT非常大时,这变得非常慢。有没有更有效的方法来做到这一点?

4 个解决方案

#1


13  

A question about efficiency and performance always deserves benchmarks...

关于效率和性能的问题总是值得基准......

The size of your data is important as growth rate makes a huge difference...

您的数据大小非常重要,因为增长率会产生巨大差异......

对名称以模式开头的所有列进行求和的最有效方法是什么? Relative Benchmark Timings between 2^4 and 2^24.
Sizes along floor( 2^logb(10^( seq( 4, 24, .5 ) ), 10 ) )

相对基准时间在2 ^ 4和2 ^ 24之间。沿地板的大小(2 ^ logb(10 ^(seq(4,24,.5)),10))

Excerpt of benchmarks at 1 million rows...

100万行的基准测试摘录......

## Unit: milliseconds
##             expr    min     lq median    uq   max neval
##    dplyr.sol(DT) 21.803 50.260 51.765 52.45 73.30   100
##  rowSums.sol(DT) 20.759 50.224 51.418 52.56 96.28   100
##   SDCols.sol(DT)  7.250  8.916 37.699 38.50 52.69   100
##     eval.sol(DT)  6.883  7.007  7.916  9.45 50.91   100

eval.sol is an answer that takes advantage of data.table's handling of expressions, in the below source...

eval.sol是一个利用data.table处理表达式的答案,在下面的源代码中......

library(compiler)
library(data.table)
suppressMessages(library(dplyr))
library(microbenchmark)

buildDT <- function(reps) {
  data.table(x=seq_len(reps*4),
               skill_a=rep(c(0,1,0,0),reps),
               skill_b=rep(c(0,1,1,0),reps),
               skill_c=rep(c(0,1,1,1),reps))
}

OP.sol <- function(DT) {
  DT[, row_idx := 1:nrow(DT)]
  DT[, count_skills := 
          sapply(1:nrow(DT), 
                 function(id) sum(DT[row_idx == id, 
                                     grepl("skill_", names(DT)), with=FALSE]))]
}

dplyr.sol <- function(DT)
  DT %.% select(starts_with("skill_")) %.% rowSums()

SDCols.sol <- function(DT)
  DT[, Reduce(`+`, .SD),
     .SDcols = grep("skill_", names(DT), value = T)]

rowSums.sol <- function(DT)
  rowSums(DT[,grep("skill_", names(DT)),with=FALSE])

eval.sol <- function(DT) {
  cmd <- parse(text=paste(colnames(DT)[grepl("^skill_", colnames(DT))],collapse='+') )
  DT[,eval(cmd)]
}

DT <- buildDT(1)
identical(OP.sol(DT)$count_skills, dplyr.sol(DT))

## [1] TRUE

identical(OP.sol(DT)$count_skills, rowSums.sol(DT))

## [1] TRUE

identical(OP.sol(DT)$count_skills, SDCols.sol(DT))

## [1] TRUE

identical(OP.sol(DT)$count_skills, eval.sol(DT))

## [1] TRUE

DT<-buildDT(2500)
nrow(DT)

## [1] 10000

microbenchmark( # OP.sol(DT), forget this method.
                dplyr.sol(DT),
                rowSums.sol(DT),
                SDCols.sol(DT),
                eval.sol(DT),
                times=100)

## Unit: microseconds
##             expr   min    lq median    uq   max neval
##    dplyr.sol(DT) 760.1 809.0  848.2 951.5  2276   100
##  rowSums.sol(DT) 580.5 605.3  627.6 745.7 28481   100
##   SDCols.sol(DT) 559.8 610.5  638.8 694.0  2016   100
##     eval.sol(DT) 636.4 677.7  692.4 740.5  2021   100

DT<-buildDT(25000)
nrow(DT)

## [1] 100000

microbenchmark( # OP.sol(DT), forget this method.
                dplyr.sol(DT),
                rowSums.sol(DT),
                SDCols.sol(DT),
                eval.sol(DT),
                times=100)

## Unit: milliseconds
##             expr   min    lq median    uq   max neval
##    dplyr.sol(DT) 2.668 3.744  4.045 4.573 33.87   100
##  rowSums.sol(DT) 2.455 3.339  3.756 4.235 34.19   100
##   SDCols.sol(DT) 1.253 1.401  2.179 2.392 31.72   100
##     eval.sol(DT) 1.294 1.427  2.116 2.484 32.02   100

DT<-buildDT(250000)
nrow(DT)

## [1] 1000000

microbenchmark( # OP.sol(DT), forget this method.
                dplyr.sol(DT),
                rowSums.sol(DT),
                SDCols.sol(DT),
                eval.sol(DT),
                times=100)

## Unit: milliseconds
##             expr    min     lq median    uq   max neval
##    dplyr.sol(DT) 21.803 50.260 51.765 52.45 73.30   100
##  rowSums.sol(DT) 20.759 50.224 51.418 52.56 96.28   100
##   SDCols.sol(DT)  7.250  8.916 37.699 38.50 52.69   100
##     eval.sol(DT)  6.883  7.007  7.916  9.45 50.91   100

identical(dplyr.sol(DT), rowSums.sol(DT))

## [1] TRUE

identical(dplyr.sol(DT), SDCols.sol(DT))

## [1] TRUE

identical(dplyr.sol(DT), eval.sol(DT))

## [1] TRUE

#2


9  

Why not to use rowSums, It is generally efficient:

为什么不使用rowSums,它通常很有效:

rowSums(DT[,grep("skill_", names(DT)),with=FALSE])

#3


8  

Here is a dplyr solution:

这是一个dplyr解决方案:

library(dplyr)

DT %>% mutate(count = DT %>% select(starts_with("skill_")) %>% rowSums())

#4


7  

Solution using data.table and .SDcols.

使用data.table和.SDcols的解决方案。

require(data.table)

DT <- data.table(x=1:4, skill_a=c(0,1,0,0), skill_b=c(0,1,1,0),
                 skill_c=c(0,1,1,1))

DT[, row_idx := 1:nrow(DT)]

DT[, count_skills := Reduce(`+`, .SD),
   .SDcols = grep("skill_", names(DT), value = T)]
DT

#1


13  

A question about efficiency and performance always deserves benchmarks...

关于效率和性能的问题总是值得基准......

The size of your data is important as growth rate makes a huge difference...

您的数据大小非常重要,因为增长率会产生巨大差异......

对名称以模式开头的所有列进行求和的最有效方法是什么? Relative Benchmark Timings between 2^4 and 2^24.
Sizes along floor( 2^logb(10^( seq( 4, 24, .5 ) ), 10 ) )

相对基准时间在2 ^ 4和2 ^ 24之间。沿地板的大小(2 ^ logb(10 ^(seq(4,24,.5)),10))

Excerpt of benchmarks at 1 million rows...

100万行的基准测试摘录......

## Unit: milliseconds
##             expr    min     lq median    uq   max neval
##    dplyr.sol(DT) 21.803 50.260 51.765 52.45 73.30   100
##  rowSums.sol(DT) 20.759 50.224 51.418 52.56 96.28   100
##   SDCols.sol(DT)  7.250  8.916 37.699 38.50 52.69   100
##     eval.sol(DT)  6.883  7.007  7.916  9.45 50.91   100

eval.sol is an answer that takes advantage of data.table's handling of expressions, in the below source...

eval.sol是一个利用data.table处理表达式的答案,在下面的源代码中......

library(compiler)
library(data.table)
suppressMessages(library(dplyr))
library(microbenchmark)

buildDT <- function(reps) {
  data.table(x=seq_len(reps*4),
               skill_a=rep(c(0,1,0,0),reps),
               skill_b=rep(c(0,1,1,0),reps),
               skill_c=rep(c(0,1,1,1),reps))
}

OP.sol <- function(DT) {
  DT[, row_idx := 1:nrow(DT)]
  DT[, count_skills := 
          sapply(1:nrow(DT), 
                 function(id) sum(DT[row_idx == id, 
                                     grepl("skill_", names(DT)), with=FALSE]))]
}

dplyr.sol <- function(DT)
  DT %.% select(starts_with("skill_")) %.% rowSums()

SDCols.sol <- function(DT)
  DT[, Reduce(`+`, .SD),
     .SDcols = grep("skill_", names(DT), value = T)]

rowSums.sol <- function(DT)
  rowSums(DT[,grep("skill_", names(DT)),with=FALSE])

eval.sol <- function(DT) {
  cmd <- parse(text=paste(colnames(DT)[grepl("^skill_", colnames(DT))],collapse='+') )
  DT[,eval(cmd)]
}

DT <- buildDT(1)
identical(OP.sol(DT)$count_skills, dplyr.sol(DT))

## [1] TRUE

identical(OP.sol(DT)$count_skills, rowSums.sol(DT))

## [1] TRUE

identical(OP.sol(DT)$count_skills, SDCols.sol(DT))

## [1] TRUE

identical(OP.sol(DT)$count_skills, eval.sol(DT))

## [1] TRUE

DT<-buildDT(2500)
nrow(DT)

## [1] 10000

microbenchmark( # OP.sol(DT), forget this method.
                dplyr.sol(DT),
                rowSums.sol(DT),
                SDCols.sol(DT),
                eval.sol(DT),
                times=100)

## Unit: microseconds
##             expr   min    lq median    uq   max neval
##    dplyr.sol(DT) 760.1 809.0  848.2 951.5  2276   100
##  rowSums.sol(DT) 580.5 605.3  627.6 745.7 28481   100
##   SDCols.sol(DT) 559.8 610.5  638.8 694.0  2016   100
##     eval.sol(DT) 636.4 677.7  692.4 740.5  2021   100

DT<-buildDT(25000)
nrow(DT)

## [1] 100000

microbenchmark( # OP.sol(DT), forget this method.
                dplyr.sol(DT),
                rowSums.sol(DT),
                SDCols.sol(DT),
                eval.sol(DT),
                times=100)

## Unit: milliseconds
##             expr   min    lq median    uq   max neval
##    dplyr.sol(DT) 2.668 3.744  4.045 4.573 33.87   100
##  rowSums.sol(DT) 2.455 3.339  3.756 4.235 34.19   100
##   SDCols.sol(DT) 1.253 1.401  2.179 2.392 31.72   100
##     eval.sol(DT) 1.294 1.427  2.116 2.484 32.02   100

DT<-buildDT(250000)
nrow(DT)

## [1] 1000000

microbenchmark( # OP.sol(DT), forget this method.
                dplyr.sol(DT),
                rowSums.sol(DT),
                SDCols.sol(DT),
                eval.sol(DT),
                times=100)

## Unit: milliseconds
##             expr    min     lq median    uq   max neval
##    dplyr.sol(DT) 21.803 50.260 51.765 52.45 73.30   100
##  rowSums.sol(DT) 20.759 50.224 51.418 52.56 96.28   100
##   SDCols.sol(DT)  7.250  8.916 37.699 38.50 52.69   100
##     eval.sol(DT)  6.883  7.007  7.916  9.45 50.91   100

identical(dplyr.sol(DT), rowSums.sol(DT))

## [1] TRUE

identical(dplyr.sol(DT), SDCols.sol(DT))

## [1] TRUE

identical(dplyr.sol(DT), eval.sol(DT))

## [1] TRUE

#2


9  

Why not to use rowSums, It is generally efficient:

为什么不使用rowSums,它通常很有效:

rowSums(DT[,grep("skill_", names(DT)),with=FALSE])

#3


8  

Here is a dplyr solution:

这是一个dplyr解决方案:

library(dplyr)

DT %>% mutate(count = DT %>% select(starts_with("skill_")) %>% rowSums())

#4


7  

Solution using data.table and .SDcols.

使用data.table和.SDcols的解决方案。

require(data.table)

DT <- data.table(x=1:4, skill_a=c(0,1,0,0), skill_b=c(0,1,1,0),
                 skill_c=c(0,1,1,1))

DT[, row_idx := 1:nrow(DT)]

DT[, count_skills := Reduce(`+`, .SD),
   .SDcols = grep("skill_", names(DT), value = T)]
DT