What is the most efficient way to sum all columns whose name starts with a pattern?

data.table, r

Solution

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

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

Relative Benchmark Timings between 2^4 and 2^24. Sizes along `floor( 2^logb(10^( seq( 4, 24, .5 ) ), 10 ) )`

Excerpt of benchmarks at 1 million rows...

## 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...

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

Problem

Original source