CODEWITHSUNDEEP
rperformancesortingmatrixapply
wake_wake
wake_wake

Reputation: 1204

The fastest way to sort the elements in each row of a matrix?

I have a matrix with a couple million rows and about 40 columns.

I want to sort the elements in each row in decreasing order. Thus, the element with the highest value of each row should be in the first column.

To do this I can use the apply function:

set.seed(1)
mm <- replicate(10, rnorm(20)) #random matrix with 20 rows and 10 columns
mm.sorted <- apply(mm,1,sort,decreasing=T)

But for a very large matrix this approach takes a very long time.

Are there different approaches to speed up the sorting of elements in rows?

Upvotes: 4

Views: 1574

Answers (3)

Karolis Koncevičius
Karolis Koncevičius

Reputation: 9666

Here is one clever way:

res <- matrix(mm[order(row(mm), -mm)], nrow = nrow(mm), byrow = TRUE)

Also faster than others:

system.time(
  res <- matrix(mm[order(row(mm), -mm)], nrow=nrow(mm), byrow=TRUE)
)
 user  system elapsed
1.910   0.254   2.170

Upvotes: 1

Roland
Roland

Reputation: 132989

You could use package data.table:

set.seed(1)
mm <- matrix(rnorm(1000000*40,0,10),ncol=40) 
library(data.table)
system.time({
  d <- as.data.table(mm)
  d[, row := .I]
  d <- melt(d, id.vars = "row") #wide to long format
  setkey(d, row, value) #sort
  d[, variable := paste0("V", ncol(mm):1)] #decreasing order

  #back to wide format and coerce to matrix
  msorted <- as.matrix(dcast(d, row ~ variable)[, row := NULL]) 
})
#user  system elapsed 
#4.96    0.59    5.62

If you could keep it as a long-format data.table (i.e., skip the last step), it would take about 2 seconds on my machine.

For comparison, timings of @qjgods' answer on my machine:

#user  system elapsed 
#3.71    2.08    8.81

Note that using apply (or parallel versions of it) transposes the matrix.

Upvotes: 6

qjgods
qjgods

Reputation: 1000

use the parallel package to speed up

library(parallel)
data<-matrix(rnorm(1000000*40,0,10),ncol=40) 
cl <- makeCluster(8)  # 8 is the number of CPU
system.time({
   parApply(cl,data,1,sort,decreasing=T)
 })
   user  system elapsed 
   9.68   10.11   29.87 
stopCluster(cl)

Upvotes: 7

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