Fastest way to replace NAs in a large data.table
Here's a solution using data.table's := operator, building on Andrie and Ramnath's answers.
require(data.table) # v1.6.6require(gdata) # v2.8.2set.seed(1)dt1 = create_dt(2e5, 200, 0.1)dim(dt1)[1] 200000 200 # more columns than Ramnath's answer which had 5 not 200f_andrie = function(dt) remove_na(dt)f_gdata = function(dt, un = 0) gdata::NAToUnknown(dt, un)f_dowle = function(dt) { # see EDIT later for more elegant solution na.replace = function(v,value=0) { v[is.na(v)] = value; v } for (i in names(dt)) eval(parse(text=paste("dt[,",i,":=na.replace(",i,")]")))}system.time(a_gdata = f_gdata(dt1)) user system elapsed 18.805 12.301 134.985 system.time(a_andrie = f_andrie(dt1))Error: cannot allocate vector of size 305.2 MbTiming stopped at: 14.541 7.764 68.285 system.time(f_dowle(dt1)) user system elapsed 7.452 4.144 19.590 # EDIT has faster than thisidentical(a_gdata, dt1) [1] TRUENote that f_dowle updated dt1 by reference. If a local copy is required then an explicit call to the copy function is needed to make a local copy of the whole dataset. data.table's setkey, key<- and := do not copy-on-write.
Next, let's see where f_dowle is spending its time.
Rprof()f_dowle(dt1)Rprof(NULL)summaryRprof()$by.self self.time self.pct total.time total.pct"na.replace" 5.10 49.71 6.62 64.52"[.data.table" 2.48 24.17 9.86 96.10"is.na" 1.52 14.81 1.52 14.81"gc" 0.22 2.14 0.22 2.14"unique" 0.14 1.36 0.16 1.56... snip ...There, I would focus on na.replace and is.na, where there are a few vector copies and vector scans. Those can fairly easily be eliminated by writing a small na.replace C function that updates NA by reference in the vector. That would at least halve the 20 seconds I think. Does such a function exist in any R package?
The reason f_andrie fails may be because it copies the whole of dt1, or creates a logical matrix as big as the whole of dt1, a few times. The other 2 methods work on one column at a time (although I only briefly looked at NAToUnknown).
EDIT (more elegant solution as requested by Ramnath in comments) :
f_dowle2 = function(DT) { for (i in names(DT)) DT[is.na(get(i)), (i):=0]}system.time(f_dowle2(dt1)) user system elapsed 6.468 0.760 7.250 # faster, tooidentical(a_gdata, dt1) [1] TRUEI wish I did it that way to start with!
EDIT2 (over 1 year later, now)
There is also set(). This can be faster if there are a lot of column being looped through, as it avoids the (small) overhead of calling [,:=,] in a loop. set is a loopable :=. See ?set.
f_dowle3 = function(DT) { # either of the following for loops # by name : for (j in names(DT)) set(DT,which(is.na(DT[[j]])),j,0) # or by number (slightly faster than by name) : for (j in seq_len(ncol(DT))) set(DT,which(is.na(DT[[j]])),j,0)}
Here's the simplest one I could come up with:
dt[is.na(dt)] <- 0
It's efficient and no need to write functions and other glue code.
Dedicated functions (nafill and setnafill) for that purpose are available in data.table package (version >= 1.12.4):
It process columns in parallel so well address previously posted benchmarks, below its timings vs fastest approach till now, and also scaled up, using 40 cores machine.
library(data.table)create_dt <- function(nrow=5, ncol=5, propNA = 0.5){ v <- runif(nrow * ncol) v[sample(seq_len(nrow*ncol), propNA * nrow*ncol)] <- NA data.table(matrix(v, ncol=ncol))}f_dowle3 = function(DT) { for (j in seq_len(ncol(DT))) set(DT,which(is.na(DT[[j]])),j,0)}set.seed(1)dt1 = create_dt(2e5, 200, 0.1)dim(dt1)#[1] 200000 200dt2 = copy(dt1)system.time(f_dowle3(dt1))# user system elapsed # 0.193 0.062 0.254 system.time(setnafill(dt2, fill=0))# user system elapsed # 0.633 0.000 0.020 ## setDTthreads(1) elapsed: 0.149all.equal(dt1, dt2)#[1] TRUEset.seed(1)dt1 = create_dt(2e7, 200, 0.1)dim(dt1)#[1] 20000000 200dt2 = copy(dt1)system.time(f_dowle3(dt1))# user system elapsed # 22.997 18.179 41.496system.time(setnafill(dt2, fill=0))# user system elapsed # 39.604 36.805 3.798 all.equal(dt1, dt2)#[1] TRUE