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) # v1.6.6
require(gdata) # v2.8.2
set.seed(1)
dt1 = create_dt(2e5, 200, 0.1)
dim(dt1)
[1] 200000 200 # more columns than Ramnath's answer which had 5 not 200
f_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 Mb
Timing 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 this
identical(a_gdata, dt1)
[1] TRUE
Note 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, too
identical(a_gdata, dt1)
[1] TRUE
I 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)
}
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 200
dt2 = 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: 0.149
all.equal(dt1, dt2)
#[1] TRUE
set.seed(1)
dt1 = create_dt(2e7, 200, 0.1)
dim(dt1)
#[1] 20000000 200
dt2 = copy(dt1)
system.time(f_dowle3(dt1))
# user system elapsed
# 22.997 18.179 41.496
system.time(setnafill(dt2, fill=0))
# user system elapsed
# 39.604 36.805 3.798
all.equal(dt1, dt2)
#[1] TRUE
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.
library(data.table)
DT = data.table(a=c(1,"A",NA),b=c(4,NA,"B"))
DT
a b
1: 1 4
2: A NA
3: NA B
DT[,lapply(.SD,function(x){ifelse(is.na(x),0,x)})]
a b
1: 1 4
2: A 0
3: 0 B
Just for reference, slower compared to gdata or data.matrix, but uses only the data.table package and can deal with non numerical entries.