Fastest way to replace NAs in a large data.table

Question

I have a large data.table, with many missing values scattered throughout its ~200k rows and 200 columns. I would like to re code those NA values to zeros as efficiently as

Answer 1

> DT = data.table(a=LETTERS[c(1,1:3,4:7)],b=sample(c(15,51,NA,12,21),8,T),key="a")
> DT
   a  b
1: A 12
2: A NA
3: B 15
4: C NA
5: D 51
6: E NA
7: F 15
8: G 51
> DT[is.na(b),b:=0]
> DT
   a  b
1: A 12
2: A  0
3: B 15
4: C  0
5: D 51
6: E  0
7: F 15
8: G 51
> 

Answer 2

To generalize to many columns you could use this approach (using previous sample data but adding a column):

z = data.table(x = sample(c(NA_integer_, 1), 2e7, TRUE), y = sample(c(NA_integer_, 1), 2e7, TRUE))

z[, names(z) := lapply(.SD, function(x) fifelse(is.na(x), 0, x))]

Didn't test for the speed though

Answer 3

Using the fifelse function from the newest data.table versions 1.12.6, it is even 10 times faster than NAToUnknown in the gdata package:

z = data.table(x = sample(c(NA_integer_, 1), 2e7, TRUE))
system.time(z[,x1 := gdata::NAToUnknown(x, 0)])

#   user  system elapsed 
#  0.798   0.323   1.173 

system.time(z[,x2:= fifelse(is.na(x), 0, x)])

#   user  system elapsed 
#  0.172   0.093   0.113 

Answer 4

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

Answer 5

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)
}

Answer 6

Here is a solution using NAToUnknown in the gdata package. I have used Andrie's solution to create a huge data table and also included time comparisons with Andrie's solution.

# CREATE DATA TABLE
dt1 = create_dt(2e5, 200, 0.1)

# FUNCTIONS TO SET NA TO ZERO   
f_gdata  = function(dt, un = 0) gdata::NAToUnknown(dt, un)
f_Andrie = function(dt) remove_na(dt)

# COMPARE SOLUTIONS AND TIMES
system.time(a_gdata  <- f_gdata(dt1))

user  system elapsed 
4.224   2.962   7.388 

system.time(a_andrie <- f_Andrie(dt1))

 user  system elapsed 
4.635   4.730  20.060 

identical(a_gdata, g_andrie)  

TRUE