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

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.

Answer 2

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.

Answer 3

For the sake of completeness, another way to replace NAs with 0 is to use

f_rep <- function(dt) {
dt[is.na(dt)] <- 0
return(dt)
}

To compare results and times I have incorporated all approaches mentioned so far.

set.seed(1)
dt1 <- create_dt(2e5, 200, 0.1)
dt2 <- dt1
dt3 <- dt1

system.time(res1 <- f_gdata(dt1))
   User      System verstrichen 
   3.62        0.22        3.84 
system.time(res2 <- f_andrie(dt1))
   User      System verstrichen 
   2.95        0.33        3.28 
system.time(f_dowle2(dt2))
   User      System verstrichen 
   0.78        0.00        0.78 
system.time(f_dowle3(dt3))
   User      System verstrichen 
   0.17        0.00        0.17 
system.time(res3 <- f_unknown(dt1))
   User      System verstrichen 
   6.71        0.84        7.55 
system.time(res4 <- f_rep(dt1))
   User      System verstrichen 
   0.32        0.00        0.32 

identical(res1, res2) & identical(res2, res3) & identical(res3, res4) & identical(res4, dt2) & identical(dt2, dt3)
[1] TRUE

So the new approach is slightly slower than f_dowle3 but faster than all the other approaches. But to be honest, this is against my Intuition of the data.table Syntax and I have no idea why this works. Can anybody enlighten me?

Answer 4

My understanding is that the secret to fast operations in R is to utilise vector (or arrays, which are vectors under the hood.)

In this solution I make use of a data.matrix which is an array but behave a bit like a data.frame. Because it is an array, you can use a very simple vector substitution to replace the NAs:

A little helper function to remove the NAs. The essence is a single line of code. I only do this to measure execution time.

remove_na <- function(x){
  dm <- data.matrix(x)
  dm[is.na(dm)] <- 0
  data.table(dm)
}

A little helper function to create a data.table of a given size.

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))
}

Demonstration on a tiny sample:

library(data.table)
set.seed(1)
dt <- create_dt(5, 5, 0.5)

dt
            V1        V2        V3        V4        V5
[1,]        NA 0.8983897        NA 0.4976992 0.9347052
[2,] 0.3721239 0.9446753        NA 0.7176185 0.2121425
[3,] 0.5728534        NA 0.6870228 0.9919061        NA
[4,]        NA        NA        NA        NA 0.1255551
[5,] 0.2016819        NA 0.7698414        NA        NA

remove_na(dt)
            V1        V2        V3        V4        V5
[1,] 0.0000000 0.8983897 0.0000000 0.4976992 0.9347052
[2,] 0.3721239 0.9446753 0.0000000 0.7176185 0.2121425
[3,] 0.5728534 0.0000000 0.6870228 0.9919061 0.0000000
[4,] 0.0000000 0.0000000 0.0000000 0.0000000 0.1255551
[5,] 0.2016819 0.0000000 0.7698414 0.0000000 0.0000000