问题
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 possible.
I see two options:
1: Convert to a data.frame, and use something like this
2: Some kind of cool data.table sub setting command
I'll be happy with a fairly efficient solution of type 1. Converting to a data.frame and then back to a data.table won't take too long.
回答1:
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)
}
回答2:
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.
回答3:
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
回答4:
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.
回答5:
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
回答6:
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?
回答7:
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
回答8:
> 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
>
回答9:
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
来源：https://stackoverflow.com/questions/7235657/fastest-way-to-replace-nas-in-a-large-data-table