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在大型数据表中替换 NA 的最快方法

发布于 2024-12-02 03:32:39 字数 467 浏览 2 评论 0原文

我有一个很大的 data.table，其中缺少许多值分散在大约 20 万行和 200 列中。我想尽可能有效地将这些 NA 值重新编码为零。

我看到两个选项：
1：转换为data.frame，并使用一些这样
2：某种很酷的 data.table 子设置命令

我会对类型 1 的相当有效的解决方案感到满意。转换为 data.frame 然后返回 data.table 不会花费太长时间。

原文

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.

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评论（11）

娇女薄笑 2024-12-09 03:32:39

这是一个使用 data.table 的 := 运算符的解决方案，建立在 Andrie 和 Ramnath 的基础上答案。

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

请注意，f_dowle 通过引用更新了 dt1。如果需要本地副本，则需要显式调用copy函数来制作整个数据集的本地副本。 data.table 的 setkey、key<- 和 := 不进行写入时复制。

接下来，让我们看看 f_dowle 都把时间花在哪里了。

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 ...

在那里，我将重点关注 na.replace 和 is.na，其中有一些矢量副本和矢量扫描。通过编写一个小的 na.replace C 函数，通过向量中的引用更新 NA ，可以很容易地消除这些问题。我认为这至少可以将 20 秒时间缩短一半。 R包中是否存在这样的函数？

f_andrie 失败的原因可能是因为它复制了整个 dt1，或者创建了一个与整个 dt1 一样大的逻辑矩阵，几个次。其他 2 种方法一次只适用于一列（尽管我只是简单地查看了 NAToUnknown）。

编辑（根据 Ramnath 在评论中要求的更优雅的解决方案）：

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

我希望我一开始就这样做！

EDIT2（一年多后，现在）

还有set()。如果有很多列被循环，这会更快，因为它避免了在循环中调用 [,:=,] 的（小）开销。 set 是一个可循环的:=。请参阅？设置。

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 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)
}
回复 原文
北凤男飞 2024-12-09 03:32:39

这是我能想到的最简单的一个：

dt[is.na(dt)] <- 0

它非常高效，不需要编写函数和其他粘合代码。

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.

回复 原文
终难愈 2024-12-09 03:32:39

用于此目的的专用函数（nafill 和 setnafill）可在 data.table 包（版本 >= 1.12.4）中找到：

它处理并行列很好地解决了之前发布的基准，低于其计时与迄今为止最快的方法，并且还使用 40 核机器进行了扩展。

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

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
回复 原文
莳間冲淡了誓言ζ 2024-12-09 03:32:39
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

仅供参考，与 gdata 或 data.matrix 相比速度较慢，但​​仅使用 data.table 包并且可以处理非数字条目。

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.

回复 原文
江南烟雨〆相思醉 2024-12-09 03:32:39

这是使用 gdata 包中的 NAToUnknown 的解决方案。我已经使用 Andrie 的解决方案创建了一个巨大的数据表，并且还包括与 Andrie 的解决方案的时间比较。

# 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

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
回复 原文
小嗷兮 2024-12-09 03:32:39

我的理解是，R 中快速运算的秘诀是利用向量（或数组，它们是底层的向量。）

在这个解决方案中，我使用了 data.matrix，它是一个 array 但行为有点像 data.frame。因为它是一个数组，所以您可以使用非常简单的向量替换来替换 NA：

一个用于删除 NA 的小辅助函数。本质是一行代码。我这样做只是为了测量执行时间。

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

一个小辅助函数，用于创建给定大小的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))
}

小样本演示：

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

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
回复 原文
王权女流氓 2024-12-09 03:32:39

为了完整起见，用 0 替换 NA 的另一种方法是使用

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

来比较结果和时间，我已经合并了到目前为止提到的所有方法。

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

因此，新方法比 f_dowle3 稍慢，但比所有其他方法更快。但说实话，这违背了我对 data.table 语法的直觉，我不知道为什么会这样。有人可以启发我吗？

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?

回复 原文
千と千尋 2024-12-09 03:32:39

使用最新 data.table 版本 1.12.6 中的 fifelse 函数，它甚至比 gdata 中的 NAToUnknown 快 10 倍包：

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

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
回复 原文
两仪 2024-12-09 03:32:39

要推广到许多列，您可以使用这种方法（使用以前的示例数据但添加一列）：

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))]

但没有测试速度

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

回复 原文
在你怀里撒娇 2024-12-09 03:32:39
> 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
> 

> 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
>
回复 原文
夏の忆 2024-12-09 03:32:39

一个快速替代方案是 collapse::replace_NA，默认情况下将 NA 替换为 0。

library(collapse)
replace_NA(df)

在具有 10 列（每行 1M 行）和 10% NA 的 data.frame 上进行微基准测试。

# Unit: milliseconds
#        expr     min       lq     mean   median       uq      max neval
#    f_dowle3 30.9308 34.44890 50.46713 47.33065 58.04780 160.4836   100
#   setnafill 10.3389 10.92065 12.34867 11.79305 13.41090  22.0207   100
#  replace_NA  9.9896 10.98030 15.19177 12.87030 17.00505  83.5094   100

代码：

library(data.table)
library(collapse)
library(microbenchmark)

set.seed(1)
df <- as.data.frame(replicate(10, runif(1e6)))
df <- na_insert(df, prop = 0.1)
dt <- df
setDT(dt)

f_dowle3 = function() {
  for (j in seq_len(ncol(dt)))
    set(dt, which(is.na(dt[[j]])),j,0)
}

mb <- 
  microbenchmark(
  f_dowle3 = f_dowle3(),
  setnafill = data.table::setnafill(dt, fill = 0),
  replace_NA = collapse::replace_NA(dt, set = TRUE)
)

A fast alternative is collapse::replace_NA, which by default replaces NAs with 0.

library(collapse)
replace_NA(df)

Microbenchmark on a data.frame with 10 columns of 1M rows and 10% NAs.

# Unit: milliseconds
#        expr     min       lq     mean   median       uq      max neval
#    f_dowle3 30.9308 34.44890 50.46713 47.33065 58.04780 160.4836   100
#   setnafill 10.3389 10.92065 12.34867 11.79305 13.41090  22.0207   100
#  replace_NA  9.9896 10.98030 15.19177 12.87030 17.00505  83.5094   100

Code:

library(data.table)
library(collapse)
library(microbenchmark)

set.seed(1)
df <- as.data.frame(replicate(10, runif(1e6)))
df <- na_insert(df, prop = 0.1)
dt <- df
setDT(dt)

f_dowle3 = function() {
  for (j in seq_len(ncol(dt)))
    set(dt, which(is.na(dt[[j]])),j,0)
}

mb <- 
  microbenchmark(
  f_dowle3 = f_dowle3(),
  setnafill = data.table::setnafill(dt, fill = 0),
  replace_NA = collapse::replace_NA(dt, set = TRUE)
)
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