如何在rdataframe中将NA值替换为0 ?

I have a data.frame and some columns have NA values. I want to replace the NAs with zeros. How I do this?

我有一个data.frame和一些列有NA值。我想用0替换NAs。我该如何去做呢?

14 个解决方案

#1

See my comment in @gsk3 answer. A simple example:

请参阅@gsk3的回复。一个简单的例子:

> m <- matrix(sample(c(NA, 1:10), 100, replace = TRUE), 10)> d <- as.data.frame(m)   V1 V2 V3 V4 V5 V6 V7 V8 V9 V101   4  3 NA  3  7  6  6 10  6   52   9  8  9  5 10 NA  2  1  7   23   1  1  6  3  6 NA  1  4  1   64  NA  4 NA  7 10  2 NA  4  1   85   1  2  4 NA  2  6  2  6  7   46  NA  3 NA NA 10  2  1 10  8   47   4  4  9 10  9  8  9  4 10  NA8   5  8  3  2  1  4  5  9  4   79   3  9 10  1  9  9 10  5  3   310  4  2  2  5 NA  9  7  2  5   5> d[is.na(d)] <- 0> d   V1 V2 V3 V4 V5 V6 V7 V8 V9 V101   4  3  0  3  7  6  6 10  6   52   9  8  9  5 10  0  2  1  7   23   1  1  6  3  6  0  1  4  1   64   0  4  0  7 10  2  0  4  1   85   1  2  4  0  2  6  2  6  7   46   0  3  0  0 10  2  1 10  8   47   4  4  9 10  9  8  9  4 10   08   5  8  3  2  1  4  5  9  4   79   3  9 10  1  9  9 10  5  3   310  4  2  2  5  0  9  7  2  5   5

There's no need to apply apply. =)

没有必要申请。=)

EDIT

编辑

You should also take a look at norm package. It has a lot of nice features for missing data analysis. =)

你也应该看看norm软件包。对于缺失的数据分析，它有很多很好的特性。=)

#2

The hybrid dplyr/Base R option: mutate_all(funs(replace(., is.na(.), 0)))) is more than twice as fast as the base R d[is.na(d)] <- 0 option. (please see benchmark analyses below.)

混合dplyr/Base R选项:mutate_all(funs(替换)。， is.na(.)， 0))))))比基本的R d[is.na(d)] <- 0选项快两倍以上。(请参阅下面的基准分析)

If you are struggling with massive dataframes, data.table is the fastest option of all: 30% less time than dplyr, and 3 times faster than the Base R approaches. It also modifies the data in place, effectively allowing you to work with nearly twice as much of the data at once.

如果您正在与大量的dataframes和data作斗争。表是所有选项中最快的:比dplyr少30%，比基本R快3倍。它还修改了适当的数据，有效地使您可以同时处理几乎两倍的数据。

---

A clustering of other helpful tidyverse replacement approaches

Locationally:

区位:

• index mutate_at(c(5:10), funs(replace(., is.na(.), 0)))
• 指数mutate_at(c(5:10)乐趣(取代。is.na(。),0)))
• direct reference mutate_at(vars(var5:var10), funs(replace(., is.na(.), 0)))
• 直接引用mutate_at(var(var5:var10)乐趣(取代。is.na(。),0)))
• fixed match mutate_at(vars(contains("1")), funs(replace(., is.na(.), 0)))
  • or in place of contains(), try ends_with(),starts_with()
  • 或者代替contains()，尝试ends_with()，starts_with()
• 固定搭配mutate_at(var(包含(" 1 ")),乐趣(取代。, is.na(.)， 0))或in place of contains()， try ends_with()，starts_with()
• pattern match mutate_at(vars(matches("\\d{2}")), funs(replace(., is.na(.), 0)))
• 模式匹配mutate_at(var(匹配(\ \ d { 2 })),乐趣(取代。is.na(。),0)))

Conditionally:
(change just numeric (columns) and leave string (columns) alone.)

有条件地:(仅更改数字(列)并单独保留字符串(列)。

• integers mutate_if(is.integer, funs(replace(., is.na(.), 0)))
• 整数mutate_if(is.integer乐趣(取代。is.na(。),0)))
• doubles mutate_if(is.numeric, funs(replace(., is.na(.), 0)))
• 双打mutate_if(。数字,乐趣(取代。is.na(。),0)))
• strings mutate_if(is.character, funs(replace(., is.na(.), 0)))
• 字符串mutate_if(。性格,乐趣(取代。is.na(。),0)))

---

The Complete Analysis -

Approaches tested:

# Base R: baseR.sbst.rssgn   <- function(x) { x[is.na(x)] <- 0; x }baseR.replace      <- function(x) { replace(x, is.na(x), 0) }baseR.for          <- function(x) { for(j in 1:ncol(x))                                    x[[j]][is.na(x[[j]])] = 0 }# tidyverse## dplyrlibrary(tidyverse)dplyr_if_else      <- function(x) { mutate_all(x, funs(if_else(is.na(.), 0, .))) }dplyr_coalesce     <- function(x) { mutate_all(x, funs(coalesce(., 0))) }## tidyrtidyr_replace_na   <- function(x) { replace_na(x, as.list(setNames(rep(0, 10), as.list(c(paste0("var", 1:10)))))) }## hybrid hybrd.ifelse     <- function(x) { mutate_all(x, funs(ifelse(is.na(.), 0, .))) }hybrd.rplc_all   <- function(x) { mutate_all(x, funs(replace(., is.na(.), 0))) }hybrd.rplc_at.idx<- function(x) { mutate_at(x, c(1:10), funs(replace(., is.na(.), 0))) }hybrd.rplc_at.nse<- function(x) { mutate_at(x, vars(var1:var10), funs(replace(., is.na(.), 0))) }hybrd.rplc_at.stw<- function(x) { mutate_at(x, vars(starts_with("var")), funs(replace(., is.na(.), 0))) }hybrd.rplc_at.ctn<- function(x) { mutate_at(x, vars(contains("var")), funs(replace(., is.na(.), 0))) }hybrd.rplc_at.mtc<- function(x) { mutate_at(x, vars(matches("\\d+")), funs(replace(., is.na(.), 0))) }hybrd.rplc_if    <- function(x) { mutate_if(x, is.numeric, funs(replace(., is.na(.), 0))) }# data.table   library(data.table)DT.for.set.nms   <- function(x) { for (j in names(x))                                    set(x,which(is.na(x[[j]])),j,0) }DT.for.set.sqln  <- function(x) { for (j in seq_len(ncol(x)))                                    set(x,which(is.na(x[[j]])),j,0) }

The code for this analysis:

library(microbenchmark)# 20% NA filled dataframe of 5 Million rows and 10 columnsset.seed(42) # to recreate the exact dataframedfN <- as.data.frame(matrix(sample(c(NA, as.numeric(1:4)), 5e6*10, replace = TRUE),                            dimnames = list(NULL, paste0("var", 1:10)),                             ncol = 10))# Running 250 trials with each replacement method # (the functions are excecuted locally - so that the original dataframe remains unmodified in all cases)perf_results <- microbenchmark(    hybrid.ifelse    = hybrid.ifelse(copy(dfN)),    dplyr_if_else    = dplyr_if_else(copy(dfN)),    baseR.sbst.rssgn = baseR.sbst.rssgn(copy(dfN)),    baseR.replace    = baseR.replace(copy(dfN)),    dplyr_coalesce   = dplyr_coalesce(copy(dfN)),    hybrd.rplc_at.nse= hybrd.rplc_at.nse(copy(dfN)),    hybrd.rplc_at.stw= hybrd.rplc_at.stw(copy(dfN)),    hybrd.rplc_at.ctn= hybrd.rplc_at.ctn(copy(dfN)),    hybrd.rplc_at.mtc= hybrd.rplc_at.mtc(copy(dfN)),    hybrd.rplc_at.idx= hybrd.rplc_at.idx(copy(dfN)),    hybrd.rplc_if    = hybrd.rplc_if(copy(dfN)),    tidyr_replace_na = tidyr_replace_na(copy(dfN)),    baseR.for        = baseR.for(copy(dfN)),    DT.for.set.nms   = DT.for.set.nms(copy(dfN)),    DT.for.set.sqln  = DT.for.set.sqln(copy(dfN)),    times = 250L)

Summary of Results

> perf_resultsUnit: milliseconds              expr       min        lq      mean    median        uq      max neval     hybrid.ifelse 5250.5259 5620.8650 5809.1808 5759.3997 5947.7942 6732.791   250     dplyr_if_else 3209.7406 3518.0314 3653.0317 3620.2955 3746.0293 4390.888   250  baseR.sbst.rssgn 1611.9227 1878.7401 1964.6385 1942.8873 2031.5681 2485.843   250     baseR.replace 1559.1494 1874.7377 1946.2971 1920.8077 2002.4825 2516.525   250    dplyr_coalesce  949.7511 1231.5150 1279.3015 1288.3425 1345.8662 1624.186   250 hybrd.rplc_at.nse  735.9949  871.1693 1016.5910 1064.5761 1104.9590 1361.868   250 hybrd.rplc_at.stw  704.4045  887.4796 1017.9110 1063.8001 1106.7748 1338.557   250 hybrd.rplc_at.ctn  723.9838  878.6088 1017.9983 1063.0406 1110.0857 1296.024   250 hybrd.rplc_at.mtc  686.2045  885.8028 1013.8293 1061.2727 1105.7117 1269.949   250 hybrd.rplc_at.idx  696.3159  880.7800 1003.6186 1038.8271 1083.1932 1309.635   250     hybrd.rplc_if  705.9907  889.7381 1000.0113 1036.3963 1083.3728 1338.190   250  tidyr_replace_na  680.4478  973.1395  978.2678 1003.9797 1051.2624 1294.376   250         baseR.for  670.7897  965.6312  983.5775 1001.5229 1052.5946 1206.023   250    DT.for.set.nms  496.8031  569.7471  695.4339  623.1086  861.1918 1067.640   250   DT.for.set.sqln  500.9945  567.2522  671.4158  623.1454  764.9744 1033.463   250

Boxplot of Results (on a log scale)

# adjust the margins to prepare for better boxplot printingpar(mar=c(8,5,1,1) + 0.1) # generate boxplotboxplot(opN, las = 2, xlab = "", ylab = "log(time)[milliseconds]")

Color-coded Scatterplot of Trials (on a log scale)

qplot(y=time/10^9, data=opN, colour=expr) +     labs(y = "log10 Scaled Elapsed Time per Trial (secs)", x = "Trial Number") +    scale_y_log10(breaks=c(1, 2, 4))

A note on the other high performers

When the datasets get larger, Tidyr''s replace_na had historically pulled out in front. With the current collection of 50M data points to run through, it performs almost exactly as well as a Base R For Loop. I am curious to see what happens for different sized dataframes.

当数据集变大时，Tidyr的replace_na历来都是放在前面。通过运行当前收集的50M数据点，它执行的几乎与基本的R For循环一样好。我很好奇不同大小的数据爆炸会发生什么。

Additional examples for the mutate and summarize _at and _all function variants can be found here: https://rdrr.io/cran/dplyr/man/summarise_all.htmlAdditionally, I found helpful demonstrations and collections of examples here: https://blog.exploratory.io/dplyr-0-5-is-awesome-heres-why-be095fd4eb8a

关于突变和汇总_at和_all函数变体的其他示例可以在这里找到:https://rdrr.io/cran/dplyr/man/abstrse_all。在这里，我找到了有用的演示和示例集合:https://blog.exploratory.io/dplyr-0-5-is-awesome-here -why-be095fd4eb8a

Attributions and Appreciations

With special thanks to:

特别感谢:

• Tyler Rinker and Akrun for demonstrating microbenchmark.
• 泰勒·林克和Akrun演示微基准。
• alexis_laz for working on helping me understand the use of local(), and (with Frank's patient help, too) the role that silent coercion plays in speeding up many of these approaches.
• alexis_laz帮助我理解local()的使用，以及(在Frank耐心的帮助下)沉默胁迫在加速这些方法中所起的作用。
• ArthurYip for the poke to add the newer coalesce() function in and update the analysis.
• ArthurYip for the poke添加新的coalesce()函数并更新分析。
• Gregor for the nudge to figure out the data.table functions well enough to finally include them in the lineup.
• 格里高尔用轻推来计算数据。表功能足够好，最终可以将它们包括在队列中。
• Base R For loop: alexis_laz
• 底R为循环:alexis_laz
• data.table For Loops: Matt_Dowle
• 数据。表圈:Matt_Dowle

(Of course, please reach over and give them upvotes, too if you find those approaches useful.)

(当然，如果你觉得这些方法有用，也请向他们伸出援手。)

Note on my use of Numerics: If you do have a pure integer dataset, all of your functions will run faster. Please see alexiz_laz's work for more information. IRL, I can't recall encountering a data set containing more than 10-15% integers, so I am running these tests on fully numeric dataframes.

注意我对数字的使用:如果您有一个纯整数数据集，那么所有的函数都将运行得更快。请参阅alexiz_laz的工作以获得更多信息。IRL，我不记得遇到一个包含超过10-15%整数的数据集，所以我在全数字的dataframes上运行这些测试。

#3

For a single vector:

为一个向量:

x <- c(1,2,NA,4,5)x[is.na(x)] <- 0

For a data.frame, make a function out of the above, then apply it to the columns.

对于一个data.frame，从上面创建一个函数，然后将其应用到列中。

Please provide a reproducible example next time as detailed here:

请下次提供一个可复制的例子，详情如下:

How to make a great R reproducible example?

如何做出一个伟大的R可再现的例子?

#4

dplyr example:

dplyr例子:

library(dplyr)df1 <- df1 %>%    mutate(myCol1 = if_else(is.na(myCol1), 0, myCol1))

Note: This works per selected column, if we need to do this for all column, see @reidjax's answer using mutate_each.

注意:这适用于每个选定的列，如果我们需要对所有列都这样做，请参阅使用mutate_each的@reidjax的答案。

#5

I know the question is already answered, but doing it this way might be more useful to some:

我知道这个问题已经得到了回答，但这样做对某些人可能更有用:

Define this function:

定义这个函数:

na.zero <- function (x) {    x[is.na(x)] <- 0    return(x)}

Now whenever you need to convert NA's in a vector to zero's you can do:

现在，当你需要将向量中的NA转换为0时你可以这样做:

na.zero(some.vector)

#6

If we are trying to replace NAs when exporting, for example when writing to csv, then we can use:

如果我们尝试在导出时替换NAs，例如写入到csv时，我们可以使用:

  write.csv(data, "data.csv", na = "0")

#7

More general approach of using replace() in matrix or vector to replace NA to 0

在矩阵或向量中使用replace()来将NA替换为0的更一般的方法

For example:

例如:

> x <- c(1,2,NA,NA,1,1)> x1 <- replace(x,is.na(x),0)> x1[1] 1 2 0 0 1 1

This is also an alternative to using ifelse() in dplyr

这也是在dplyr中使用ifelse()的一种替代方法

df = data.frame(col = c(1,2,NA,NA,1,1))df <- df %>%   mutate(col = replace(col,is.na(col),0))

#8

With dplyr 0.5.0, you can use coalesce function which can be easily integrated into %>% pipeline by doing coalesce(vec, 0). This replaces all NAs in vec with 0:

使用dplyr 0.5.0，您可以使用coalesce函数，它可以轻松集成到%>%的管道中(vec, 0)。

Say we have a data frame with NAs:

假设我们有一个带NAs的数据框架:

library(dplyr)df <- data.frame(v = c(1, 2, 3, NA, 5, 6, 8))df#    v# 1  1# 2  2# 3  3# 4 NA# 5  5# 6  6# 7  8df %>% mutate(v = coalesce(v, 0))#   v# 1 1# 2 2# 3 3# 4 0# 5 5# 6 6# 7 8

#9

Another example using imputeTS package:

另一个使用imputeTS package的例子:

library(imputeTS)na.replace(yourDataframe, 0)

#10

If you want to replace NAs in factor variables, this might be useful:

如果您想在因子变量中替换NAs，这可能是有用的:

n <- length(levels(data.vector))+1data.vector <- as.numeric(data.vector)data.vector[is.na(data.vector)] <- ndata.vector <- as.factor(data.vector)levels(data.vector) <- c("level1","level2",...,"leveln", "NAlevel") 

It transforms a factor-vector into a numeric vector and adds another artifical numeric factor level, which is then transformed back to a factor-vector with one extra "NA-level" of your choice.

它将一个因子向量转换为一个数字向量，并添加另一个人工的数值因子级别，然后将其转换为一个因子向量，并提供一个额外的“NA-level”供您选择。

#11

Would've commented on @ianmunoz's post but I don't have enough reputation. You can combine dplyr's mutate_each and replace to take care of the NA to 0 replacement. Using the dataframe from @aL3xa's answer...

我会评论@ianmunoz的帖子，但我没有足够的声誉。你可以把dplyr的mutate_each和replace替换为take care of the NA to 0 replacement。从@aL3xa的答案中使用dataframe…

> m <- matrix(sample(c(NA, 1:10), 100, replace = TRUE), 10)> d <- as.data.frame(m)> d    V1 V2 V3 V4 V5 V6 V7 V8 V9 V101   4  8  1  9  6  9 NA  8  9   82   8  3  6  8  2  1 NA NA  6   33   6  6  3 NA  2 NA NA  5  7   74  10  6  1  1  7  9  1 10  3  105  10  6  7 10 10  3  2  5  4   66   2  4  1  5  7 NA NA  8  4   47   7  2  3  1  4 10 NA  8  7   78   9  5  8 10  5  3  5  8  3   29   9  1  8  7  6  5 NA NA  6   710  6 10  8  7  1  1  2  2  5   7> d %>% mutate_each( funs_( interp( ~replace(., is.na(.),0) ) ) )    V1 V2 V3 V4 V5 V6 V7 V8 V9 V101   4  8  1  9  6  9  0  8  9   82   8  3  6  8  2  1  0  0  6   33   6  6  3  0  2  0  0  5  7   74  10  6  1  1  7  9  1 10  3  105  10  6  7 10 10  3  2  5  4   66   2  4  1  5  7  0  0  8  4   47   7  2  3  1  4 10  0  8  7   78   9  5  8 10  5  3  5  8  3   29   9  1  8  7  6  5  0  0  6   710  6 10  8  7  1  1  2  2  5   7

We're using standard evaluation (SE) here which is why we need the underscore on "funs_." We also use lazyeval's interp/~ and the . references "everything we are working with", i.e. the data frame. Now there are zeros!

我们在这里使用标准评估(SE)，这就是为什么我们需要在“funs_”上下划线。我们还使用了lazyeval的interp/~和the。引用“我们正在处理的一切”，即数据框架。现在是零!

#12

You can use replace()

您可以使用替代()

For example:

例如:

> x <- c(-1,0,1,0,NA,0,1,1)> x1 <- replace(x,5,1)> x1[1] -1  0  1  0  1  0  1  1> x1 <- replace(x,5,mean(x,na.rm=T))> x1[1] -1.00  0.00  1.00  0.00  0.29  0.00 1.00  1.00

#13

Another dplyr pipe compatible option with tidyrmethod replace_na that works for several columns:

另一个与tidyrmethod replace_na兼容的dplyr管道选项，适用于几个列:

require(dplyr)require(tidyr)m <- matrix(sample(c(NA, 1:10), 100, replace = TRUE), 10)d <- as.data.frame(m)myList <- setNames(lapply(vector("list", ncol(d)), function(x) x <- 0), names(d))df <- d %>% replace_na(myList)

You can easily restrict to e.g. numeric columns:

你可以很容易地限制在数字列:

d$str <- c("string", NA)myList <- myList[sapply(d, is.numeric)]df <- d %>% replace_na(myList)

#14

This simple function extracted from Datacamp could help:

从Datacamp中提取的这个简单函数可以帮助:

replace_missings <- function(x, replacement) {  is_miss <- is.na(x)  x[is_miss] <- replacement  message(sum(is_miss), " missings replaced by the value ", replacement)  x}

Then

然后

replace_missings(df, replacement = 0)