How do I replace NA values with zeros in an R dataframe? - r

Question

I have a data frame and some columns have NA values.
How do I replace these NA values with zeroes?

Answer 1

See my comment in #gsk3 answer. A simple example:
> 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 V10
1 4 3 NA 3 7 6 6 10 6 5
2 9 8 9 5 10 NA 2 1 7 2
3 1 1 6 3 6 NA 1 4 1 6
4 NA 4 NA 7 10 2 NA 4 1 8
5 1 2 4 NA 2 6 2 6 7 4
6 NA 3 NA NA 10 2 1 10 8 4
7 4 4 9 10 9 8 9 4 10 NA
8 5 8 3 2 1 4 5 9 4 7
9 3 9 10 1 9 9 10 5 3 3
10 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 V10
1 4 3 0 3 7 6 6 10 6 5
2 9 8 9 5 10 0 2 1 7 2
3 1 1 6 3 6 0 1 4 1 6
4 0 4 0 7 10 2 0 4 1 8
5 1 2 4 0 2 6 2 6 7 4
6 0 3 0 0 10 2 1 10 8 4
7 4 4 9 10 9 8 9 4 10 0
8 5 8 3 2 1 4 5 9 4 7
9 3 9 10 1 9 9 10 5 3 3
10 4 2 2 5 0 9 7 2 5 5
There's no need to apply apply. =)
EDIT
You should also take a look at norm package. It has a lot of nice features for missing data analysis. =)

Answer 2

The dplyr hybridized options are now around 30% faster than the Base R subset reassigns. On a 100M datapoint dataframe mutate_all(~replace(., is.na(.), 0)) runs a half a second faster than the base R d[is.na(d)] <- 0 option. What one wants to avoid specifically is using an ifelse() or an if_else(). (The complete 600 trial analysis ran to over 4.5 hours mostly due to including these approaches.) Please see benchmark analyses below for the complete results.
If you are struggling with massive dataframes, data.table is the fastest option of all: 40% faster than the standard Base R approach. It also modifies the data in place, effectively allowing you to work with nearly twice as much of the data at once.
A clustering of other helpful tidyverse replacement approaches
Locationally:
index mutate_at(c(5:10), ~replace(., is.na(.), 0))
direct reference mutate_at(vars(var5:var10), ~replace(., is.na(.), 0))
fixed match mutate_at(vars(contains("1")), ~replace(., is.na(.), 0))
or in place of contains(), try ends_with(),starts_with()
pattern match mutate_at(vars(matches("\\d{2}")), ~replace(., is.na(.), 0))
Conditionally:
(change just single type and leave other types alone.)
integers mutate_if(is.integer, ~replace(., is.na(.), 0))
numbers mutate_if(is.numeric, ~replace(., is.na(.), 0))
strings mutate_if(is.character, ~replace(., is.na(.), 0))
##The Complete Analysis -
Updated for dplyr 0.8.0: functions use purrr format ~ symbols: replacing deprecated funs() arguments.
###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
## dplyr
dplyr_if_else <- function(x) { mutate_all(x, ~if_else(is.na(.), 0, .)) }
dplyr_coalesce <- function(x) { mutate_all(x, ~coalesce(., 0)) }
## tidyr
tidyr_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, ~ifelse(is.na(.), 0, .)) }
hybrd.replace_na <- function(x) { mutate_all(x, ~replace_na(., 0)) }
hybrd.replace <- function(x) { mutate_all(x, ~replace(., is.na(.), 0)) }
hybrd.rplc_at.idx<- function(x) { mutate_at(x, c(1:10), ~replace(., is.na(.), 0)) }
hybrd.rplc_at.nse<- function(x) { mutate_at(x, vars(var1:var10), ~replace(., is.na(.), 0)) }
hybrd.rplc_at.stw<- function(x) { mutate_at(x, vars(starts_with("var")), ~replace(., is.na(.), 0)) }
hybrd.rplc_at.ctn<- function(x) { mutate_at(x, vars(contains("var")), ~replace(., is.na(.), 0)) }
hybrd.rplc_at.mtc<- function(x) { mutate_at(x, vars(matches("\\d+")), ~replace(., is.na(.), 0)) }
hybrd.rplc_if <- function(x) { mutate_if(x, is.numeric, ~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) }
DT.nafill <- function(x) { nafill(df, fill=0)}
DT.setnafill <- function(x) { setnafill(df, fill=0)}
###The code for this analysis:
library(microbenchmark)
# 20% NA filled dataframe of 10 Million rows and 10 columns
set.seed(42) # to recreate the exact dataframe
dfN <- as.data.frame(matrix(sample(c(NA, as.numeric(1:4)), 1e7*10, replace = TRUE),
dimnames = list(NULL, paste0("var", 1:10)),
ncol = 10))
# Running 600 trials with each replacement method
# (the functions are excecuted locally - so that the original dataframe remains unmodified in all cases)
perf_results <- microbenchmark(
hybrd.ifelse = hybrd.ifelse(copy(dfN)),
dplyr_if_else = dplyr_if_else(copy(dfN)),
hybrd.replace_na = hybrd.replace_na(copy(dfN)),
baseR.sbst.rssgn = baseR.sbst.rssgn(copy(dfN)),
baseR.replace = baseR.replace(copy(dfN)),
dplyr_coalesce = dplyr_coalesce(copy(dfN)),
tidyr_replace_na = tidyr_replace_na(copy(dfN)),
hybrd.replace = hybrd.replace(copy(dfN)),
hybrd.rplc_at.ctn= hybrd.rplc_at.ctn(copy(dfN)),
hybrd.rplc_at.nse= hybrd.rplc_at.nse(copy(dfN)),
baseR.for = baseR.for(copy(dfN)),
hybrd.rplc_at.idx= hybrd.rplc_at.idx(copy(dfN)),
DT.for.set.nms = DT.for.set.nms(copy(dfN)),
DT.for.set.sqln = DT.for.set.sqln(copy(dfN)),
times = 600L
)
###Summary of Results
> print(perf_results)
Unit: milliseconds
expr min lq mean median uq max neval
hybrd.ifelse 6171.0439 6339.7046 6425.221 6407.397 6496.992 7052.851 600
dplyr_if_else 3737.4954 3877.0983 3953.857 3946.024 4023.301 4539.428 600
hybrd.replace_na 1497.8653 1706.1119 1748.464 1745.282 1789.804 2127.166 600
baseR.sbst.rssgn 1480.5098 1686.1581 1730.006 1728.477 1772.951 2010.215 600
baseR.replace 1457.4016 1681.5583 1725.481 1722.069 1766.916 2089.627 600
dplyr_coalesce 1227.6150 1483.3520 1524.245 1519.454 1561.488 1996.859 600
tidyr_replace_na 1248.3292 1473.1707 1521.889 1520.108 1570.382 1995.768 600
hybrd.replace 913.1865 1197.3133 1233.336 1238.747 1276.141 1438.646 600
hybrd.rplc_at.ctn 916.9339 1192.9885 1224.733 1227.628 1268.644 1466.085 600
hybrd.rplc_at.nse 919.0270 1191.0541 1228.749 1228.635 1275.103 2882.040 600
baseR.for 869.3169 1180.8311 1216.958 1224.407 1264.737 1459.726 600
hybrd.rplc_at.idx 839.8915 1189.7465 1223.326 1228.329 1266.375 1565.794 600
DT.for.set.nms 761.6086 915.8166 1015.457 1001.772 1106.315 1363.044 600
DT.for.set.sqln 787.3535 918.8733 1017.812 1002.042 1122.474 1321.860 600
###Boxplot of Results
ggplot(perf_results, aes(x=expr, y=time/10^9)) +
geom_boxplot() +
xlab('Expression') +
ylab('Elapsed Time (Seconds)') +
scale_y_continuous(breaks = seq(0,7,1)) +
coord_flip()
Color-coded Scatterplot of Trials (with y-axis on a log scale)
qplot(y=time/10^9, data=perf_results, colour=expr) +
labs(y = "log10 Scaled Elapsed Time per Trial (secs)", x = "Trial Number") +
coord_cartesian(ylim = c(0.75, 7.5)) +
scale_y_log10(breaks=c(0.75, 0.875, 1, 1.25, 1.5, 1.75, seq(2, 7.5)))
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 100M 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.
Additional examples for the mutate and summarize _at and _all function variants can be found here: https://rdrr.io/cran/dplyr/man/summarise_all.html
Additionally, I found helpful demonstrations and collections of examples here: https://blog.exploratory.io/dplyr-0-5-is-awesome-heres-why-be095fd4eb8a
Attributions and Appreciations
With special thanks to:
Tyler Rinker and Akrun for demonstrating microbenchmark.
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.
ArthurYip for the poke to add the newer coalesce() function in and update the analysis.
Gregor for the nudge to figure out the data.table functions well enough to finally include them in the lineup.
Base R For loop: alexis_laz
data.table For Loops: Matt_Dowle
Roman for explaining what is.numeric() really tests.
(Of course, please reach over and give them upvotes, too if you find those approaches useful.)
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.
Hardware Used
3.9 GHz CPU with 24 GB RAM

Answer 3

For a single vector:
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.
Please provide a reproducible example next time as detailed here:
How to make a great R reproducible example?

Answer 4

dplyr example:
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.

Answer 5

If we are trying to replace NAs when exporting, for example when writing to csv, then we can use:
write.csv(data, "data.csv", na = "0")

Answer 6

It is also possible to use tidyr::replace_na.
library(tidyr)
df <- df %>% mutate_all(funs(replace_na(.,0)))
Edit (dplyr > 1.0.0):
df %>% mutate(across(everything(), .fns = ~replace_na(.,0)))

Answer 7

I know the question is already answered, but doing it this way might be more useful to some:
Define this function:
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:
na.zero(some.vector)

Answer 8

More general approach of using replace() in matrix or vector to replace NA to 0
For example:
> 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
df = data.frame(col = c(1,2,NA,NA,1,1))
df <- df %>%
mutate(col = replace(col,is.na(col),0))

Answer 9

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:
Say we have a data frame with NAs:
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 8
df %>% mutate(v = coalesce(v, 0))
# v
# 1 1
# 2 2
# 3 3
# 4 0
# 5 5
# 6 6
# 7 8

Answer 10

To replace all NAs in a dataframe you can use:
df %>% replace(is.na(.), 0)

Answer 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...
> 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 V10
1 4 8 1 9 6 9 NA 8 9 8
2 8 3 6 8 2 1 NA NA 6 3
3 6 6 3 NA 2 NA NA 5 7 7
4 10 6 1 1 7 9 1 10 3 10
5 10 6 7 10 10 3 2 5 4 6
6 2 4 1 5 7 NA NA 8 4 4
7 7 2 3 1 4 10 NA 8 7 7
8 9 5 8 10 5 3 5 8 3 2
9 9 1 8 7 6 5 NA NA 6 7
10 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 V10
1 4 8 1 9 6 9 0 8 9 8
2 8 3 6 8 2 1 0 0 6 3
3 6 6 3 0 2 0 0 5 7 7
4 10 6 1 1 7 9 1 10 3 10
5 10 6 7 10 10 3 2 5 4 6
6 2 4 1 5 7 0 0 8 4 4
7 7 2 3 1 4 10 0 8 7 7
8 9 5 8 10 5 3 5 8 3 2
9 9 1 8 7 6 5 0 0 6 7
10 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!

Answer 12

Another example using imputeTS package:
library(imputeTS)
na.replace(yourDataframe, 0)

Answer 13

Dedicated functions, nafill and setnafill, for that purpose is in data.table.
Whenever available, they distribute columns to be computed on multiple threads.
library(data.table)
ans_df <- nafill(df, fill=0)
# or even faster, in-place
setnafill(df, fill=0)

Answer 14

If you want to replace NAs in factor variables, this might be useful:
n <- length(levels(data.vector))+1
data.vector <- as.numeric(data.vector)
data.vector[is.na(data.vector)] <- n
data.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.

Answer 15

dplyr >= 1.0.0
In newer versions of dplyr:
across() supersedes the family of "scoped variants" like summarise_at(), summarise_if(), and summarise_all().
df <- data.frame(a = c(LETTERS[1:3], NA), b = c(NA, 1:3))
library(tidyverse)
df %>%
mutate(across(where(anyNA), ~ replace_na(., 0)))
a b
1 A 0
2 B 1
3 C 2
4 0 3
This code will coerce 0 to be character in the first column. To replace NA based on column type you can use a purrr-like formula in where:
df %>%
mutate(across(where(~ anyNA(.) & is.character(.)), ~ replace_na(., "0")))

Answer 16

No need to use any library.
df <- data.frame(a=c(1,3,5,NA))
df$a[is.na(df$a)] <- 0
df

Answer 17

You can use replace()
For example:
> 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

Answer 18

The cleaner package has an na_replace() generic, that at default replaces numeric values with zeroes, logicals with FALSE, dates with today, etc.:
library(dplyr)
library(cleaner)
starwars %>% na_replace()
na_replace(starwars)
It even supports vectorised replacements:
mtcars[1:6, c("mpg", "hp")] <- NA
na_replace(mtcars, mpg, hp, replacement = c(999, 123))
Documentation: https://msberends.github.io/cleaner/reference/na_replace.html

Answer 19

Another dplyr pipe compatible option with tidyrmethod replace_na that works for several columns:
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:
d$str <- c("string", NA)
myList <- myList[sapply(d, is.numeric)]
df <- d %>% replace_na(myList)

Answer 20

This simple function extracted from Datacamp could help:
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
replace_missings(df, replacement = 0)

Answer 21

An easy way to write it is with if_na from hablar:
library(dplyr)
library(hablar)
df <- tibble(a = c(1, 2, 3, NA, 5, 6, 8))
df %>%
mutate(a = if_na(a, 0))
which returns:
a
<dbl>
1 1
2 2
3 3
4 0
5 5
6 6
7 8

Answer 22

Replace is.na & NULL in data frame.
data frame with colums
A$name[is.na(A$name)]<-0
OR
A$name[is.na(A$name)]<-"NA"
with all data frame
df[is.na(df)]<-0
with replace na with blank in data frame
df[is.na(df)]<-""
replace NULL to NA
df[is.null(df)] <- NA

Answer 23

if you want to assign a new name after changing the NAs in a specific column in this case column V3, use you can do also like this
my.data.frame$the.new.column.name <- ifelse(is.na(my.data.frame$V3),0,1)

Answer 24

I wan to add a next solution which using a popular Hmisc package.
library(Hmisc)
data(airquality)
# imputing with 0 - all columns
# although my favorite one for simple imputations is Hmisc::impute(x, "random")
> dd <- data.frame(Map(function(x) Hmisc::impute(x, 0), airquality))
> str(dd[[1]])
'impute' Named num [1:153] 41 36 12 18 0 28 23 19 8 0 ...
- attr(*, "names")= chr [1:153] "1" "2" "3" "4" ...
- attr(*, "imputed")= int [1:37] 5 10 25 26 27 32 33 34 35 36 ...
> dd[[1]][1:10]
1 2 3 4 5 6 7 8 9 10
41 36 12 18 0* 28 23 19 8 0*
There could be seen that all imputations metadata are allocated as attributes. Thus it could be used later.

Answer 25

This is not exactly a new solution, but I like to write inline lambdas that handle things that I can't quite get packages to do. In this case,
df %>%
(function(x) { x[is.na(x)] <- 0; return(x) })
Because R does not ever "pass by object" like you might see in Python, this solution does not modify the original variable df, and so will do quite the same as most of the other solutions, but with much less need for intricate knowledge of particular packages.
Note the parens around the function definition! Though it seems a bit redundant to me, since the function definition is surrounded in curly braces, it is required that inline functions are defined within parens for magrittr.

Answer 26

This is a more flexible solution. It works no matter how large your data frame is, or zero is indicated by 0 or zero or whatsoever.
library(dplyr) # make sure dplyr ver is >= 1.00
df %>%
mutate(across(everything(), na_if, 0)) # if 0 is indicated by `zero` then replace `0` with `zero`

Answer 27

Another option using sapply to replace all NA with zeros. Here is some reproducible code (data from #aL3xa):
set.seed(7) # for reproducibility
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 V10
#> 1 9 7 5 5 7 7 4 6 6 7
#> 2 2 5 10 7 8 9 8 8 1 8
#> 3 6 7 4 10 4 9 6 8 NA 10
#> 4 1 10 3 7 5 7 7 7 NA 8
#> 5 9 9 10 NA 7 10 1 5 NA 5
#> 6 5 2 5 10 8 1 1 5 10 3
#> 7 7 3 9 3 1 6 7 3 1 10
#> 8 7 7 6 8 4 4 5 NA 8 7
#> 9 2 1 1 2 7 5 9 10 9 3
#> 10 7 5 3 4 9 2 7 6 NA 5
d[sapply(d, \(x) is.na(x))] <- 0
d
#> V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
#> 1 9 7 5 5 7 7 4 6 6 7
#> 2 2 5 10 7 8 9 8 8 1 8
#> 3 6 7 4 10 4 9 6 8 0 10
#> 4 1 10 3 7 5 7 7 7 0 8
#> 5 9 9 10 0 7 10 1 5 0 5
#> 6 5 2 5 10 8 1 1 5 10 3
#> 7 7 3 9 3 1 6 7 3 1 10
#> 8 7 7 6 8 4 4 5 0 8 7
#> 9 2 1 1 2 7 5 9 10 9 3
#> 10 7 5 3 4 9 2 7 6 0 5
Created on 2023-01-15 with reprex v2.0.2
Please note: Since R 4.1.0 you can use \(x) instead of function(x).

Answer 28

in data.frame it is not necessary to create a new column by mutate.
library(tidyverse)
k <- c(1,2,80,NA,NA,51)
j <- c(NA,NA,3,31,12,NA)
df <- data.frame(k,j)%>%
replace_na(list(j=0))#convert only column j, for example
result
k j
1 0
2 0
80 3
NA 31
NA 12
51 0

Answer 29

I used this personally and works fine :
players_wd$APPROVED_WD[is.na(players_wd$APPROVED_WD)] <- 0