Set NA to 0 in R - r

Question

After merging a dataframe with another im left with random NA's for the occasional row. I'd like to set these NA's to 0 so I can perform calculations with them.
Im trying to do this with:
bothbeams.data = within(bothbeams.data, {
bothbeams.data$x.x = ifelse(is.na(bothbeams.data$x.x) == TRUE, 0, bothbeams.data$x.x)
bothbeams.data$x.y = ifelse(is.na(bothbeams.data$x.y) == TRUE, 0, bothbeams.data$x.y)
})
Where $x.x is one column and $x.y is the other of course, but this doesn't seem to work.

Answer 1

You can just use the output of is.na to replace directly with subsetting:
bothbeams.data[is.na(bothbeams.data)] <- 0
Or with a reproducible example:
dfr <- data.frame(x=c(1:3,NA),y=c(NA,4:6))
dfr[is.na(dfr)] <- 0
dfr
x y
1 1 0
2 2 4
3 3 5
4 0 6
However, be careful using this method on a data frame containing factors that also have missing values:
> d <- data.frame(x = c(NA,2,3),y = c("a",NA,"c"))
> d[is.na(d)] <- 0
Warning message:
In `[<-.factor`(`*tmp*`, thisvar, value = 0) :
invalid factor level, NA generated
It "works":
> d
x y
1 0 a
2 2 <NA>
3 3 c
...but you likely will want to specifically alter only the numeric columns in this case, rather than the whole data frame. See, eg, the answer below using dplyr::mutate_if.

Answer 2

A solution using mutate_all from dplyr in case you want to add that to your dplyr pipeline:
library(dplyr)
df %>%
mutate_all(funs(ifelse(is.na(.), 0, .)))
Result:
A B C
1 0 0 0
2 1 0 0
3 2 0 2
4 3 0 5
5 0 0 2
6 0 0 1
7 1 0 1
8 2 0 5
9 3 0 2
10 0 0 4
11 0 0 3
12 1 0 5
13 2 0 5
14 3 0 0
15 0 0 1
If in any case you only want to replace the NA's in numeric columns, which I assume it might be the case in modeling, you can use mutate_if:
library(dplyr)
df %>%
mutate_if(is.numeric, funs(ifelse(is.na(.), 0, .)))
or in base R:
replace(is.na(df), 0)
Result:
A B C
1 0 0 0
2 1 <NA> 0
3 2 0 2
4 3 <NA> 5
5 0 0 2
6 0 <NA> 1
7 1 0 1
8 2 <NA> 5
9 3 0 2
10 0 <NA> 4
11 0 0 3
12 1 <NA> 5
13 2 0 5
14 3 <NA> 0
15 0 0 1
Update
with dplyr 1.0.0, across is introduced:
library(dplyr)
# Replace `NA` for all columns
df %>%
mutate(across(everything(), ~ ifelse(is.na(.), 0, .)))
# Replace `NA` for numeric columns
df %>%
mutate(across(where(is.numeric), ~ ifelse(is.na(.), 0, .)))
Data:
set.seed(123)
df <- data.frame(A=rep(c(0:3, NA), 3),
B=rep(c("0", NA), length.out = 15),
C=sample(c(0:5, NA), 15, replace = TRUE))

Answer 3

You can use replace_na() from tidyr package
df %>% replace_na(list(column1 = 0, column2 = 0)

Answer 4

To add to James's example, it seems you always have to create an intermediate when performing calculations on NA-containing data frames.
For instance, adding two columns (A and B) together from a data frame dfr:
temp.df <- data.frame(dfr) # copy the original
temp.df[is.na(temp.df)] <- 0
dfr$C <- temp.df$A + temp.df$B # or any other calculation
remove('temp.df')
When I do this I throw away the intermediate afterwards with remove/rm.

Answer 5

If you only want to replace NAs with 0s for a few select columns you also use an lapply solution, e.g:
data = data.frame(
one = c(NA,0),
two = c(NA,NA),
three = c(1,2),
four = c("A",NA)
)
data[1:2] = lapply(data[1:2],function(x){
x[is.na(x)] = 0
return(x)
})
data

Answer 6

Why not try this
na.zero <- function (x) {
x[is.na(x)] <- 0
return(x)
}
na.zero(df)