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
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))