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How to replace NAs with 0 in an R list that contains data.table objects?
To replace NAs with 0 in an R list that contains data.table objects, we can follow the below steps −
First of all, create a list of data.table objects that contain some NAs.
Then, use lapply function and is.na function to replace NAs with 0 in the list.
Example
Create the list
Let’s create a list as shown below −
library(data.table) DT1<- data.table(x1=sample(c(NA,rpois(5,5)),15,replace=TRUE),x2=sample(c(NA,rpois(5,1)),1 5,replace=TRUE)) DT2< data.table(Y1=sample(c(NA,rpois(2,2)),15,replace=TRUE),Y2=sample(c(NA,rpois(2,5)), 15,replace=TRUE)) List<-list(DT1,DT2) List
Output
On executing, the above script generates the below output(this output will vary on your system due to randomization) −
[[1]]
x1 x2
1: NA 0
2: 6 0
3: 7 0
4: 10 2
5: 7 0
6: 10 0
7: NA 0
8: 7 2
9: 10 NA
10: 7 0
11: 6 0
12: 10 NA
13: 7 0
14: 7 2
15: 6 NA
[[2]]
Y1 Y2
1: NA NA
2: 4 4
3: 4 4
4: NA 3
5: 4 NA
6: NA NA
7: 4 NA
8: 1 4
9: 1 3
10: 1 NA
11: NA 4
12: 1 4
13: 1 NA
14: NA 3
15: NA NAReplace NAs with 0
Using lapply function and is.na function to replace NAs with 0 in List as shown below −
library(data.table)
DT1<-
data.table(x1=sample(c(NA,rpois(5,5)),15,replace=TRUE),x2=sample(c(NA,rpois(5,1)),1
5,replace=TRUE))
DT2<-
data.table(Y1=sample(c(NA,rpois(2,2)),15,replace=TRUE),Y2=sample(c(NA,rpois(2,5)),
15,replace=TRUE))
List<-list(DT1,DT2)
lapply(List, function(x) {x[is.na(x)] <- 0; x})Output
[[1]]
x1 x2
1: 0 0
2: 6 0
3: 7 0
4: 10 2
5: 7 0
6: 10 0
7: 0 0
8: 7 2
9: 10 0
10: 7 0
11: 6 0
12: 10 0
13: 7 0
14: 7 2
15: 6 0
[[2]]
Y1 Y2
1: 0 0
2: 4 4
3: 4 4
4: 0 3
5: 4 0
6: 0 0
7: 4 0
8: 1 4
9: 1 3
10: 1 0
11: 0 4
12: 1 4
13: 1 0
14: 0 3
15: 0 0
Updated on: 16-Nov-2021
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