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How to create sequential index value for binary column assigning 0 to FALSE values in data.table object in R?
To create sequential index value for binary column assigning 0 to FALSE values in data.table object in R, we can follow the below steps: −
First of all, create a data.table object with binary column.
Then, use rle function along with sequence and lengths function to create sequential index column.
Example
Create the data.table object
Let’s create a data.table object as shown below −
library(data.table) x<-sample(c(TRUE,FALSE),25,replace=TRUE) DT<-data.table(x) DT
Output
On executing, the above script generates the below output(this output will vary on your system due to randomization) −
x
1: FALSE
2: TRUE
3: FALSE
4: FALSE
5: FALSE
6: TRUE
7: FALSE
8: FALSE
9: FALSE
10: FALSE
11: TRUE
12: TRUE
13: FALSE
14: FALSE
15: TRUE
16: TRUE
17: FALSE
18: TRUE
19: TRUE
20: TRUE
21: TRUE
22: TRUE
23: TRUE
24: FALSE
25: TRUE
xCreate sequential index column
Using rle function along with sequence and lengths function to create sequential index column for column x in data.table object DT −
library(data.table) x<-sample(c(TRUE,FALSE),25,replace=TRUE) DT<-data.table(x) DT$Index<-with(rle(DT$x),sequence(lengths)*DT$x) DT
Output
x Index
1: FALSE 0
2: TRUE 1
3: FALSE 0
4: FALSE 0
5: FALSE 0
6: TRUE 1
7: FALSE 0
8: FALSE 0
9: FALSE 0
10: FALSE 0
11: TRUE 1
12: TRUE 2
13: FALSE 0
14: FALSE 0
15: TRUE 1
16: TRUE 2
17: FALSE 0
18: TRUE 1
19: TRUE 2
20: TRUE 3
21: TRUE 4
22: TRUE 5
23: TRUE 6
24: FALSE 0
25: TRUE 1
x Index
Updated on: 09-Nov-2021
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