Applying a function to each row of a data.table
data.table, r
Solution
How about :
x
a b
1: 1 12 13
2: 2 14 15
3: 3 16 17
4: 1 18 19
x[,list(a=rep(a,each=2), V1=unlist(strsplit(b," ")))]
a V1
1: 1 12
2: 1 13
3: 2 14
4: 2 15
5: 3 16
6: 3 17
7: 1 18
8: 1 19
Generalized solution given comment :
x[,{s=strsplit(b," ");list(a=rep(a,sapply(s,length)), V1=unlist(s))}]
Problem
I looking for a way to efficiently apply a function to each row of data.table. Let's consider the following data table: ``` library(data.table) library(stringr) x <- data.table(a = c(1:3, 1), b = c('12 13', '14 15', '16 17', '18 19')) > x a b 1: 1 12 13 2: 2 14 15 3: 3 16 17 4: 1 18 19 ``` Let's say I want to split each element of column `b` by space (thus yielding two rows for each row in the original data) and join the resulting data tables. For the example above, I need the following result: ``` a V1 1: 1 12 2: 1 13 3: 2 14 4: 2 15 5: 3 16 6: 3 17 7: 1 18 8: 1 19 ``` The following would work if column `a` has only unique values: ``` x[, list(str_split(b, ' ')[[1]]), by = a] ``` The following almost works (unless there are some identical rows in the original data table), but is ugly when `x` has many columns and copies column b to the result, which I would like to avoid. ``` > x[, list(str_split(b, ' ')[[1]]), by = list(a,b)] a b V1 1: 1 12 13 12 2: 1 12 13 13 3: 2 14 15 14 4: 2 14 15 15 5: 3 16 17 16 6: 3 16 17 17 7: 1 18 19 18 8: 1 18 19 19 ``` What would be the most efficient and idiomatic way to solve this problem?