padding sequence in R - r

Question

I would like to create left padding column depending on the max number of element of a column in an efficient way.
Let me detail the input and output
My input data is this:
input <- data.frame(path=c("2","4,3,4","3,1"))
input
path
1 2
2 4,3,4
3 3,1
expected output:
V1 V2 V3
1 0 0 2
2 4 3 4
3 0 3 1
The logic base on the input is:
1 - We look at the row having the maximum element ( on this example it is 3 because we have 4,3,4 in row number 2) we now know that we will need to create 3 column as output
2 - for the row that doesn't have 3 element we backfill with 0.
Since the first row has only 1 element i need to put two column with 0
for the second row i can directly fill all columns because we have three elements,for the last row i have 2 elements so i need to backfill one 0.
My attempt which does the work but is dirty
input$path <- as.character(input$path)
lst <- strsplit(input$path, ",")
column_to_create <- max(lengths(lst))
output <- list()
i <- 1
for(i in 1:length(lst)){
if (length(lst[[i]]) < column_to_create) {
nb_create <- column_to_create - length(lst[[i]])
output[[i]] <- c(rep(0,nb_create),lst[[i]])
}
else{
output[[i]] <- lst[[i]]
}
}
output <- lapply(output,as.numeric)
do.call(rbind,output)
[,1] [,2] [,3]
[1,] 0 0 2
[2,] 4 3 4
[3,] 0 3 1

Answer 1

Here's a tidyverse solution
library(tidyverse)
input %>%
separate(path, into=c("V1","V2","V3"), ",", fill="left") %>%
replace(is.na(.), 0)
separate by , into three columns, fill to the left if not enough pieces, replace NA with 0s
Output
V1 V2 V3
1 0 0 2
2 4 3 4
3 0 3 1
----------------------------------------------------------------------------------------
Generalizing for unknown number of columns
Iterate through each row, determine length of vector after strsplit, and save the max as num.cols. paste new column names
num.cols <- max(sapply(1:nrow(input), function(x) length(unlist(strsplit(as.character(input$path[x]), ",")))))
new.cols <- paste0("V", 1:num.cols)
Now you can use new.cols to define your column names
input %>%
separate(path, into=new.cols, ",", fill="left") %>%
replace(is.na(.), 0)