How to summarize a list of combination

Question

I have a list of 2 elements' combination like below.

cbnl <- list(
  c("A", "B"), c("B", "A"), c("C", "D"), c("E", "D"), c("F", "G"), c("H", "I"),
  c("J", "K"), c("I", "H"), c("K", "J"), c("G", "F"), c("D", "C"), c("E", "C"),
  c("D", "E"), c("C", "E")
)

I'd like to summarize above list. Expected result is like below list. Order of element in a vector doesn't matter here.

[[1]]
[1] "A" "B"

[[2]]
[1] "C" "D" "E"

[[3]]
[1] "F" "G"

[[4]]
[1] "H" "I"

[[5]]
[1] "J" "K"

(Rule 1) {A, B} is equivalent to {B, A}. To correspond this I think I can do this.

cbnl <- unique(lapply(cbnl, function(i) { sort(i) }))

(Rule 2) {A, B}, {B, C} (One of element is common) then take a union of two sets. It results {A, B, C}. I don't have clear nice idea to do this.

Any efficient way to do this?

Answer 1

I know this answer is more like a traditional programming rather than "R like" but it solves the issue.

cbnl <- unique(lapply(cbnl, sort))

i <- 1
count <- 1
out <- list()

while (i <= length(cbnl) - 1) {
  if (sum(cbnl[[i]] %in% cbnl[[i + 1]]) == 0) {
    out[[count]] <- cbnl[[i]]
    } else {
      out[[count]] <- sort(unique(c(cbnl[[i]], cbnl[[i + 1]])))
      i <- i + 1        
    }
  count <- count + 1
  i <- i + 1 
}

out

gives,

[[1]]
[1] "A" "B"

[[2]]
[1] "C" "D" "E"

[[3]]
[1] "F" "G"

[[4]]
[1] "H" "I"

[[5]]
[1] "J" "K"