Interleave lists in R

While investigating a similar question, I came across this beautiful solution by Gabor Grothendieck (i.e. @GGrothendieck?) for certain cases:

c(rbind(a,b))

This works equally well when a and b are both lists, or when a and b are both vectors. It's not a precise solution to OP's question, because when a and b have different lengths, it will recycle the elements of the shorter sequence, printing a warning. However, since this solution is simple and elegant, and provides an answer to a very similar question--a question of some people (like me) who find their way to this page as a result--it seemed worth adding as an answer.

Here's one way:

idx <- order(c(seq_along(a), seq_along(b)))
unlist(c(a,b))[idx]

# [1] "a.1" "b.1" "a.2" "b.2" "a.3" "b.3" "b.4"

As @James points out, since you need a list back, you should do:

(c(a,b))[idx]

interleave(a, b)

# unlist(interleave(a, b))
# [1] "a.1" "b.1" "a.2" "b.2" "a.3" "b.3" "b.4"


interleave <- function(a, b) { 

  shorter <- if (length(a) < length(b)) a else b
  longer  <- if (length(a) >= length(b)) a else b

  slen <- length(shorter)
  llen <- length(longer)


  index.short <- (1:slen) + llen
  names(index.short) <- (1:slen)

  lindex <- (1:llen) + slen
  names(lindex) <- 1:llen


  sindex <- 1:slen
  names(sindex) <- 1:slen

  index <- c(sindex, lindex)
  index <- index[order(names(index))]

  return(c(a, b)[index])

}