Merge/join seq of seqs

Question

Slowly getting the hang of List matching and tail recursion, I needed a function which 'stitches' a list of lists together leaving off intermediate values (easier to show than explain):

merge [[1;2;3];[3;4;5];[5;6;7]] //-> [1;2;3;4;5;6;7]

The code for the List.merge function looks like this:

///Like concat, but removes first value of each inner list except the first one
let merge lst = 
    let rec loop acc lst = 
        match lst with
        | [] -> acc
        | h::t -> 
            match acc with
            | [] -> loop (acc @ h) t
            | _ -> loop (acc @ (List.tl h)) t //first time omit first value
    loop [] lst

(OK, it's not quite like concat, because it only handles two levels of list)

Question: How to do this for a Seq of Seqs (without using a mutable flag)?

UPDATE (re comment from Juliet): My code creates 'paths' composed of 'segments' which are based on an option type:

type SegmentDef = Straight of float | Curve of float * float
let Project sampleinterval segdefs = //('clever' code here)

When I do a List.map (Project 1.) ListOfSegmentDefs, I get back a list where each segment begins on the same point where the previous segment ends. I want to join these lists together to get a Path, keeping only the 'top/tail' of each overlap - but I don't need to do a 'Set', because I know that I don't have any other duplicates.

Answer 1

This is essentially the same as your first solution, but a little more succinct:

let flatten l =
    seq {
        yield Seq.hd (Seq.hd l) (* first item of first list *)
        for a in l do yield! (Seq.skip 1 a) (* other items *)
    }

[Edit to add]:

If you need a List version of this code, use append |> Seq.to_list at the end of your method:

let flatten l =
    seq {
        yield Seq.hd (Seq.hd l) (* first item of first list *)
        for a in l do yield! (Seq.skip 1 a) (* other items *)
    } |> Seq.to_list