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Why does Seq.tl not exist when Seq.hd does?

I can do:

let tail (xs : seq<'a>) = xs |> Seq.to_list |> List.tl |> List.to_seq;;

but that just seems inefficient as there's probably an O(n) hit for to_list/to_seq.

On the other hand, there seems to be no way of writing:

let tail (xs : list 'a) =
match xs with
| xs when (xs == []) -> failwith "tail on an empty list"
| head :: tail -> tail;;

Something like:

let tail (xs : seq 'a) =

match xs with

| xs when (Seq.is_empty xs) -> failwith "tail on an empty seq"

| head @@ tail -> tail;;

might work provided one knew @@ is.

What should one do?

--Sanatan

Note that there is no Seq.tl due to performance/semantics, see

and

for more details.

By brianmcn on 1/9/2009 12:05 PM (permalink)

Hi Sanatan,

You can use Seq.skip. Here's an example:

let s = Seq.init_finite 10 (fun x -> x)

let t = s |> Seq.skip 1

You can also create your own Seq.tl function:

module Seq = let tl (s:'a seq) = Seq.skip 1 s

And then use it like so:

let yy = Seq.tl s

Hope this helps.

-Murat

By Murat on 1/9/2009 10:42 AM (permalink)

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