Re: [Caml-list] Easy solution in OCaml?

[malc <malc@pulsesoft.com>]

On Fri, 25 Apr 2003, Siegfried Gonzi wrote:

> Hi:
>
> First off: It is not homework. I am 29 and writing my PhD in physics.
> Often I am contemplating whether it would be possible to use OCaml in
> combination with my beloved Bigloo to perform statistical evaluations. I
> am not sure whether there are any out there who /can/ do this
> evaluations with OCaml what you normally would do with Matlab. The
> problem what arises: type system and working against the compiler. In
> Scheme changing a solution from lets say integer-array to double-array
> is easy, but in Clean for example you would have to change all your
> dependencies.
> I often skim over the libraries and came to the conclusion: C, C++,
> OCaml impossible for me to see any elegance; Clean a bit better;
> Bigloo/Scheme: I am not sure here, because everything looks the same
> maybe this is cheating, but I think it looks the most elegant and less
> intimitating from all.
>
>
> Rationale: given a list of 12 month.  I would like to calculate the
> quarterly means and skip any nan. Easy? Yes it is but only on paper and
> in Scheme:
>
> e.g: [1,2,4,-1,45,56,45,56,8]
>
> nan=-1.0
>
> result: [(1+2+3)/3, (45+56)/2, (45+56+8)/3]

let nan = -1

let mean a b c = (a + b + c) / 3

let rec qmean = function
    a :: b :: c :: rest -> mean a b c :: qmean rest
  | [] -> []
  | otherwise -> failwith "malformed list"

let _ =
  let months = [1; 2; 4; -1; 45; 56; 45; 56; 8] in
  let denan = List.map (fun x -> if x = nan then 0 else x) months in
  qmean denan


-- 
mailto:malc@pulsesoft.com

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