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Date: | 2006-07-25 (12:01) |
From: | Christoph Bauer <christoph.bauer@l...> |
Subject: | Re: [Caml-list] generic Hashtbl.to_array |
Hi, > let to_array t = > let init = ref None in > begin try Hashtbl.iter (fun k v -> init := Some (k,v); > raise Exit) t > with Exit -> () > end; > match !init with > | None -> [| |] > | Some i -> > let a = Array.make (Hashtbl.length t) i in > ignore (Hashtbl.fold (fun k v i -> a.(i) <- (k, v); i + 1) t 0); > a > ;; it's curious, but this solution is slower than the others! [skaller's solution seems to be the same, so I include only this one in the "benchmark"] Rate to_array_4 to_array_3 to_array_1b to_array_2 to_array_1 to_array_4 407+-0/s -- -16% -16% -17% -17% to_array_3 486+-2/s 19% -- [-0%] [-1%] -1% to_array_1b 487+-0/s 20% [0%] -- [-0%] -1% to_array_2 489+-2/s 20% [1%] [0%] -- -1% to_array_1 491+-0/s 21% 1% 1% 1% -- from http://ocaml-benchmark.sourceforge.net/doc/Benchmark.html Benchmark.tablulate results prints a comparison table for a list of results obtained by Benchmark.latencyN or Benchmark.throughputN with each function compared to all the others. The table is of the type Rate name1 name2 ... OR s/iter name1 name2 ... name1 #/s -- r12 name1 # -- r12 name2 #/s r21 -- name2 # r21 -- ... ... where name1, name2,... are the labels of the tests sorted from slowest to fastest and rij says how much namei is faster (or slower if < 0) than namej (technically it is equal to (ri - rj) expressed in percents of rj where ri and rj are the rates of namei and namej respectively). If several results are associated to a given name, they are used to compute a Student's statistic to check whether the rates are significantly different. If ri and rj are not believed to be different, rij will be printed between brackets. (* compile with ocamlopt -o to_array -I benchmark-0.7 unix.cmxa benchmark-0.7/benchmark.cmx to_array.ml *) open Benchmark let to_array_1 t = let dummy = Array.init 0 (fun _ -> raise Not_found) in fst (Hashtbl.fold (fun k v (a, i) -> if i = 0 then let a = Array.make (Hashtbl.length t) (k, v) in (a, 0) else (a.(i) <- (k, v); (a, i + 1))) t (dummy, 0)) let to_array_2 t = let init _ = fun () -> raise Not_found in let a = Array.init (Hashtbl.length t) init in ignore (Hashtbl.fold (fun k v i -> a.(i) <- (fun () -> (k, v)); i+1) t 0); Array.map (fun f -> f ()) a let to_array_3 t = Array.of_list (Hashtbl.fold (fun a b c -> (a, b) :: c) t []) let to_array_1b t = let a = ref (Array.init 0 (fun _ -> raise Not_found)) in ignore (Hashtbl.fold (fun k v i -> if i = 0 then (a := Array.make (Hashtbl.length t) (k, v); i) else ((!a).(i) <- (k, v); i + 1)) t 0); !a let to_array_4 t = let init = ref None in begin try Hashtbl.iter (fun k v -> init := Some (k,v); raise Exit) t with Exit -> () end; match !init with | None -> [| |] | Some i -> let a = Array.make (Hashtbl.length t) i in ignore (Hashtbl.fold (fun k v i -> a.(i) <- (k, v); i + 1) t 0); a let h () = let h = Hashtbl.create 100000 in for i = 0 to (Hashtbl.length h) do Hashtbl.add h (Random.int max_int) (Random.int max_int); done; h let main () = let h = h () in let res = throughputN ~repeat:5 1 [("to_array_1", to_array_1, h); ("to_array_1b", to_array_1b, h); ("to_array_2", to_array_2, h); ("to_array_3", to_array_3, h); ("to_array_4", to_array_4, h); ] in tabulate res let () = main ()