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MathGroup Archive 2005

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Re: letrec/named let

  • To: mathgroup at smc.vnet.net
  • Subject: [mg56829] Re: letrec/named let
  • From: DrBob <drbob at bigfoot.com>
  • Date: Sat, 7 May 2005 04:16:43 -0400 (EDT)
  • References: <069a01c55260$afdad760$6400a8c0@Main>
  • Reply-to: drbob at bigfoot.com
  • Sender: owner-wri-mathgroup at wolfram.com

That's a very nice "rescue" of the Ordering@Ordering solution.

Here's a test with (probably) no repeated elements, however, that shows treat ahead. Perhaps there's an intermediate situation where carl wins, or it matters whether the data is integer or real?

data=Table[Random[],{10^6}];
Timing[one=treat@data;]
Timing[three=carl@data;]
one==three

{10.109 Second,Null}
{15.938 Second,Null}
True

Bobby

On Fri, 6 May 2005 13:26:03 -0400, Carl K. Woll <carlw at u.washington.edu> wrote:

> "DrBob" <drbob at bigfoot.com> wrote in message
> news:d5f5ka$6dn$1 at smc.vnet.net...
>> Our solutions agree on lists of strictly positive integers, but timings
>> depend a great deal on the minimum value:
>>
>> Clear[treat, andrzej]
>> treat[s_List] := Module[{u = Union@s}, s /. Dispatch@Thread[u -> Range@
>>     Length@u]]
>> andrzej[s_List] :=
>>      First@NestWhile[Apply[If[FreeQ[##], {#1 /. x_ /;
>>             x > #2 :> x - 1, #2}, {#1, #2 + 1}] &, #] &, {s, 1},
>>     Last[#] < Max[First[#]] &]
>>
>> data=1+RandomArray[PoissonDistribution[2],{10^6}];
>> Timing[one=treat@data;]
>> Timing[two=andrzej@data;]
>> one == two
>>
>> {0.593 Second,Null}
>> {0.657 Second,Null}
>> True
>>
>
> [snip]
>
> Another possibility modeled after DrBob's first answer is the following:
>
> carl[s_] := Module[{ord,t},
>  ord = Ordering[s];
>  t = FoldList[Plus, 1, Sign[Abs[ListCorrelate[{1, -1}, s[[ord]]]]]];
>  t[[Ordering[ord]]]
> ]
>
> If there are a lot of repeated elements in the data, then treat seems to be
> faster. On the other hand, if there aren't a lot of repeated elements, then
> carl seems to be faster. It seems like it ought to be possible to compute
> Ordering[Ordering[data]] more quickly since
> Ordering[Ordering[Ordering[data]]] equals Ordering[data], but I couldn't
> think of a way.
>
> Carl Woll
>
>
>
>
>
>
>



-- 
DrBob at bigfoot.com


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