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MathGroup Archive 2002

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RE: Re: Pattern Matching in Lists

  • To: mathgroup at smc.vnet.net
  • Subject: [mg35609] RE: [mg35586] Re: Pattern Matching in Lists
  • From: "DrBob" <majort at cox-internet.com>
  • Date: Mon, 22 Jul 2002 02:11:12 -0400 (EDT)
  • Reply-to: <drbob at bigfoot.com>
  • Sender: owner-wri-mathgroup at wolfram.com

Selwyn's code is almost always a little faster than Allan's (though not
much).  The difference is around 4% for lists of 2,000,000 entries.

Here are results for thirty trials of the two arithmetic methods, with 2
million entries each time, using a paired t-test:

Needs["Statistics`HypothesisTests`"]
f1[w_]:=Count[Partition[w, 2,1],{1,0}]
f2[w_]:=Count[Drop[w,-1] - Drop[w,1],1]
f3[w_]:=Count[Drop[w+2RotateRight[w],1],2]
trial:=(n=2000000;w = Table[Random[Integer], {n}];
First@Timing[#[w];]/Second&/@{f2,f3}
    )

{t1, t2} = Transpose[(trial & ) /@ Range[30]]; 
Mean /@ {t1, t2}
r = MeanTest[t1 - t2, 0, FullReport -> True]
meanDiff = (FullReport /. r)[[1,1,1]]; 
Print["% difference = ", meanDiff/Mean[t1]]

{0.5864333333333358, 0.5621333333333306}

{FullReport -> TableForm[
    {{"Mean", "TestStat", 
      "Distribution"}, 
     {0.024300000000005186, 
      3.3246821117809677, 
      StudentTDistribution[
       29]}}, 
    TableHeadings -> 
     {None, {"Mean", 
       "TestStat", 
       "Distribution"}}], 
  OneSidedPValue -> 
   0.0012041503843005326}

"% difference = "0.0414369351

That's a 4% difference and a p-value of about 0.1%.

Bobby Treat

-----Original Message-----
From: Allan Hayes [mailto:hay at haystack.demon.co.uk] 
To: mathgroup at smc.vnet.net
Subject: [mg35609] [mg35586] Re: Pattern Matching in Lists 

[second posting in view of reported technical problem]

Anthony,
Take
    w = Table[Random[Integer], {200000}

My first thought was, and several posts used this,

    Count[Partition[w, 2,1],{1,0}]//Timing

        {3.24 Second,49851}

Later it occured to me to use arithmetic, which turned out to be twice
as
fast:

     Count[ Drop[w,-1] - Drop[w,1],1]//Timing

        {1.49 Second,49851}

This is close to Selwyn Hollis's code

    Count[Drop[w+2RotateRight[w],1],2]//Timing

        {1.6 Second,49851}

--
Allan

---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565


"Anthony Mendes" <amendes at zeno.ucsd.edu> wrote in message
news:ah5qce$59o$1 at smc.vnet.net...
> Hello,
>
> Suppose w={1,1,1,0,0,1,0,1,0,0,1,0,0}.
>
> How can I count the number of occurrences of a 1 in w immediately
> followed by a 0 in w?
>
> I have tried every incarnation of Count[] I can think of; for example,
>
> Count[w,{___,1,0,___}]
>
> does not seem to work.  In general, how can I count the number of
> occurrences of a 1 followed by a 0 in a list of 1's and 0's?  Thank
you!
>
>
> --
> Tony
> _____________________
> amendes at math.ucsd.edu
>
>










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