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MathGroup Archive 1999

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Re: Re: Fast List-Selection

  • To: mathgroup at smc.vnet.net
  • Subject: [mg19944] Re: [mg19925] Re: [mg19880] Fast List-Selection
  • From: "Carl K.Woll" <carlw at fermi.phys.washington.edu>
  • Date: Wed, 22 Sep 1999 04:11:22 -0400
  • Organization: Department of Physics
  • References: <1046AA21E07@gauss.cam.wits.ac.za>
  • Sender: owner-wri-mathgroup at wolfram.com

Hi all,

I've come up with another variation of my (and Rob Pratt's) idea:

rep[li_,n_]:=Flatten at Position[SameQ @@@ Partition[li, n, 1], True]

Compared to Pratt's version, it is nearly twice as fast on my machine.

On a side note, I don't believe Knophmacher's function returns the correct result for
all cases. At least his result disagreed with that returned by the others on my test
cases.

Carl Woll
Physics Dept
U of Washington

Arnold Knopfmacher wrote:

> Some Timings for Andrzejs functions applied to the list
>
> l=Table[Random[Integer,{0,9}],{100000}];
>
> findsequence[3][l] // Timing
> {3.51 Second,{{32432}}}
>
> findsequence[4][l] // Timing
> {2.58 Second,32432}
>
> findsequence[5][l] // Timing
> {5.27 Second,32432}
>
> Carl Wolls function:
>
> rep[ls_,n_]:=Position[Partition[ls,n,1],{x_ ..}]
>
> rep[l,6]//Timing
> {2.47 Second,{{32432}}}
>
> Rob Pratts function
>
> Consec[l_,n_]:=
>   Flatten[Position[Partition[l,n,1],Table[x_,{n}]]]
>
> Consec[l,6]//Timing
> {2.14 Second,{32432}}
>
> My function  is still faster (at least for this test list and under Mathematica
> 3)
>
> dif[s_]:=Drop[s,1]-Drop[s,-1];
> nconsecB[s_,n_]:=
>   Module[{ss=Flatten[Position[dif[s],0]],ans={}},
>     Do[If[ss[[i+n-2]]-ss[[i]]==n-2,ans={ans,ss[[i]]}],{i,Length[ss]-n+2}];ans]
>
> nconsecB[l,6]//Timing
> {1.81 Second,{32432}}
>
> Perhaps my function can be rewritten in a more elegant form?
>
> Arnold Knopfmacher
> Witwatersrand University
> South Africa
>
> > Date:          Tue, 21 Sep 1999 02:22:54 -0400
> > Subject: [mg19944]       [mg19925] Re: [mg19880] Fast List-Selection
> > From:          "Andrzej Kozlowski" <andrzej at tuins.ac.jp>
To: mathgroup at smc.vnet.net
> > To:            mathgroup at smc.vnet.net
>
> > I have managed to produce four additional examples, three  of which seem to
> > be faster than yours. I will change the search to 6 rather than 7
> > consecutive identical elements because I want to use for my testing a famous
> > example:
> >
> > In[1]:=
> > l = RealDigits[N[Pi, 10000]][[1]];
> >
> > Hear are the functions I will test, beginning with yours:
> >
> > In[2]:=
> > findsequence[1][l_] := Do[If[Count[t = Take[l, {i, i + 5}], t[[1]]] == 6,
> > Print[i]], {i, 1, Length[l] - 5}]
> >
> > In[3]:=
> > findsequence[2][l_] :=
> >   Position[l /. {a___, y_, y_, y_, y_, y_, y_, b___} -> {a, mark, b}, mark,
> > 1]
> >
> > In[4]:=
> > findsequence[3][l_] :=
> >   Module[{m = Split[l], mark},
> >     Position[Flatten[m /. Cases[m, _?(Length[#] == 6 &)][[1]] -> mark],
> > mark,
> >       1]]
> >
> > In[5]:=
> > findsequence[4][l_] :=
> >   Module[{m = Split[l]},
> >     Length[Flatten[
> >           Take[m, Position[m, Select[m, Length[#] == 6 &][[1]]][[1, 1]] -
> >               1]]] + 1]
> >
> > In[6]:=
> > f[x_, x_, x_, x_, x_, x_] := 0;
> > f[y__] := 1;
> > g[l_, i_] := f[Apply[Sequence, Take[l, {i, i + 5}]]];
> > findsequence[5][l_] := Scan[If[g[l, #] == 0, Return[#]] &,
> > Range[Length[l]]];
> >
> > Now the test:
> >
> > In[9]:=
> > Table[findsequence[i][l] // Timing, {i, 1, 5}]
> > 763
> > Out[9]=
> > {{1.38333 Second,
> >     Null}, {2.18333 Second, {{763}}}, {0.783333 Second, {{763}}}, {0.683333
> > Second, 763}, {0.183333 Second, 763}}
> >
> > The last one wins by a big margin The programs using Split may do better on
> > other machines (I am using Mac PowerBook G3 ,233 mghz) because there seems
> > to be something wrong with Split on the Mac where it doesn't scale linearly
> > with the size of the input.
> >
> > Finally: In[10]:=
> > Take[l, {763, 763 + 5}]
> > Out[10]=
> > {9, 9, 9, 9, 9, 9}
> > --
> > Andrzej Kozlowski
> > Toyama International University
> > JAPAN
> > http://sigma.tuins.ac.jp
> > http://eri2.tuins.ac.jp
> >
> >
> > ----------
> > >From: Hans Havermann <haver at total.net>
To: mathgroup at smc.vnet.net
> > To: mathgroup at smc.vnet.net
> > >To: mathgroup at smc.vnet.net
> > >Subject: [mg19944] [mg19925] [mg19880] Fast List-Selection
> > >Date: Mon, Sep 20, 1999, 7:47 AM
> > >
> >
> > > I have a list 's' composed of a large number of (small) integers. I wish to
> > > search this list for instances of 7 consecutive, identical elements.
> > >
> > > My approach is:
> > >
> > > Do[If[Count[t = Take[s, {i, i + 6}], t[[1]]] == 7,
> > >     Print[i]], {i, 1, Length[s] - 6}]
> > >
> > > Can anyone think of a *faster* way of doing this?
> > >
> > >
> > >
> >
> > --
> > Andrzej Kozlowski
> > Toyama International University
> > JAPAN
> > http://sigma.tuins.ac.jp
> > http://eri2.tuins.ac.jp
> >
> >
> > ----------
> > >From: Hans Havermann <haver at total.net>
To: mathgroup at smc.vnet.net
> > To: mathgroup at smc.vnet.net
> > >To: mathgroup at smc.vnet.net
> > >Subject: [mg19944] [mg19925] [mg19880] Fast List-Selection
> > >Date: Sun, 19 Sep 1999 18:47:32 -0400
> > >
> >
> > > I have a list 's' composed of a large number of (small) integers. I wish to
> > > search this list for instances of 7 consecutive, identical elements.
> > >
> > > My approach is:
> > >
> > > Do[If[Count[t = Take[s, {i, i + 6}], t[[1]]] == 7,
> > >     Print[i]], {i, 1, Length[s] - 6}]
> > >
> > > Can anyone think of a *faster* way of doing this?
> > >
> > >
> > >
> >
> >



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