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MathGroup Archive 2004

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Re:Re: Counting Runs

  • To: mathgroup at smc.vnet.net
  • Subject: [mg52212] Re:[mg52170] Re: Counting Runs
  • From: "Fred Simons" <f.h.simons at tue.nl>
  • Date: Sun, 14 Nov 2004 20:15:09 -0500 (EST)
  • References: <200411130940.EAA01037@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Dear Carl,

Yes, I remember the problem how to count how many 0's are followed by a 1.
Allan Hayes and you had some wonderful ideas how to solve that.

With respect to your question, it turns out that using 1-Abs[Sign[data-int]]
is slightly faster. But when we use Abs[Sign[data-int] and then count how
may 1's are followed by a 0 there is a considerable gain.

In[1]:=
runs[int_,data_]:=
  Module[{modlist},modlist=Abs@Quotient[#,#+1,1]&@Abs[data-int];
Tr[BitXor[modlist,RotateRight[modlist]]]/2+BitAnd[modlist[[1]],modlist[[-1]]
]]

In[2]:=
runs1[int_,data_]:=Module[{modlist},modlist=1-Abs[Sign[data-int]];
Tr[BitXor[modlist,RotateRight[modlist]]]/2+BitAnd[modlist[[1]],modlist[[-1]]
]]

In[3]:=
runs2[int_,data_] :=
  With[{lst=Abs[Sign[data-int]]},
    Total[ BitXor[lst ,RotateLeft[lst]]]/2+
      If[lst[[1]]\[Equal]1, 0,If[lst[[-1]]\[Equal]0, 1,0]] ]

In[4]:=
data=Table[Random[Integer,20], {2 10^7}];

In[5]:=
n=5;{Do[runs[3, data], {n}]// Timing, Do[runs1[3, data], {n}]// Timing,
  Do[runs2[3, data], {n}]// Timing}
Out[5]=
{{7.516 Second,Null},{6.328 Second,Null},{4.109 Second,Null}}

Regards,

Fred Simons
Eindhoven University of Technology


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