Re: Re: Counting Runs

*To*: mathgroup at smc.vnet.net*Subject*: [mg51995] Re: [mg51934] Re: [mg51890] Counting Runs*From*: János <janos.lobb at yale.edu>*Date*: Sat, 6 Nov 2004 02:08:47 -0500 (EST)*References*: <200411040650.BAA18131@smc.vnet.net> <200411050717.CAA06890@smc.vnet.net> <opsg0lmi1fiz9bcq@monster.cox-internet.com> <opsg0pe8woiz9bcq@monster.cox-internet.com>*Sender*: owner-wri-mathgroup at wolfram.com

It must be machine or OS dependent. I re-discovered Hanlon3 method :) and ran it with Bobby's newest. I don't have Bobby's data so I generated random didgits in the 0-9 range Here are the results: In[28]:= v = Table[Random[Integer, {0, 9}], {i, 1, 10^7}]; In[29]:= Timing[({First[#1], Length[#1]} & ) /@ Split[Sort[First /@ Split[v]]]] Out[29]= {35.58*Second, {{0, 898901}, {1, 899397}, {2, 901191}, {3, 899449}, {4, 900824}, {5, 900262}, {6, 899338}, {7, 900293}, {8, 900196}, {9, 901311}}} In[32]:= Timing[({First[#1], Length[#1]} & ) /@ Split[Sort[Split[v][[All, 1]]]]] Out[32]= {38.67999999999998*Second, {{0, 898901}, {1, 899397}, {2, 901191}, {3, 899449}, {4, 900824}, {5, 900262}, {6, 899338}, {7, 900293}, {8, 900196}, {9, 901311}}} My machine is a 1.25Ghz G4 with 2G Ram and with OSX 10.3.5. János On Nov 5, 2004, at 7:38 PM, DrBob wrote: > I found an even faster (rather obvious) solution: > > hanlonTreat[v_] := {First@#, Length@#} & /@ Split@Sort[Split[v][[All, > 1]]] > > It about 80% faster than hanlon4. > > Bobby > > On Fri, 05 Nov 2004 17:16:56 -0600, DrBob <drbob at bigfoot.com> wrote: > >> I timed the posted methods except Andrzej's -- it's the only one that >> works only for +1/-1 data -- plus a couple of my own that I haven't >> posted. David Park's method seems the same as the fastest method, >> hanlon3. I modified all methods to return a pair {x, number of runs >> in x} for each x in the data. >> >> Two of Bob Hanlon's methods beat all the rest of us -- but one of his >> is the slowest method, too. >> >> I've posted a notebook at the Run Counts link at: >> >> http://eclecticdreams.net/DrBob/mathematica.htm >> >> Bobby >> >> On Fri, 5 Nov 2004 02:17:54 -0500 (EST), Selwyn Hollis >> <sh2.7183 at misspelled.erthlink.net> wrote: >> >>> Hi Greg, >>> >>> The following seems to work pretty well: >>> >>> runscount[lst_?VectorQ] := >>> Module[{elems, flips, counts}, >>> elems = Union[lst]; >>> flips = Cases[Partition[lst, 2, 1], {x_, y_} /; x =!= y]; >>> counts = {#, Count[Most[flips], {#, _}]} & /@ elems; >>> {x1, x2} = Last[flips]; >>> counts /. {{x1, y_} -> {x1, y+1}, {x2, y_} -> {x2, y+1}}] >>> >>> Example: >>> >>> Table[Random[Integer, {1, 5}], {20}] >>> runscount[%] >>> >>> {2, 2, 3, 1, 3, 2, 2, 3, 1, 1, 2, 3, 1, 1, 3, 1, 1, 2, 2, 2} >>> >>> {{1, 4}, {2, 4}, {3, 5}} >>> >>> >>> ----- >>> Selwyn Hollis >>> http://www.appliedsymbols.com >>> (edit reply-to to reply) >>> >>> >>> On Nov 4, 2004, at 1:50 AM, Gregory Lypny wrote: >>> >>>> Looking for an elegant way to count runs to numbers in a series. >>>> Suppose I have a list of ones and negative ones such as >>>> v={1,1,1,-1,1,1,1,1,1,-1,-1,-1,-1,1}. >>>> I'd like to create a function that counts the number of runs of 1s >>>> and >>>> -1s, which in this case is 3 and 2. >>>> >>>> Greg >>>> >>>> >>> >>> >>> >>> >> >> >> > > > > -- > DrBob at bigfoot.com > www.eclecticdreams.net > > ------------------------------------------------------------------- János Löbb Yale University School of Medicine Department of Pathology Phone: 203-737-5204 Fax: 203-785-7303 E-mail: janos.lobb at yale.edu

**References**:**Counting Runs***From:*Gregory Lypny <gregory.lypny@videotron.ca>

**Re: Counting Runs***From:*Selwyn Hollis <sh2.7183@misspelled.erthlink.net>