Re: Re: Counting Runs

*To*: mathgroup at smc.vnet.net*Subject*: [mg51994] Re: [mg51934] Re: [mg51890] Counting Runs*From*: DrBob <drbob at bigfoot.com>*Date*: Sat, 6 Nov 2004 02:08:43 -0500 (EST)*References*: <200411040650.BAA18131@smc.vnet.net> <200411050717.CAA06890@smc.vnet.net> <opsg0lmi1fiz9bcq@monster.cox-internet.com>*Reply-to*: drbob at bigfoot.com*Sender*: owner-wri-mathgroup at wolfram.com

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

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

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