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>
- Counting Runs