Re: Re: Re: Counting Runs

```But it's very easy to adapt my method to general data:

v = Table[Random[Integer,{0,3}],{20}]

{0,2,2,1,2,1,1,1,1,1,3,2,3,2,3,3,0,3,3,1}

With[{u=Split[v],w=Union[v]},Map[Count[u,{Repeated[#]}]&,w]]

{2,3,4,4}

Andrzej

On 6 Nov 2004, at 16:08, DrBob 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
>>
>>
>> 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
>

```

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