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MathGroup Archive 2004

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Re: Re: newbie is looking foracustomDistributionfunction

  • To: mathgroup at smc.vnet.net
  • Subject: [mg50437] Re: [mg50411] Re: [mg50383] newbie is looking foracustomDistributionfunction
  • From: DrBob <drbob at bigfoot.com>
  • Date: Fri, 3 Sep 2004 03:36:27 -0400 (EDT)
  • References: <200409010549.BAA29848@smc.vnet.net> <003d01c49152$37f20450$3f57fea9@Tomas>
  • Reply-to: drbob at bigfoot.com
  • Sender: owner-wri-mathgroup at wolfram.com

No question, a different mapping into the reals gives a different distribution function. But if you've chosen a mapping, you know it both ways, so you know the order you speak of.

Anyway, my solution doesn't depend on any of that. Since the OP wanted to sample from a list based on frequency in the list, this works:

random[a:{__}]:=a[[Random[Integer,{1,Length@a} ]]]

Bobby

On Thu, 02 Sep 2004 20:06:18 -0500, Tomas Garza <tgarza01 at prodigy.net.mx> wrote:

> Quite right. However, implicit in the mapping of a List into the real line
> is the problem of order. If the mapping takes, say, the element {a, a} to 0,
> {a, b} to 1, and {a, c} to 2, etc., what would be the meaning of the
> probability that the result of the experiment takes a value less than or
> equal to 2? (Yes, I realize that once you define a fixed order in twocombs,
> such as that produced by Distribute, this would be interpreted as the
> probability of  {{a, a}, {a, b}, {a,c}}. But this goes well above my head).
>
> Tomas Garza
> ----- Original Message -----
> From: "DrBob" <drbob at bigfoot.com>
To: mathgroup at smc.vnet.net
> To: "Tomas Garza" <tgarza01 at prodigy.net.mx>; <mathgroup at smc.vnet.net>
> Sent: Thursday, September 02, 2004 12:20 PM
> Subject: [mg50437] Re: [mg50411] Re: [mg50383] newbie is looking for
> acustomDistributionfunction
>
>
>> More generally, any mapping of a finite measure space into the extended
> real line implies a measure on the Borel sets, giving rise to a distribution
> function. It isn't difficult to map the values of a List into the reals--to
> Range[Length[theList]], for instance. Given such an embedding, any
> probability measure on the original set implies a distribution function, and
> vice-versa.
>>
>> Anyway, the OP apparently wanted a sampling function, like Random[], which
> doesn't require computing a distribution function at all. He already knew
> (and described in his post) how to calculate relative frequencies.
>>
>> Bobby
>>
>> On Thu, 2 Sep 2004 04:34:45 -0400 (EDT), Tomas Garza
> <tgarza01 at prodigy.net.mx> wrote:
>>
>> >> From a formal point of view, a distribution function is defined on the
>> > reals. You can't, therefore, define it on lists of two, three, or more
>> > symbols. A frequency function, on the other hand, may be defined on any
> set
>> > of events. But, then, it is rather trivial for your problem: once you
> have
>> > your list distpro, just divide each value in the list by the sum of all
> the
>> > values, and that's it. Or, maybe, I misunderstood your question?
>> >
>> > Tomas Garza
>> > Mexico City
>> > ----- Original Message -----
>> > From: "János" <janos.lobb at yale.edu>
To: mathgroup at smc.vnet.net
>> > To: mathgroup at smc.vnet.net
>> > Subject: [mg50437] [mg50411] [mg50383] newbie is looking for a customDistribution
> function
>> >
>> >
>> >> Hi,
>> >>
>> >> I looked for it in the archives, but found none.  I am looking for ways
>> >> to create a custom distribution, which I can call as a function.  Here
>> >> is an example for illustration.  Let's say I have a list created from a
>> >> 4 elements alphabet  {a,b,c,d}:
>> >>
>> >> In[1]:=
>> >> lst={a,a,b,c,a,d,a,c,c,a}
>> >>
>> >> Out[1]=
>> >> {a,a,b,c,a,d,a,c,c,a}
>> >>
>> >> Distribute gives me - thanks David Park - all the two element
>> >> combinations of {a,b,c,d}
>> >>
>> >> In[11]:=
>> >> twocombs=Distribute[Table[{a,b,c,d},{2}],List]
>> >>
>> >> Out[11]=
>> >>
> {{a,a},{a,b},{a,c},{a,d},{b,a},{b,b},{b,c},{b,d},{c,a},{c,b},{c,c},{c,d}
>> >> ,{
>> >>    d,a},{d,b},{d,c},{d,d}}
>> >>
>> >> I can count the occurrence of an element of twocombs in lst with the
>> >> following function:
>> >>
>> >> occuranceCount[x_List] := Count[Partition[lst, 2, 1], x]
>> >>
>> >> Mapping this function over twocombs gives me the number of occurances
>> >> of elements of twocombs in lst:
>> >>
>> >> In[12]:=
>> >> distro=Map[occuranceCount,twocombs]
>> >>
>> >> Out[12]=
>> >> {1,1,1,1,0,0,1,0,2,0,1,0,1,0,0,0}
>> >>
>> >> It shows that for example {c,a} occurs twice, {d,a} occurs once and
>> >> {d,c} or {d,d} never occur.
>> >>
>> >> Now, I would like to create a distribution function called
>> >> twocombsLstDistribution which I could call and it would give me back
>> >> elements of twocombs with the probability as they occur in distro, that
>> >> is for on average I would get twice as much {c,a}s as {d,a}s and never
>> >> get {d.c} or {d,d}.
>> >>
>> >> How can I craft that ?
>> >>
>> >> /Of course I need it for an arbitrary but finite length string lst over
>> >> a fixed length alphabet {a,b,c,d,....} for k-length elements of kcombs,
>> >> and it has to be super fast  :).  My real lst is between 30,000 and
>> >> 70,000 element long over a four element alphabet and I am looking for k
>> >> between 5 and a few hundred. /
>> >>
>> >> Thanks ahead,
>> >> János
>> >> -------------------------------------------------
>> >> People never lie so much as after a
>> >> hunt, during a war or before an election
>> >> - Otto von Bismarck -
>> >>
>> >>
>> >
>> >
>> >
>> >
>>
>>
>>
>> --
>> DrBob at bigfoot.com
>> www.eclecticdreams.net
>>
>
>
>
>
>



-- 
DrBob at bigfoot.com
www.eclecticdreams.net


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