Re: Re: newbie is looking

• To: mathgroup at smc.vnet.net
• Subject: [mg50436] Re: [mg50411] Re: [mg50383] newbie is looking
• From: Tomas Garza <tgarza01 at prodigy.net.mx>
• Date: Fri, 3 Sep 2004 03:36:23 -0400 (EDT)
• References: <200409010549.BAA29848@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```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
Subject: [mg50436] 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: [mg50436] [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. /
> >>
> >> 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
>

```

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