Re: Re: newbie is looking for a customDistribution function

*To*: mathgroup at smc.vnet.net*Subject*: [mg50452] Re: [mg50435] Re: newbie is looking for a customDistribution function*From*: DrBob <drbob at bigfoot.com>*Date*: Sat, 4 Sep 2004 01:43:33 -0400 (EDT)*References*: <ch3o86$t96$1@smc.vnet.net> <ch6nlk$2d5$1@smc.vnet.net> <200409030736.DAA15603@smc.vnet.net>*Reply-to*: drbob at bigfoot.com*Sender*: owner-wri-mathgroup at wolfram.com

>> However, if you really want to use the distribution instead of the data >> that gave rise to it then you should look into the "Alias Method" of >> generating random observations from an arbitary discrete distribution. How would we look into that? Is that in the Mathematica help files, math books, what? Bobby On Fri, 3 Sep 2004 03:36:14 -0400 (EDT), Ray Koopman <koopman at sfu.ca> wrote: > koopman at sfu.ca (Ray Koopman) wrote in message > news:<ch6nlk$2d5$1 at smc.vnet.net>... >> János <janos.lobb at yale.edu> wrote in message news:<ch3o86$t96$1 at smc.vnet.net>... >> [...] >>> 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. / >> >> For a 4-element alphabet, kcombs will have 4^k terms. >> If k = "a few hundred", kcombs will be too big. >> Why not just sort and count the k-sequences in the data? >> >> In[1]:= data = Table[Random[Integer,{1,4}],{100}] >> >> Out[1]= {2,4,3,3,3,4,3,2,3,3,1,3,2,2,4,1,4,4,4,1,2,3,3,4,1, >> 2,1,4,1,1,2,2,4,3,3,1,2,4,2,3,4,2,2,2,3,4,3,4,3,2, >> 2,3,3,3,1,3,3,1,3,1,1,1,1,4,2,2,3,4,2,4,3,4,3,1,4, >> 4,3,4,4,1,3,2,1,2,4,2,4,1,1,2,3,2,4,3,1,4,3,4,4,1} >> >> In[2]:= With[{k = 3}, Reverse /@ Reverse@Sort@Map[{Length[#],#[[1]]}&, >> Split@Sort[FromDigits/@Partition[data,k,1]]]] >> >> Out[2]= {{434, 4}, {343, 4}, {331, 4}, {243, 4}, {441, 3}, {313, 3}, >> {234, 3}, {233, 3}, {223, 3}, {433, 2}, {432, 2}, {431, 2}, >> {424, 2}, {422, 2}, {412, 2}, {411, 2}, {344, 2}, {342, 2}, >> {334, 2}, {333, 2}, {322, 2}, {314, 2}, {242, 2}, {241, 2}, >> {224, 2}, {144, 2}, {132, 2}, {124, 2}, {123, 2}, {112, 2}, >> {111, 2}, {444, 1}, {443, 1}, {423, 1}, {414, 1}, {413, 1}, >> {341, 1}, {324, 1}, {323, 1}, {321, 1}, {312, 1}, {311, 1}, >> {232, 1}, {222, 1}, {214, 1}, {212, 1}, {143, 1}, {142, 1}, >> {141, 1}, {133, 1}, {131, 1}, {122, 1}, {121, 1}, {114, 1}} > > Having read the other replies, I see that I missed your question, > which is how to generate a random observation from the distribution > of k-tuples in the observed data. By far the easiest way is to take > a random k-tuple from the original data: > > Take[data,{1,k}+Random[Integer,Length@data-k]] > > However, if you really want to use the distribution instead of the data > that gave rise to it then you should look into the "Alias Method" of > generating random observations from an arbitary discrete distribution. > > > -- DrBob at bigfoot.com www.eclecticdreams.net

**References**:**Re: newbie is looking for a customDistribution function***From:*koopman@sfu.ca (Ray Koopman)