Re: newbie is looking for a customDistribution function

*To*: mathgroup at smc.vnet.net*Subject*: [mg50418] Re: newbie is looking for a customDistribution function*From*: koopman at sfu.ca (Ray Koopman)*Date*: Thu, 2 Sep 2004 04:35:10 -0400 (EDT)*References*: <ch3o86$t96$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

János <janos.lobb at yale.edu> wrote in message news:<ch3o86$t96$1 at smc.vnet.net>... > 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. / 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}}