Re: Trying to eliminate a loop

*To*: mathgroup at smc.vnet.net*Subject*: [mg50604] Re: [mg50595] Trying to eliminate a loop*From*: DrBob <drbob at bigfoot.com>*Date*: Sun, 12 Sep 2004 04:42:15 -0400 (EDT)*References*: <200409111045.GAA19926@smc.vnet.net>*Reply-to*: drbob at bigfoot.com*Sender*: owner-wri-mathgroup at wolfram.com

I haven't gotten to the loop yet, but Timing[pr=Flatten[First[Extract[freq[[All,2]],Split[Position[freq,Max[freq[[\ All,1]]]][[All,1]]]]]]] {0.031 Second,{a,c,g,t,t,t,t}} could be replaced with Timing[pr=First@freq[[Ordering[freq,-1],-1]]] {0. Second,{t,t,a,t,a,t,c}} When I generated a simple, there were 11 ties for the maximum; your code finds the first, and mine finds the last. Given that the sample is random, I doubt you care which one you work with, and Ordering is faster. Bobby On Sat, 11 Sep 2004 06:45:09 -0400 (EDT), JÃ¡nos <janos.lobb at yale.edu> wrote: > Hi, > > I have a four letter alphabet: > In[3]:= > baseAlphabet={a,t,c,g} > > I can create an arbitrary length list from it with: > > generateRandomStrand[alphabet_List, len_Integer] := \ > Table[alphabet[[Random[Integer, {1, Length[alphabet]}] ]], {len}]; > > for example: > > ls = generateRandomStrand[baseAlphabet, 23460]; > > I want to know what kind of primerLength=7 sub-strands are in it, so I > partition it: > > lspr = Partition[ls, primerLength, 1]; > > All the possible sub-strands form a set: > > primerSet=Distribute[Table[baseAlphabet, {primerLength}], List]; > > << Statistics`DataManipulation` > freq=Frequencies[lspr]; > gives me a frequency distribution, that is what elements of primerSet > occur at what frequency in lspr: > > To know which sublist occurs the most I do: > In[64]:= > pr=Flatten[First[Extract[freq[[All, 2]], Split[Position[freq, > Max[freq[[All, > 1]]]][[All, 1]] ]] ]] > > Out[64]= > {a,a,c,t,g,c,g} > > and it is at positions > > In[65]:= > prpos=Position[lspr,pr] > > Out[65]= > {{2860},{4336},{6791},{11387},{12164},{17472},{17833},{17954}} > > in lspr and in ls. > > At those positions in ls I want to attach the complement of this pr > sublist, so I create the following rules: > > complementRule = {a -> t, c -> g, t -> a, g -> c}; > replaceWithComplement = {a -> {a, t}, c -> {c, g}, t -> {t, a}, g -> > {g, c}} > > I created a For loop which at prpos replaces the elements there with > the double elements indicated by the rule above on primerLength > intervals of ls: > > For[i = 1, i Â² Length[pr], i++, ls = ReplacePart[ls, > replaceWithComplement, prpos + i - 1, \ > Flatten[Position[replaceWithComplement, {Part[Extract[ls, > prpos + i - 1], 1], Part[Extract[ > ls, prpos + i - 1], 1] /. complementRule}]]] ] > > Mathematica does this For loop in about 0.046678 Second on my machine. > With primerLength=9 it is 0.050748 Second - pr had just three positions > on the strand. > > I have the feeling that it can be done faster with Map or MapAt, so the > For loop could go away. I also do not like that I have to rewrite ls > in every cycle. ReplacePart does not look good to me in this > situation, but I have not find yet the way to apply the > replaceWithComplement rule directly to the primerLength long intervals > of ls at prpos positions. > > Any good tip ? > > Thanks ahead, > > Jâ?¡nos > > > ---------------------------------------------- > Trying to argue with a politician is like lifting up the head of a > corpse. > (S. Lem: His Master Voice) > > > -- DrBob at bigfoot.com www.eclecticdreams.net

**References**:**Trying to eliminate a loop***From:*János <janos.lobb@yale.edu>