Re: list manipulation

*To*: mathgroup at smc.vnet.net*Subject*: [mg100138] Re: [mg100122] list manipulation*From*: Leonid Shifrin <lshifr at gmail.com>*Date*: Tue, 26 May 2009 05:08:45 -0400 (EDT)*References*: <200905191101.HAA07313@smc.vnet.net> <gv0gn0$a19$1@smc.vnet.net>

Hi Scot, I would not call your solutions clumsy. I think they are brief, reasonably fast and accomplish their goal. Note that your second solution won't work in the way you put it - you also need Replace, otherwise the matching will happen on a higher level - f will be applied to the 4-th and 5-th sublists of your main list (you did not observe it since you only had 2 sublists). Also, I think you need there RuleDelayed rather than Rule: In[1] = data = {{element11, element12, element1N, target11, target12, element1N3, element1N4}, {element21, element22, element2N, target21, target22, element2N3, element2N4}}; In[2] = Replace[data, {a : Repeated[_, {3}], b : Repeated[_, {2}], c : __} :> {a, f[b], c}, {1}] Out[2] = {{element11, element12, element1N, f[target11, target12], element1N3, element1N4}, {element21, element22, element2N, f[target21, target22], element2N3, element2N4}} Note also that they are not equivalent: your first one creates f[{a,b}] while the second one just f[a,b]. Here is a little more obscure but faster (about 4 x times, according to my benchmarks) way to accomplish your goal (assuming the format of your second solution): In[3] = Module[{copy = data}, copy[[All, 4]] = f @@@ copy[[All, {4, 5}]]; copy[[All, 5]] = Sequence[]; copy] Out[3] = {{element11, element12, element1N, f[target11, target12], element1N3, element1N4}, {element21, element22, element2N, f[target21, target22], element2N3, element2N4}} Hope this helps. Regards, Leonid On Mon, May 25, 2009 at 3:18 AM, Scot T. Martin <smartin at seas.harvard.edu>wrote: > I'm wondering if anyone has bright ideas on the puzzle below. I'm trying > to take a very large list of the following form: > > {{.., .., a, b, ..},{.., .., c, d, ..},..} > > into the form > > {{.., .., f[a, b], ..},{.., .., f[c, d], ..},..} > > It seems to me that there must be some clever and beautiful way to do this > that I haven't thought of. My code is below with two clumsy solutions. > > Suggestions on how to do better? > > My code: > > In[1]:= (*shown here as minimal to define problem but actual list has \ > large N and many entries*) > > data = {{element11, element12, element1N, target11, target12, > element1N3, element1N4}, {element21, element22, element2N, > target21, target22, element2N3, element2N4}}; > > > > In[2]:= (* first clumsy approach *) > > Replace[data, list : _List :> {Sequence @@ list[[1 ;; 3]], > Sequence[f[list[[4 ;; 5]]]], Sequence @@ list[[6 ;;]]}, {1}] > > > Out[2]= {{element11, element12, element1N, f[{target11, target12}], > element1N3, element1N4}, {element21, element22, element2N, > f[{target21, target22}], element2N3, element2N4}} > > > > In[3]:= (* second clumsy approach *) > > data /. {a : Repeated[_, {3}], b : Repeated[_, {2}], > c : __} -> {a, f[b], c} > > Out[3]= {{element11, element12, element1N, f[target11, target12], > element1N3, element1N4}, {element21, element22, element2N, > f[target21, target22], element2N3, element2N4}} > >

**References**:**Are you wolfing tonight?***From:*magma <maderri2@gmail.com>