Re: the faster way to find repeated sublists

*To*: mathgroup at smc.vnet.net*Subject*: [mg60963] Re: [mg60923] the faster way to find repeated sublists*From*: János <janos.lobb at yale.edu>*Date*: Wed, 5 Oct 2005 02:28:06 -0400 (EDT)*References*: <200510040524.BAA17873@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

On Oct 4, 2005, at 1:24 AM, giampiero wrote: > i'm newbie > with a stupid problem > > a function for find repetead sublist in faster way > > ex > f[{1,2,1,2,1,2},{1,2}]-> True cause {1,2} is repated three times > f[{1,2,3,1,2,3},{1,2,3}]-> True cause {1,2,3} is repeated two times > f[{1,2,1,2,3},{1,2}]->False cause after two {1,2} there is another > symbol. > > True if the second list in containes many times exactly in first list > False otherwise. > > > bye everyone and sorry for my stupidity. > > giampiero Here is a newbie approach: In[2]:= lst = Flatten[Table[{a, b}, {i, 1, 5}]] Out[2]= {a, b, a, b, a, b, a, b, a, b} In[3]:= sublst = {a, b} Out[3]= {a, b} In[12]:= f[l_List, s_List] := If[Length[l]/Length[s] == Length[Position[Partition[ l, Length[s], 1], s]], True, False] In[13]:= f[lst, sublst] Out[13]= True In[20]:= tst = lst Out[20]= {a, b, a, b, a, b, a, b, a, b} In[22]:= lst = Join[tst, {a}] Out[22]= {a, b, a, b, a, b, a, b, a, b, a} In[27]:= f[lst, sublst] Out[27]= False János ---------------------------------------------- Trying to argue with a politician is like lifting up the head of a corpse. (S. Lem: His Master Voice)

**References**:**the faster way to find repeated sublists***From:*"giampiero" <giampiero196019@yahoo.it>