Re: rule based program for "Deleting repeated members of a list."

*To*: mathgroup at smc.vnet.net*Subject*: [mg81990] Re: rule based program for "Deleting repeated members of a list."*From*: Peter Pein <petsie at dordos.net>*Date*: Mon, 8 Oct 2007 02:06:22 -0400 (EDT)*References*: <fe7j44$qga$1@smc.vnet.net>

mumat schrieb: > Hi, > > I have a list s={a,b,c,c,d,e,e,f,g,g,g,h,a,b}; > > I need to write a program so that it reads the list and ignors > repeated elements; so it outputs the following: > > RepeatRemover[s]={a,b,c,d,e,f,g,h,a,b}; > > I am looking for a rule-based program to do this! > > Any help would be greatly appreciated. > > C.S. > > Hi numat, if you have to do this with patterns, In[2]:= s //. {x___, y_, y_, z___} :> {x, y, z} Out[2]= {a, b, c, d, e, f, g, h, a, b} will do the task. If you've got longer lists, this could last a while: In[3]:= SeedRandom[2007]; s = Flatten[Table[Random[Integer, {1, 8}], {k, Random[Integer, {1, 5}]}, {1234}]]; In[5]:= AbsoluteTiming[s1 = s //. {a___, x_, x_, b___} :> {a, x, b}; ][[1]] Out[5]= 74.0625 Second Compare this to In[6]:= AbsoluteTiming[s2 = First /@ Split[s]; ][[1]] Out[6]= 0. Second or (to put at least a pattern into the solution): In[7]:= AbsoluteTiming[ s3 = First /@ DeleteCases[Partition[s, 2, 1, {1, 1}, "foo"], {x_, x_}]; ][[1]] Out[7]= 0. Second and the reasults are the same: In[8]:= s1 === s2 === s3 Out[8]= True Regards, Peter