Re: irritating little problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg32939] Re: [mg32906] irritating little problem*From*: Chris Johnson <cjohnson at shell.faradic.net>*Date*: Wed, 20 Feb 2002 01:26:22 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

This may not be too elegent, but it gets the job done. Actually, it only catches pairs, so if something appears 3 times you get the 3 permutations of matches. A little extra work might catch higher order matches. matches[list_List] := Union[ Select[ Position[ Map[ If[First[#] == Last[#], #, Null] &, Outer[List, list, list], {2}] , x : {_, _}] /. {x_, y_} :> {y, x} /; y < x, First[#] != Last[#] &]]; In[60]:= list=Table[Random[Integer,{0,9}],{10}] Out[60]= {4,3,3,3,4,6,1,5,5,2} In[61]:= matches[list] Out[61]= {{1,5},{2,3},{2,4},{3,4},{8,9}} It took me about 10 seconds to do a List of length 1000. Probably could be sped up by cleaning up the duplicates of the Outer earlier. Good luck! Chris On Tue, 19 Feb 2002, KIMIC Weijnitz Peter wrote: > I have a simple vector > and I want to find the position of elements that are equal. > > I.e I want to test the vector and find all cases of similar elements. > > Brute force is not what I want, it can be a long vector. > Best regards > Petr W >