Re: DeleteCases

*To*: mathgroup at smc.vnet.net*Subject*: [mg98167] Re: [mg98144] DeleteCases*From*: Fred Simons <f.h.simons at tue.nl>*Date*: Wed, 1 Apr 2009 05:56:04 -0500 (EST)*References*: <200903310919.EAA03943@smc.vnet.net>

Filippo Miatto wrote: > Dear all, > Given this example list: > > {{0, 0, 1, 1}, {0, 0, 2, 2}, {0, 0, 3, 3}, {1, 1, 0, 0}, {1, 1, 2, 2}, > {1, 1, 3, 3}, {2, 2, 0, 0}, {2, 2, 1, 1}, {2, 2, 3, 3}, {3, 3, 0, 0}, > {3, 3, 1, 1}, {3, 3, 2, 2}} > > I need the elements {a,a,b,b} and {b,b,a,a} to be the same, i.e. i > want to delete one of the two occurrences from the list. (or all but > one if the condition is more complicated, say {a,b,c,d} and {b,a,c,d} > or {a,b,c,d} and {a,b,d,c}) > How can i do that? Can i use DeleteCases? if so, what conditional > should I use to compare different elements in the list? > I tried without success for all the night.. (it's 6.15AM now!) > Thank you, > > Filippo > > > Filippo, If I understand you right, you want to take only one occurence ot those elements from your list having the same union. That can be done as follows: In[1]:= lst={{0,0,1,1},{0,0,2,2},{0,0,3,3},{1,1,0,0},{1,1,2,2},{1,1,3,3},{2,2,0,0},{2,2,1,1},{2,2,3,3},{3,3,0,0},{3,3,1,1},{3,3,2,2}}; In[3]:= GatherBy[lst, Union][[All,1]] Out[3]= {{0,0,1,1},{0,0,2,2},{0,0,3,3},{1,1,2,2},{1,1,3,3},{2,2,3,3}} Regards, Fred Simons Eindhoven University of Technology