Re: Need a functional process for this.

*To*: mathgroup at smc.vnet.net*Subject*: [mg55618] Re: [mg55589] Need a functional process for this.*From*: János <janos.lobb at yale.edu>*Date*: Thu, 31 Mar 2005 01:24:22 -0500 (EST)*References*: <200503300822.DAA21909@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

On Mar 30, 2005, at 3:22 AM, Steve Gray wrote: > I have two lists, aa and bb. aa has the general form > {2,5,7,9,11,...} > (or{{2},{5},{7},{9},{11}, > }}), and bb has the general form {{6,4},{9,2},{5,6},{3,8}, > }. If either > the first or second element in any sublist (pair of integers) of bb > matches any element in aa, I > want to delete that sublist from bb. In the above example, neither > member of {6,4} or {3,8} belongs > to aa, while at least one element of {9,2} and {5,6} belongs to aa, > so bb becomes {{6,4},{3,8}}. If > aa had only one element, for example 7, I could do bb=Delete > [bb,Position[bb,{x_,7}|{7,y_}]], but I > don't know how to do it for several values instead of just "7" > without using a procedural loop. > What is a good functional way to do this? > Thank you for any tips. > > Steve Gray I do not know if it is good or not but it is somewhat functional and 100% pedagogical :) In[1]:= aa = Table[Random[Integer, {0, 9}], {i, 1, 10}] Out[1]= {8, 6, 9, 7, 9, 8, 5, 1, 9, 1} In[2]:= bb = Table[{Random[Integer, {0, 9}], Random[Integer, {0, 9}]}, {i, 1, 10}] Out[2]= {{9, 0}, {3, 3}, {9, 7}, {2, 8}, {3, 3}, {3, 4}, {3, 1}, {8, 5}, {4, 9}, {5, 2}} In[42]:= Intersection[Pick[bb, Table[Position[aa, bb[[All,2]][[i]]] == {}, {i, 1, Length[ aa]}]], Pick[bb, Table[Position[aa, bb[[All,1]][[i]]] == {}, {i, 1, Length[aa]}]]] Out[42]= {{3, 3}, {3, 4}} János ---------------------------------------------- Trying to argue with a politician is like lifting up the head of a corpse. (S. Lem: His Master Voice)

**References**:**Need a functional process for this.***From:*Steve Gray <stevebg@adelphia.net>