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  • $\begingroup$ Generalization: how would this work with a slightly different data set, where the marker for deletion isn't the last member of the list element: say, lis =lis = {{"a", "b", "c", "z"}, {"d", "e", "f", "z"}, {"d", "e", "x", "z"}, {"g", "h", "i", "z"}, {"w", "x", "y", "z"}, {"w", "x", "x", "z"}} $\endgroup$ Commented Sep 19, 2021 at 3:33
  • $\begingroup$ please ignore the above comment, the comment edit timed out before I finished my question.. Here is the correct comment: Generalization : how would this work with a slightly different data set, where the \ marker for deletion isn' t the last member of the list element : say, {{"a", "b", "c", "z"}, {"d", "e", "f", "z"}, {"d", "e", "x", "z"}, {"g", "h", "i", "z"}, {"q", "r", "s", "t"}, {"q", "r", "x", "t"}} with the desired res to be : {{"a", "b", "c", "z"}, {"d", "e", "f", "z"}, {"g", "h", "i", "z"}, {"q", "r", "s", "t"}} $\endgroup$ Commented Sep 19, 2021 at 3:44
  • $\begingroup$ @Suite401, maybe something like: ClearAll[f]; f[p_] := DeleteDuplicates[SortBy[#[[p]] == "x" &]@#, Drop[#, {p}] == Drop[#2, {p}] && MemberQ[{##}[[All, p]], "x"] &] &; f[3]@lis2? $\endgroup$ Commented Sep 19, 2021 at 3:53
  • $\begingroup$ OK will give it a try. $\endgroup$ Commented Sep 19, 2021 at 3:56