I have a large ordered list of lists of integers like this


and I would like to filter out the nested list based on two conditions. The conditions are two types of shorter nested lists

  • pairs of consecutive numbers that are not allowed anywhere in the nested list
  • pairs of consecutive numbers that must occur somewhere in the list



The output in this case would be

  • $\begingroup$ why do you need both notallowed and also allowed? Would not just using notallowed be enought to remove those unwanted? $\endgroup$
    – Nasser
    Feb 19, 2018 at 23:45
  • $\begingroup$ No because allowed must be there all present. I can remove all notallowed but that does not guarantee there are all alowed (can be {3,4} but not {5,6}). $\endgroup$ Feb 20, 2018 at 0:01
  • $\begingroup$ For your use case, what would the maximum integer be? Are the allowed and disallowed sets always pairs of consecutive numbers? Are your sublists always ordered? $\endgroup$
    – Carl Woll
    Feb 20, 2018 at 0:54

1 Answer 1


You can use SequencePosition -- if it returns a nonempty list, then the given sublist is present:

 // Select[Length@SequencePosition[#, Alternatives @@ notallowed, 1] == 0 &]
 // Select@Function[list, 
       Length@SequencePosition[list, #, 1] > 0 & /@ allowed // Apply@And]
{{3, 4, 5, 6, 7, 9}, {3, 4, 5, 6, 11, 12}}
  • $\begingroup$ This is brilliant. Thank you. $\endgroup$ Feb 20, 2018 at 2:34

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