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I have a list of lists of integers that are each sorted numerically. Here is a sample of it: {{2, 8, 9}, {3, 7, 9}, {4, 6, 9}, ...}. I wish to only pick out members which meet the following criteria:

  1. No two elements within it are consecutive:ContainsNone[Union@Differences[#], {1}] &
  2. The length of the list is itself a member of the list:ContainsAny[#, {Length@#}] &]

Each test independently works fine:

In=Select[mylist, ContainsNone[Union@Differences[#], {1}] &]
Out={{3, 7, 9}, {4, 6, 9}, {1, 3, 6, 9}, {1, 4, 6, 8}}

In=Select[mylist, ContainsAny[#, {Length@#}] &]
Out={{3, 7, 9}, {1, 4, 5, 9}, {1, 4, 6, 8}, {2, 4, 5, 8}, {2, 4, 6, 7}, {3, 4, 5, 7}, {1, 2, 3, 5, 8}, {1, 2, 4, 5, 7}, {1, 3, 4, 5, 6}}

However when I attempt to use an And[] function with Select[] to only pick out lists that meet both criteria I am presented with an empty result when clearly at least {3,7,9} meets both criteria:

In=Select[mylist,And[ContainsNone[Union@Differences[#], {1}] &,ContainsAny[#, {Length@#}] &]]
Out={}

Is it that And[] is not working as I intend it inside this Select[] function? Surely if Select[] passes each member one by one, the And[] will return True when it gets to {3,7,9}, no? Or is it some other reason? I know I can just output each and do an Intersection[] but I just want to figure out how to use Select to do what I intend.

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  • 1
    $\begingroup$ Use myList // Select[And[pred1, pred2, ...] /* Through]. $\endgroup$
    – Shredderroy
    Commented Sep 16, 2023 at 17:24

1 Answer 1

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list = {{2, 8, 9}, {3, 7, 9}, {4, 6, 9}};

Select[list, 
 ContainsNone[Union @ Differences[#], {1}] && 
   ContainsAny[#, {Length @ #}] &]

{{3, 7, 9}}

Or, if you want the And:

Select[list, 
 And[ContainsNone[Union@Differences[#], {1}], 
   ContainsAny[#, {Length@#}]] &]

{{3, 7, 9}}

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  • $\begingroup$ So I understand correctly, it was me treating each test as a separate function that was causing the issue? I see here you're merging them both into one function? $\endgroup$
    – user2686410
    Commented Sep 16, 2023 at 16:23
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    $\begingroup$ Yes, one could say so $\endgroup$
    – eldo
    Commented Sep 16, 2023 at 16:24
  • $\begingroup$ Do you know what it is about Select's innerworkings that would reject an And[] evaluation of two different functions onto one list member? $\endgroup$
    – user2686410
    Commented Sep 16, 2023 at 16:25
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    $\begingroup$ The predicate you give to Select must evaluate to either True or False. Your only mistake was, that you duplicated the & and put the ] in the wrong positions (at the end) $\endgroup$
    – eldo
    Commented Sep 16, 2023 at 16:33

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