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I have a nested list of the form

{{{{a, 0},{b, 1},{c, 2}}, {{d, 4},{e, 5},{f, 6}}} ... }

where the $...$ just means it goes on like that. On the innermost level, as you can see, I have pairs of the form {string, number}. All I want to do is to delete from this nested list the pairs whose number entry is 0.

For instance, in the above example, I want to delete only the pair {a, 0}.

I've spent the last two hours reading up on answers on this website and elsewhere on the internet, without managing to successfully apply them to this case. Most commands just return the empty list. I also find it very hard to understand the syntax of pretty much all of the list-related commands.

I should also say that I don't really care about the nested structure being preserved or about speed, since the list isn't too big.

Thank you for your time.

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    $\begingroup$ list/.{_, 0} :> Sequence[] $\endgroup$
    – Basheer Algohi
    Commented Feb 23, 2016 at 1:08
  • $\begingroup$ @Algohi it works, thank you! $\endgroup$
    – Guest
    Commented Feb 23, 2016 at 1:10
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    $\begingroup$ DeleteCases[list, {_, 0}, ∞] also works. $\endgroup$
    – bbgodfrey
    Commented Feb 23, 2016 at 1:23
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    $\begingroup$ @bbgodfrey This DeleteCases command is the first that actually makes sense to me, so thank you very much. $\endgroup$
    – Guest
    Commented Feb 23, 2016 at 3:39
  • $\begingroup$ Also Select[#[[-1]] != 0 &][Partition[Flatten[list], 2]] - without nested structure and not fast :)) $\endgroup$
    – garej
    Commented Feb 23, 2016 at 10:59

2 Answers 2

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For the record, so that this question have answers in an Answer block, correct answers given in the comments to the OP include

  • list/.{_, 0} :> Sequence[]
  • DeleteCases[list, {_, 0}, ∞]

In addition, here are two that are similar to the ones above but slightly more specific to the requirements stated in the OP:

  • list/.{_, 0} -> Sequence[] (Rule instead of RuleDelayed)
  • DeleteCases[list, {_, 0}, {3}] (deletes only sublists at level 3)
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Let's use a more complex expression than given above to stress the solution methods a bit more.

data = 
  {{{{a, 0}, {b, 1}, {c, 2}}, {{d, 4}, {e, 5}, {f, 6}}},
  {{{{{a, 0}}}, {b, 1}, {c, 2}}, {{d, 4}, {{e, 0}, {f, 6}}}}};

I think the following have advantages over the solutions already given.

data /. {_, 0} -> Nothing
DeleteCases[data, {_, 0}, {-2}]

Both give

{{{{b, 1}, {c, 2}}, {{d, 4}, {e, 5}, {f, 6}}}, 
 {{{{}}, {b, 1}, {c, 2}}, {{d, 4}, {{f, 6}}}}}

The first method I think more easily understood than /. {_, 0} :> Sequence[] and second simply is more robust -- DeleteCases[data, {_, 0}, {3}] fails on this more complex data list.

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