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If I have a list of lists with different lengths:

{{1, 2, 3}, {1}, {1, 2, 3, 4}, {x}, {1, x}, {2}, {2, 3}}

How can I extract all the sub-lists with no symbols:

{{1, 2, 3}, {1}, {1, 2, 3, 4}, {2}, {2, 3}}

There must be an easy way to do this but I can't find it. Unfortunately, I'm not so good with Mathematica to be able to modify similar cases that I found(for example Link1 and Link2) to the problem at hand.

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3 Answers 3

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Cases[{{1, 2, 3}, {1}, {1, 2, 3, 4}, {x}, {1, x}, {2}, {2, 3}}, {__?NumericQ}]
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    $\begingroup$ Or 3 underscores if empty lists should also be kept. $\endgroup$
    – Martin Ender
    Commented Mar 17, 2016 at 21:42
  • $\begingroup$ Perfect! Thank you. $\endgroup$
    – MathX
    Commented Mar 17, 2016 at 21:44
  • $\begingroup$ Or DeleteCases[ {{1, 2, 3}, {1}, {1, 2, 3, 4}, {x}, {1, x}, {2}, {2, 3}}, {___, _Symbol, ___}] $\endgroup$
    – xyz
    Commented Mar 18, 2016 at 1:58
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Another approach is to use VectorQ:

lis = {{1, 2, 3}, {1}, {1, 2, 3, 4}, {x}, {1, x}, {2}, {2, 3}};
Select[lis, VectorQ[#, NumericQ] &]

(* {{1, 2, 3}, {1}, {1, 2, 3, 4}, {2}, {2, 3}} *)

Variations:

Select[lis, AllTrue[#, NumericQ] &]
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I think @march's answer is probably the best way. Here's another for giggles...

Select[{{1, 2, 3}, {1}, {1, 2, 3, 4}, {x}, {1, x}, {2}, {2, 3}}, ! MemberQ[#, _Symbol] &]
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  • $\begingroup$ Interesting. What is the name of that _Symbol part? I want to search for other properties that have the same usage. $\endgroup$
    – MathX
    Commented Mar 17, 2016 at 23:24
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    $\begingroup$ @BehzadNazari That's shortcut for Blank[Symbol] which is a pattern match for a certain Head. Head[x] (if x isn't defined) evaluates to Symbol. Try Head /@ {x, 1.0, 1, "test"} for examples. $\endgroup$
    – kale
    Commented Mar 18, 2016 at 2:10

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