Selective list deletions

Question

I have a list:

lis = {{a,b,c},{d},{e,f},{g},{h,i,j}}

I would like to remove each element that consists of only one subelement from the list to get:

res = {{a,b,c},{e,f},{h,i,j}}

This seems to be for SequenceReplace, but I'm having trouble with structuring the command.

Syed · Accepted Answer · 2021-10-07 12:39:40Z

6

Cases[lis, Except[{_}]] should be good.

OR

Select[lis, Length[#] > 1 &]

Pick[lis, Length[#] > 1 & /@ lis]

DeleteCases[lis, {_}]

lis /. {_} -> Nothing

EDIT a few more

Select[lis, Rest[#] != {} &]

Select[lis, Most[#] != {} &]

Select[lis, Last@TakeDrop[#, 1] != {} &]

answered Oct 6, 2021 at 18:27

Syed

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$\begingroup$ Cases, DeleteCases, and Select can all be used here. But as a purely aesthetic point, I would prefer to use either Cases or Select so that I can know with certainty the structure of the elements that I have retained in the next step. With DeleteCases, I would need to use the structure up to that point, and the pattern I have chosen to delete, in order to come to a conclusion about the structure of the result. $\endgroup$
– Shredderroy
Oct 7, 2021 at 17:17

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kglr · Accepted Answer · 2021-10-06 18:46:10Z

8

Cases[{_, __}] @ lis

{{a, b, c}, {e, f}, {h, i, j}}

answered Oct 6, 2021 at 18:46

kglr

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2 Answers 2