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I have a list filled with a large number of elements, eg. {1,2,3,4,5,6,7}

I wanna partition the list by a reference list. The reference list shows how those elements need to be grouped. eg. {{1,3},{4,6},{7,2,5},{9,10,11}...}. The reference list also contains the unwanted element/sublist.

My current methods are:

list1 = {1, 2, 3, 4, 5, 6, 7};
reflist = {{1, 3}, {4, 6}, {7, 2, 5}, {9, 10, 11}};

Flatten[DeleteDuplicates[
  Table[Select[reflist, MemberQ[#, list1[[i]]] &], {i, 1, 
    Length@list1}]], 1]

This methods works but is not efficient. my list1 could be more than half-million unique elements. I am seeking a more effective way to do it.

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  • 2
    $\begingroup$ Is this more efficient? Select[reflist, Length@Intersection[list1, #] > 0 &]. $\endgroup$
    – Rohit Namjoshi
    Jan 4, 2022 at 19:10

2 Answers 2

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Select[reflist, ContainsAll[list1, #] &]

{{1, 3}, {4, 6}, {7, 2, 5}}

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  • $\begingroup$ The 8 shouldn't be there, it's not in list1. $\endgroup$
    – flinty
    Jan 4, 2022 at 19:11
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    $\begingroup$ @flinty The output is the same with op ask $\endgroup$
    – yode
    Jan 4, 2022 at 19:13
  • $\begingroup$ yeah that confused me: it says "partition the list by a reference list" so why would the list1 which does not contain an 8 get partitioned into a list-of-lists which does? $\endgroup$
    – flinty
    Jan 4, 2022 at 19:15
  • $\begingroup$ @flinty By the name, he should be a Chinese. Not very good at expressing in English. But my results are the same as his. $\endgroup$
    – yode
    Jan 4, 2022 at 19:17
  • $\begingroup$ Sorry, I have made some mistakes typing. It has been corrected. $\endgroup$
    – Teng
    Jan 4, 2022 at 19:27
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$\begingroup$ 
Cases[{___, Alternatives @@ list1 , ___}] @ reflist

{{1, 3}, {4, 6}, {7, 2, 5}}
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