How to keep elements only if their first key is part of another list?

Question

I have two huge lists: list1 and list2.

I would like to keep elements of list1, only if their first key is part of list2.

Example:

list1 = {{aa, 2, 5}, {be, 1, 5}, {ce, 5, 2}, {de, 1, 20}, {ee, 5, 2}}

list2 = {aa, ce, de}

Expected resulted:

list1 = {{aa, 2, 5}, {ce, 5, 2}, {de, 1, 20}}

I tried Pick, but I couldn't find a way. I think I have to look for a pattern that says "Pick element only if element belongs to list2"? What could it be?

My lists are very large so I can't do it manually.

Welcome to Mathematica.SE! I suggest that: 1) You take the introductory Tour now! 2) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! 3) As you receive help, try to give it too, by answering questions in your area of expertise. — bbgodfrey, May 14 '15 at 19:08
How "huge" is "huge"? None of the answers so far is particularly fast on "huge" lists (i.e., they can b beaten by order of magnitude +) - please specify cardinality expected of each list, key universe, etc... — ciao, May 14 '15 at 22:11
@ciao: list 1 has dimensions {2866206,3} and list 2 {100221,1} — Cancan, May 15 '15 at 08:23

score 2 · Accepted Answer · answered May 14 '15 at 19:06

Some additional methods

list1 = {{aa, 2, 5}, {be, 1, 5}, {ce, 5, 2}, {de, 1, 20}, {ee, 5, 2}};

list2 = {aa, ce, de};

Select[list1, MemberQ[list2, #[[1]]] &]

{{aa, 2, 5}, {ce, 5, 2}, {de, 1, 20}}

Cases[list1, _?(MemberQ[list2, #[[1]]] &)]

{{aa, 2, 5}, {ce, 5, 2}, {de, 1, 20}}

Pick[list1, MemberQ[list2, #[[1]]] & /@ list1]

{{aa, 2, 5}, {ce, 5, 2}, {de, 1, 20}}

DeleteCases[list1, _?(! MemberQ[list2, #[[1]]] &)]

{{aa, 2, 5}, {ce, 5, 2}, {de, 1, 20}}

DeleteCases[list1, _?(FreeQ[list2, #[[1]]] &)]

{{aa, 2, 5}, {ce, 5, 2}, {de, 1, 20}}

Thank you very much @bob hanlon!! – Cancan May 15 '15 at 08:28 — Cancan, May 15 '15 at 08:28

score 1 · Answer 2 · edited May 14 '15 at 19:26

1

Select[list1, MemberQ[list2, #[[1]]] &]

edited May 14 '15 at 19:26

bbgodfrey

61,439
17
89
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answered May 14 '15 at 19:03

sonright

141
5

How to keep elements only if their first key is part of another list?

2 Answers2