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Given:

list1 = {{"a",1},{"b",2},{"c",3}} 

I would like to delete any pair in list1 that does not have a first element that belongs to list2.

Example 1: If list2 = {"b"}, Output = {{"b",2}}
Example 2: If list2 = {"A","b"}, Output = {{"a",1},{"b",2}}

Here's what I have so far:

DeleteCases[list1,{#,_}/; !StringMatchQ[#,list2, IgnoreCase -> True]&].

I know this won't work (because StringMatchQ doesn't accept a list of values to compare to as a parameter, which leads me to thinking I need to Map the function above to list2.

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You've almost got it:

list1 = {{"a", 1}, {"b", 2}, {"c", 3}};
list2 = {"A", "b"};

test = StringMatchQ[#, Alternatives @@ list2, IgnoreCase -> True] &;

Cases[list1, {_?test, _}]
{{"a", 1}, {"b", 2}}

A key element is clearly Alternatives.
I used Cases rather than DeleteCases as that seemed simpler to me.


The first part of my post is in direct answer to your question. Summary: use Alternatives, not Map.

Borrowing from belisarius' answer, if one is going to use ToLowerCase I propose this:

Pick[list1, list1[[All, 1]], Alternatives @@ ToLowerCase @ list2]

If some of the elements of list1 are not all lower case then:

Pick[list1, ToLowerCase @ list1[[All, 1]], Alternatives @@ ToLowerCase @ list2]
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Also:

list1 = {{"a", 1}, {"b", 2}, {"c", 3}};
f[list1_, list2_] := Select[list1, MemberQ[ToLowerCase /@ list2, #[[1]]] &]
f[list1, {"A", "b"}]

(* {{"a", 1}, {"b", 2}} *)
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  • $\begingroup$ That's nice too. I'd +1 but I'm out of votes for the day. (Remind me tomorrow.) $\endgroup$ – Mr.Wizard Feb 21 '13 at 17:27
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    $\begingroup$ @Mr.Wizard Thanks :)I turned it to CW so neither of us will need to remember the nuisance :) $\endgroup$ – Dr. belisarius Feb 21 '13 at 17:29
  • $\begingroup$ @belisarius - Very nice, thanks! $\endgroup$ – tjm167us Feb 21 '13 at 17:43

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