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Re: Intersection problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg126046] Re: Intersection problem
  • From: Barrie Stokes <Barrie.Stokes at newcastle.edu.au>
  • Date: Fri, 13 Apr 2012 04:56:50 -0400 (EDT)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <201204100627.CAA10654@smc.vnet.net>

Hi Fred

Of the many offerings to this question, I like yours as much as any of them, as

Pick [list1, list1[[All, 1]], Alternatives @@ list2 ] is quite transparent.

Pick [list1, First /@ list1, Alternatives @@ list2 ]

avoids the need to use "[[" and "]]", and needs even fewer characters.

Looking through the recent postings in reply to Cisco's query, I think this question might be a contender for the title of "Question with the Most Syntactically Different Valid Answers"?  (Steven M. Christensen?)

Sort of companion to the well-known "One-Liner" competition ...

Cheers

Barrie

>>> On 12/04/2012 at 8:16 am, in message <201204112216.SAA02663 at smc.vnet.net>, Fred
Simons <f.h.simons at tue.nl> wrote:
> Op 10-4-2012 8:27, Cisco Lane schreef:
>> I need to select out the elements of the first list whose first elements are 
> contained in the second list. For example
>>
>> first list: {{a,x},{b,y},{c,z}}
>>
>> second list: {b,c,d}
>>
>> result: {{b,y},{c,z}}
>>
>> Is there a quick way to do this? Intersection, maybe somehow?
>>
>>
> I am not sure if this is quick, but it is short:
> 
> Pick[list1, list1[[All, 1]], Alternatives @@ list2]
> 
> 
> Regards,
> 
> Fred Simons
> Eindhoven University of Technology




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