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MathGroup Archive 2006

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Re: Using Select within Map

  • To: mathgroup at smc.vnet.net
  • Subject: [mg71794] Re: Using Select within Map
  • From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
  • Date: Wed, 29 Nov 2006 02:56:36 -0500 (EST)
  • Organization: Uni Leipzig
  • References: <ekh70s$s7p$1@smc.vnet.net>
  • Reply-to: kuska at informatik.uni-leipzig.de

Hi,

a) the parser will parse if correct as to see from
    Hold[Map[Select[#, a < #[[1]] < b &], x]] // FullForm

b) if you use

Map[Select[#, Function[{q},a < q[[1]] < b],x]

    you will not mix up the two Slot[]s

Regards
   Jens


Mark Teagarden wrote:
> Hi,
> 
> I have a data set which comprises several paired lists, like so (I apologize
> if the tabs do not come out correctly in your mail client):
> 
> x = 
> {
>     {
>         { , },
>         { , },
>         { , }
>     },
>     {
>         { , },
>         { , },
>         { , }
>     },
>     {
>         { , },
>         { , },
>         { , }
>     }
> }
> 
> What I would like to do is to select from each of those paired lists, only
> those pairs where the first value in the pair falls within some specified
> range, and then obtain a mean.  This would be simple enough if I was
> operating on a single paired list, for example:
> 
> Select[x[[1]],a < #[[1]] < b&]
> 
> However, I would like to Map over x so that I would end up with means for
> each of the level 1 lists within x; therein lies the problem.  Both Map and
> Select use the Slot operator (#), and I don't know how to distinguish
> between the slot operator used by Map, and the slot operator used by Select:
> 
> Select[#,a < #[[1]] < b&]/@x
> 
> Or if you prefer:
> 
> Map[Select[#, a < #[[1]] < b&,x]
> 
> This problem has been plaguing me for some time, and if I could punch
> through it I would be very happy indeed.  Any ideas?  I have had no luck
> looking through the archives or the Help Browser.  On a similar note, I
> would eventually like to modify this solution so that the criteria, a and b,
> could vary with each level 1 list in x, but one thing at a time...
> 
> Thanks in advance,
> Mark
> 


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