Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2006
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2006

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Using Select within Map

  • To: mathgroup at smc.vnet.net
  • Subject: [mg71792] Re: Using Select within Map
  • From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
  • Date: Wed, 29 Nov 2006 02:56:34 -0500 (EST)
  • Organization: The Open University, Milton Keynes, UK
  • References: <ekh70s$s7p$1@smc.vnet.net>

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]

Map[Select[#, (a < #[[1]] < b) &] &, x]

should do what you are looking for.

> 
> 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
> 

In[1]:=
x = Table[Random[Integer, 10], {3}, {3}, {2}]

Out[1]=
{{{1,5},{8,6},{0,4}},{{2,6},{4,9},{0,4}},{{5,6},{0,9},{10,4}}}

In[2]:=
With[{a = 2, b = 8},
   (Select[#1, a < #1[[1]] < b & ] & ) /@ x]

Out[2]=
{{},{{4,9}},{{5,6}}}

In[3]:=
With[{a = 2, b = 8},
   (Select[#1, a < #1[[1]] < b & ] & ) /@ x]

Out[3]=
{{},{{4,9}},{{5,6}}}

Regards,
Jean-Marc


  • Prev by Date: Re: Using Select within Map
  • Next by Date: Re: Using Select within Map
  • Previous by thread: Re: Using Select within Map
  • Next by thread: Re: Using Select within Map