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MathGroup Archive 2004

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Re: Selecting by first element of each list

  • To: mathgroup at smc.vnet.net
  • Subject: [mg48155] Re: Selecting by first element of each list
  • From: "Carlo Teubner" <.Teubner at t-online.de (FirstLetterOfMyFirstName)>
  • Date: Fri, 14 May 2004 20:59:13 -0400 (EDT)
  • Organization: Oxford University, England
  • References: <c81i3m$509$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hi,

try

Split[list1, #1[[1]] == #2[[1]] &].

This will give you what you want, provided first lists that belong together
appear consecutively.

Carlo

<camartin at snet.net> schrieb im Newsbeitrag news:c81i3m$509$1 at smc.vnet.net...
> Hi,
>
> I'm trying to learn to handle a flat file kind of database using
> Mathematica. An example is the list of lists below:
>
> {{2,3,4,5,77},{2,4,5,66,77},{3,4,7,8,90,6},{3,5,6,7,8,0},{3,45,67,77,12}}
>
> It's actually more complicated than this but good enough for
> illustrative purposes. The first element in each list is actually an
> identifier, like an account number. I want to create a list of lists
> with each account, that is,
>
> {{{2,3,4,5,77},{2,4,5,66,77}},
> {{3,4,7,8,90,6},{3,5,6,7,8,0},{3,45,67,77,12}}}.
>
> When I use Select with an anonymous function such as
>
> Select[list1,#1[[1]] = = #2[[1]]&]
>
> I get an error because it stops (of course) after the first two lists. I
> get the right grouping for the first list but it doesn't finish. I don't
> understand how to use the anonymous function to go through my list (it's
> got several thousand entries) and select and group by the first element
> in each group. I've been through the archive but nothing there quite
helps.
>
> I would appreciate some direction.
>
> Thanks
>
> Cliff
>
>



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