Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2005
*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 2005

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

Search the Archive

Re: simple set operations

  • To: mathgroup at smc.vnet.net
  • Subject: [mg57643] Re: simple set operations
  • From: "Jens-Peer Kuska" <kuska at informatik.uni-leipzig.de>
  • Date: Fri, 3 Jun 2005 05:33:24 -0400 (EDT)
  • Organization: Uni Leipzig
  • References: <d7mk8b$c5m$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hi,

1) Table[FromCharacterCode[i], {i, 97, 122}]
2) what may MemberQ[] do ?

Regards
  Jens

"Edward Peschko" <esp5 at pge.com> schrieb im 
Newsbeitrag news:d7mk8b$c5m$1 at smc.vnet.net...
> hey all,
>
> I'm hesitant to ask these questions (because 
> they are so simple) but after a
> 15 minute search through the docs I'm getting 
> nowhere, so here goes:
>
>    1) what's the easiest way to generate a list 
> of elements? ie:
>
> 'a' .. 'h' == { a,b,c,d,e,f,g,h }
>
>    2) Is there a quick way to check whether an 
> element is in a set?
>
> if ('a' == (any('a','b','c','d')) { print "a is 
> in a,b,c,d"; }
>
> The first one I see could possibly be done by 
> 'Array', but I don't see how -
> the '#' refers to the generation of numbers, but 
> there seems to be no
> corresponding 'letter' symbol.
>
>
> As for #2, the easiest way would be through an 
> overloading of the '==' operator,
> but again, that doesn't seem to work..
>
> Thanks much for any help,
>
> Ed
> 



  • Prev by Date: Re: FourierTransform
  • Next by Date: Re: simple set operations
  • Previous by thread: Re: simple set operations
  • Next by thread: Re: simple set operations