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MathGroup Archive 2007

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Re: Passing a list as seperate parameters?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg73230] Re: [mg73194] Passing a list as seperate parameters?
  • From: gardyloo <gardyloo at mail.wsu.edu>
  • Date: Thu, 8 Feb 2007 03:38:46 -0500 (EST)
  • References: <200702071006.FAA00361@smc.vnet.net>

Hi, 4zumanga,

    I'm not *entirely* sure what you're wanting here; you want {a,b} to 
be treated as an element on its own, so that, for example, {1,3} U {1,3} 
= {1,3}, and {1,3} U {2,3} = {{1,3}, {2, 3}} ? In that case, the Union 
provided by Mathematica 5.2 and 5.1 seem to work the way you want (and, 
in particular, the "nested list" as input seems to work fine). In your 
example, Union[ {{a,b}, {c,d}}] just returns {{a,b}, {c,d}}, but if you 
assign values to the letters (I did it in a With statement), and the 
values happen to be the same in the same positions in the subsets, then 
Union really does contract the list properly.
   Which version of Mathematica are you using? According to my 
documentation, Union was last changed in v. 3.

         Cheers,
                   C.O.

Azumanga wrote:
> I have a recuring problem in using mathematica, which I'm sure there
> must be a simple fix for, but I've looked through the manual and can't
> find it.
>
> As an example, I build up the list { {a,b},{c,d} }, and I want to find
> the union of it's elements.
>
> At first glance, I want Union. However, Union wants it's input as
> Union[{a,b},{c,d}], Union[{{a,b},{c,d}}] doesn't do anything.
>
> In general I get around this by fold, but it seems there should be a
> simpler method?
>
>
>
>   

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