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MathGroup Archive 2002

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Re: Stumped again on a simple list

  • To: mathgroup at smc.vnet.net
  • Subject: [mg32817] Re: Stumped again on a simple list
  • From: "Allan Hayes" <hay at haystack.demon.co.uk>
  • Date: Thu, 14 Feb 2002 01:43:35 -0500 (EST)
  • References: <a4auba$bu0$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Steven,
Is this the sort of thing that you want?

    ii=Table[z,{z,50,140,5}]

        {50,55,60,65,70,75,80,85,90,95,100,105,110,115,120,125,130,135,140}

    ii/.n_Integer\[RuleDelayed] If[n<100,50,200]

        {50,50,50,50,50,50,50,50,50,50,200,200,200,200,200,200,200,200,200}

    ii/.{n_Integer/;n<100\[RuleDelayed] 50,n_Integer\[RuleDelayed] 200}

        {50,50,50,50,50,50,50,50,50,50,200,200,200,200,200,200,200,200,200}

--
Allan

---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565


"1.156" <rob at piovere.com> wrote in message news:a4auba$bu0$1 at smc.vnet.net...
> Greetings cypherers all,
>
> I make a simple list ii and I find I can make other lists easily by
algebraic moves like
> jj=ii/2, etc.  But when I want make a new list which splits two different
formula, I'm stumped.
>
> Here's a stripped down version to illustrate my problem.  I want to use
the value in the original
> array to determine how to make the new array.  The If[] is the only thing
I can find to make
> this decision.  But, it seems I'm not even close when I try:
>
>
> ii=Table[z,{z,50,1400,5}];
>
> jj=If[ii<100,50,200]
>
> I've looked around a good bit but have not stumbled into any topic that
appears to bear on this problem.
>
> Can I please ask a kind soul to (again) point me in the right direction?
>
> Thanks, Rob
>
>




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