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MathGroup Archive 2002

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Re: Operating on every k-th element of list?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg37089] Re: Operating on every k-th element of list?
  • From: "Allan Hayes" <hay at haystack.demon.co.uk>
  • Date: Wed, 9 Oct 2002 05:25:38 -0400 (EDT)
  • References: <anufl8$ap2$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

lst= Range[14]

{1,2,3,4,5,6,7,8,9,10,11,12,13,14}

A list of positions in lst ( for your purpose Range[1, Length[lst], 3] will
do)

ps= {3,5,10};

The following applies h to each ps element in lst and adds the result to the
following element

(lst[[ps+1]]=h/@lst[[ps]]+lst[[ps+1]];lst)

{1,2,3,4+h[3],5,6+h[5],7,8,9,10,11+h[10],12,13,14}

--
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


"AES" <siegman at stanford.edu> wrote in message
news:anufl8$ap2$1 at smc.vnet.net...
> I want to apply a function to every  k-th  element of a long list and
> add the result to the  k+1  element.
>
> [Actually  k = 3  and I just want to multiply  myList[[k]]  by a
> constant (independent of k)  and add the result to  myList[[k+1]]  for
> every value of  k  that's divisible by 3.]
>
> Is there a way to do this -- or in general to get at every  k-th
> element of a list -- that's faster and more elegant than writing a brute
> force  Do[]  loop or using  Mod[]  operators, and that will take
> advantage of native List operators,  but still not be too recondite?
>
> I've been thinking about multiplying a copy of  myList  by a "mask list"
> {0,0,1,0,0,1,..} to generate a "masked copy" and approaches like that.
> Better ways???
>




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