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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: [mg37094] Re: Operating on every k-th element of list?
  • From: stagat at mrcsb.com (Bob Stagat)
  • Date: Wed, 9 Oct 2002 05:25:46 -0400 (EDT)
  • References: <anufl8$ap2$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Take[list, am, n, sa] gives elements m through n in steps of s. 



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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