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Re: working with lists

  • To: mathgroup at smc.vnet.net
  • Subject: [mg113211] Re: working with lists
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Tue, 19 Oct 2010 05:54:15 -0400 (EDT)
  • Reply-to: hanlonr at cox.net

Your sample output does not appear to agree with your description of the desired operation. The following implements the description.

f[x_?VectorQ] := Module[
  {n = 0},
  If[Mod[#, 3] == 0,
     n = 1; 2 #, # - n] & /@ x]

f[{1, 2, 3, 5, 7}]

{1, 2, 6, 4, 6}


Bob Hanlon

---- Sam Takoy <sam.takoy at yahoo.com> wrote: 

=============
Hi,

I'm not very good at working with lists. May I ask for someone to work 
out an example which has several elements of what I need to do.

What's the best way to write a function f[list] that goes through each 
element of the lest, doubles each element divisible by three and reduces 
each of the following elements by 1. That is


f[{ 1 2 3 5 7}] is { 1 2 6 4 12 }

Many thanks in advance,

Sam



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