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Re: List of multiple elements

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  • Subject: [mg112047] Re: List of multiple elements
  • From: Bill Rowe <readnews at>
  • Date: Fri, 27 Aug 2010 04:06:35 -0400 (EDT)

On 8/26/10 at 6:48 AM, weh at (Dr. Wolfgang Hintze) wrote:

>Given the list


>we can reduce multiple instances of elements using

>b=Union[a] {1,2}

>The question is now how to compute the list of multiple elements. In
>our example this would be m={1,1,2}.

>A possible solution is

>m[x_]:= Flatten[Take[#, {1, Length[#] - 1}] & /@ Select[Split[x],
>Length[#] > 1 &]]

>m[a] {1,1,2}

>I'm sure there is a much more elegant solution. Can you suggest one?

The function you have included cannot possibly give the result
you indicate. With the particular example you start with Split
will only create one sub-list of length greater than 1 which is
{1,1}. Since you only select sub-lists with length greater than
1, you effectively drop all elements not equal to 1 from the
original list. Consequently, you cannot end up with what you
have indicated.

The following will obtain the result you have indicated you want.

In[13]:= b = Union[a];
=46latten@DeleteDuplicates[a //. {x__, Sequence @@ b, y___} :> {x,
b, y}]

Out[14]= {1,1,2}

But, it is far from clear to me this will scale in an
intelligent way to deal with a more complex case.

The simplest way I can think of to get the desired end result
would be something like




but these clearly aren't general solutions

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