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MathGroup Archive 2007

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Re: List representation using element position

  • To: mathgroup at smc.vnet.net
  • Subject: [mg72515] Re: List representation using element position
  • From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
  • Date: Wed, 3 Jan 2007 05:36:42 -0500 (EST)
  • Organization: The Open University, Milton Keynes, UK
  • References: <enfgem$ss1$1@smc.vnet.net>

Dr. Wolfgang Hintze wrote:
> Hello group,
> happy new year to all of you!
> 
> This one was put up in a slightly different form by me in March 2006.
> It is now more general and it is lossless with respect to information:
> 
> Given a list of integers which may repeat, e.g.
> 
> lstIn = {2,3,4,4,2,1,1,5,4}
> 
> provide a list of the different values and their respective positions in 
> the original list. In the example,
> 
> LstOut= {
> 	{1,{6,7}},
> 	{2,{2,5}},
------------^
Above, I am sure you meant {2,{1,5}}

> 	{3,{2}},
> 	{4,{3,4,9}},
> 	{5,{8}}
> 	}
> 
> Who finds the shortest function doing this task in general?

Here is my function:

In[1]:=
myfun[lst_] :=
   ({#1, Flatten[Position[lst, #1]]} & ) /@ Union[lst]

Below, we check that the list returned by myfun is matches the expected 
format.

In[2]:=
lstIn = {2, 3, 4, 4, 2, 1, 1, 5, 4};

In[3]:=
LstOut = {{1, {6, 7}}, {2, {1, 5}}, {3, {2}},
     {4, {3, 4, 9}}, {5, {8}}};

In[4]:=
myfun[lstIn]

Out[4]=
{{1, {6, 7}}, {2, {1, 5}}, {3, {2}}, {4, {3, 4, 9}},
   {5, {8}}}

In[5]:=
% == LstOut

Out[5]=
True

That works fine!

> My solution appears 15 lines below

By the way, your solution works "correctly" only for a list of positive 
integers. (Otherwise, the elements of the original list are replace by 
their index number, index number starting from one.) Compare,

fPos[{a,b,a,a,e,f,e,d}]

--> {{1,{1,3,4}},{2,{2}},{3,{8}},{4,{5,7}},{5,{6}}}

vs
myfun[{a,b,a,a,e,f,e,d}]

--> {{a,{1,3,4}},{b,{2}},{d,{8}},{e,{5,7}},{f,{6}}}

and
fPos[{-3,1,-1,0,0,2,2.1,2.1}]

--> {{1,{1}},{2,{3}},{3,{4,5}},{4,{2}},{5,{6}},{6,{7,8}}}

vs
myfun[{-3,1,-1,0,0,2,2.1,2.1}]

--> {{-3,{1}},{-1,{3}},{0,{4,5}},{1,{2}},{2,{6}},{2.1,{7,8}}}

> Thanks.
> 
> Best regards,
> Wolfgang
> 1
> 
> 
> 
> 5
> 
> 
> 
> 
> 10
> 
> 
> 
> 
> fPos[lstIn_] := Module[{f = Flatten /@ (Position[lstIn, #1] & ) /@ 
> Union[lstIn]}, ({#1, f[[#1]]} & ) /@ Range[Length[f]]]
> 
> In[15]:=
> fPos[lstIn]
> 
> Out[15]=
> {{1, {6, 7}}, {2, {1, 5}}, {3, {2}}, {4, {3, 4, 9}}, {5, {8}}}
> 


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