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
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> 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}}}
>