Re: List representation using element position

• To: mathgroup at smc.vnet.net
• Subject: [mg72526] Re: List representation using element position
• From: "Dr. Wolfgang Hintze" <weh at snafu.de>
• Date: Thu, 4 Jan 2007 07:08:25 -0500 (EST)
• Organization: privat
• References: <enfgem\$ss1\$1@smc.vnet.net> <enfvj8\$6vc\$1@smc.vnet.net>

```
Ray Koopman wrote:

> 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}},
>>	{3,{2}},
>>	{4,{3,4,9}},
>>	{5,{8}}
>>	}
>>
>>Who finds the shortest function doing this task in general?
>>
>>My solution appears 15 lines below
>>
>>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}}}
>>
>
> fPoz[lstIn_] := {#[[1,1]],Last/@#}& /@ Split[
> Sort@MapIndexed[Flatten@{##}&,lstIn], #1[[1]]==#2[[1]] ]
>
>
Hello Ray,

I'm getting
{{1, {6}}, {1, {7}}, {2, {1}}, {2, {5}}, {3, {2}}, {4, {3}}, {4, {4}},
{4, {9}}, {5, {8}}}
from your function. Did I miss something here?
Regards
Wolfgang

```

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