MathGroup Archive 2007

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Re: List representation using element position


Ray Koopman wrote:
> fPoz[lstIn_] := {#[[1,1]],Last/@#}& /@ Split[
> Sort@MapIndexed[Flatten@{##}&,lstIn], #1[[1]]==#2[[1]] ]

That should be

  fPoz[lstIn_] := {#[[1,1]],Last/@#}& /@ Split[
  Sort@MapIndexed[Flatten@{##}&,lstIn], #1[[1]]==#2[[1]]&]

which will still be slow but will at least get what was wanted.


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