Re: List of multiple elements
- To: mathgroup at smc.vnet.net
- Subject: [mg112085] Re: List of multiple elements
- From: Dana DeLouis <dana.del at gmail.com>
- Date: Sun, 29 Aug 2010 02:51:38 -0400 (EDT)
> Given the list
> a={1,1,2,1,2};
> ... how to compute the list of multiple elements.
> {1,1,2}
Hi. One additional way might be to just delete each unique item from the original list.
DupsOnly[v_List] := Module[{u, t},
u = Union[v];
t = Fold[DeleteCases[#1, #2, {1}, 1] &, v, u];
Sort[t]
]
DupsOnly[{1, 1, 2, 1, 2}]
{1, 1, 2}
v = RandomChoice[Range[5], 10]
{4, 2, 2, 3, 5, 5, 3, 3, 1, 1}
Sort[v]
{1, 1, 2, 2, 3, 3, 3, 4, 5, 5}
DupsOnly[v]
{1, 2, 3, 3, 5}
I may be wrong, but the suggestion on =93DuplicateElements=94 seems a little slower in the following case:
v = RandomChoice[Range[5], 10000];
Timing[DupsOnly[v]] // First
0.003875
Timing[DuplicateElements[v]] // First
0.5468
= = = = = = = = = =
HTH : >)
Dana DeLouis
On Aug 26, 6:48 am, "Dr. Wolfgang Hintze" <w... at snafu.de> wrote:
> Given the list
>
> a={1,1,2,1,2};
>
> 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?