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?