Re: List of multiple elements

*To*: mathgroup at smc.vnet.net*Subject*: [mg112088] Re: List of multiple elements*From*: Ray Koopman <koopman at sfu.ca>*Date*: Mon, 30 Aug 2010 06:18:33 -0400 (EDT)*References*: <i5d019$bs8$1@smc.vnet.net>

On Aug 28, 11:51 pm, Dana DeLouis <dana.... at gmail.com> wrote: >> 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 DuplicateElements 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 dupels2 is a faster version of DuplicateElements: dupels2[v_List] := Module[{f,dups={}}, f[n_] := (f[n] := (dups = {dups,n})); f/@v; Flatten@dups] DupsOnlee is an unsorted version of DupsOnly: DupsOnlee[v_List] := Fold[DeleteCases[#1, #2, {1}, 1] &, v, Union@v] DupsOnlee slows as Length@Union@v/Length@v increases; dupels2 does not. The crossover seems to be at about 5/1000. (The dummy call to Timing is necessary to avoid the problem noted in http://groups.google.ca/group/comp.soft-sys.math.mathematica/msg/1125a9dbeae0d5db ) {m,n} = {40,1*^4} Timing[w = Table[Random[Integer,{1,m}],{n}]; Do[Null,{1*^7}]]; First@AbsoluteTiming[t1 = dupels2[w];] First@AbsoluteTiming[t2 = DupsOnlee[w];] t1 === t2 {40,10000} 0.068693 Second 0.059314 Second True {m,n} = {50,1*^4} Timing[w = Table[Random[Integer,{1,m}],{n}]; Do[Null,{1*^7}]]; First@AbsoluteTiming[t1 = dupels2[w];] First@AbsoluteTiming[t2 = DupsOnlee[w];] t1 === t2 {50,10000} 0.069162 Second 0.072681 Second True {m,n} = {60,1*^4} Timing[w = Table[Random[Integer,{1,m}],{n}]; Do[Null,{1*^7}]]; First@AbsoluteTiming[t1 = dupels2[w];] First@AbsoluteTiming[t2 = DupsOnlee[w];] t1 === t2 {60,10000} 0.066558 Second 0.100195 Second True > > 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?