Re: List of multiple elements

• To: mathgroup at smc.vnet.net
• Subject: [mg112139] Re: List of multiple elements
• From: Raffy <adraffy at gmail.com>
• Date: Wed, 1 Sep 2010 06:29:13 -0400 (EDT)
• References: <i5d019\$bs8\$1@smc.vnet.net> <i5g0hu\$gos\$1@smc.vnet.net> <i5idoj\$jr6\$1@smc.vnet.net>

```On Aug 31, 1:16 am, Ray Koopman <koop... at sfu.ca> wrote:
> On Aug 30, 3:18 am, Ray Koopman <koop... at sfu.ca> wrote:
>
>
>
> > dupels2 is a faster version of DuplicateElements:
>
> > dupels2[v_List] := Module[{f,dups={}},
> > f[n_] := (f[n] := (dups = {dups,n})); f/@v; Flatten@dups]
>
> dupels1 is twice as fast as dupels2.
>
> dupels1[v_List] :=  Module[{f}, f[x_] := ((f[x] = x); {});
>                                 Flatten[f=
/@v]]
>
> {m,n} = {50,1*^4}
> Timing[w = Table[Random[Integer,{1,m}],{n}]; Do[Null,{1*^7}]];
> First@AbsoluteTiming[t1 = dupels1[w];]
> First@AbsoluteTiming[t2 = dupels2[w];]
> t1 === t2
>
> {50,10000}
> 0.030357 Second
> 0.060681 Second
> True

I'm pretty sure the solution I posted above is the fastest unless more
Most of the solutions listed require multiple searches over the
elements or require sorting neither of which are necessary.

dup3[v_] := Join @@ (ConstantArray[#1, #2 - 1] & @@@ Tally[v]);

v = RandomInteger[{1, 100}, 10000000];
dup3[v]; // Timing ==> 0.031
sordups[v]; // Timing ==> 2.87

```

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