Re: irritating little problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg32937] Re: irritating little problem*From*: Thomas Burton <tburton at brahea.com>*Date*: Wed, 20 Feb 2002 01:26:19 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

Hello, Given a vector v, v = {b, a, b, b, c, d, e, d, f, d, c, b, c, g}; I would recommend the following method. First, append the positions to the values: vv = Transpose[{v, Range[Length[v]]}] Next, sort and split into sublists of equal values. vs = Split[Sort[vv], First[#1] == First[#2] & ] What you do next depends upon details I don't know. Hope this helps. Tom Burton On 2/18/02 11:36 PM, in article a4sv9c$ido$1 at smc.vnet.net, "KIMIC Weijnitz Peter" <micweij at eka.ericsson.se> wrote: > I have a simple vector > and I want to find the position of elements that are equal. > > I.e I want to test the vector and find all cases of similar elements. > > Brute force is not what I want, it can be a long vector. > Best regards > Petr W >