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
>