Thanks everyone for chiming in!
For now, I think that I will take Alex's heuristic (?) solution (also
described here:
http://www.sci.utah.edu/~weiliu/research/clustering_fmri/Zhong_sphericalKmeans.pdf).
I also like CP's projection idea - might work for my kind of data, which
has antipodal symmetry. I will play with this a bit and report back on how
well that's working for me.
Thanks again,
Ariel
On Thu, Jan 24, 2013 at 5:35 AM, Bertrand Thirion <bertrand.thir...@inria.fr
> wrote:
> As alluded previously, you need to find a way to compute the centroid by
> minimizing the sum of squared distances to a given set of points within
> each cluster. However, it is true that re-projecting the euclidean mean to
> the sphere would approximate well the theoretical solution in most cases.
>
> A standard alternative to k-means is Von Mises-Fisher distribution.
>
> Bertrand
>
> ------------------------------
>
> *De: *"Vince Fernando" <y...@vincefernando.co.uk>
> *À: *math...@mblondel.org, scikit-learn-general@lists.sourceforge.net
> *Envoyé: *Jeudi 24 Janvier 2013 09:55:46
> *Objet: *Re: [Scikit-learn-general] K means on a sphere
>
>
> Are there any theoretical problems if one uses the great circle
> (orthodromic) distance on a sphere in k-means or any other clustering
> algorithm?
> vince
>
>
> On 24 January 2013 07:11, Mathieu Blondel <math...@mblondel.org> wrote:
>
>> On Thu, Jan 24, 2013 at 9:24 AM, Gael Varoquaux
>> <gael.varoqu...@normalesup.org> wrote:
>>
>> > Yes, there is a massive difference in amount of work and performance
>> when
>> > you try to replace the Euclidean distance. Amongst other problems, the
>> > mean is no longer the sum divided by the number of points, but the
>> > Frechet mean, which requires solving an optimization problem.
>>
>> Indeed, if you replace the Euclidean distance, you also need to change
>> the averaging.
>> If you use the Manhattan distance, the averaging becomes the median.
>>
>> Mathieu
>>
>>
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