hi ariel,
if you can precompute affinity between your vectors, you could also try
spectral clustering.
cheers,
satra
On Thu, Jan 24, 2013 at 1:18 PM, Ariel Rokem <aro...@gmail.com> wrote:
> 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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>>
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