Hi Satra,
Thanks for the hint. "Affinity", as in an affine transformation? Isn't the
angle between any two unit vectors that transformation?
Cheers,
Ariel
On Thu, Jan 24, 2013 at 1:01 PM, Satrajit Ghosh <sa...@mit.edu> wrote:
> 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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