Hey Wei,
On Mon, Jan 28, 2013 at 5:33 AM, Wei LI <li...@ee.cuhk.edu.hk> wrote:
> Hi Ariel:
>
> There is one matlab implementation for spherical kmeans:
> http://www.mathworks.com/matlabcentral/fileexchange/28902-spherical-k-means
> for
> spherical kmeans and you can have a look at it :) That seems quite simple
> and may be easily transformed into a python function.
Nice and simple - thanks for pointing this out!
Ariel
> Best,
> Wei
>
> On Fri, Jan 25, 2013 at 6:29 AM, Satrajit Ghosh <sa...@mit.edu> wrote:
>
>> hi ariel,
>>
>> on the unit sphere, the dot product of the vectors would be exactly that
>> except for the range. you would want to scale it so that -1 to 1 maps to 0
>> to 1 and then run spectral clustering on that matrix. if you have too many
>> vectors you can create a sparse matrix, but on my mbp i can handle a dense
>> 10000 x 10000 in about a few minutes.
>>
>> dummy example:
>>
>> In [1]: import numpy as np
>> In [2]: from sklearn.cluster import SpectralClustering
>> In [4]: foo = np.random.rand(10000, 10000)
>> In [7]: sc = SpectralClustering(n_clusters=2, affinity='precomputed',
>> assign_labels='discretize').fit(foo)
>>
>> cheers,
>>
>> satra
>>
>> On Thu, Jan 24, 2013 at 5:07 PM, Ariel Rokem <aro...@gmail.com> wrote:
>>
>>> 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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