Is the 'spherical' GaussianMixture Model of sklearn the same as performing k-means?
The GaussianMixture() implementation in scikit-learn offers four different types of covariance matrices when fitting the model. One of those is the 'spherical' type, in which each component has its own single variance.
My question, isn't this the same as doing k-means on a dataset?
python scikit-learn gaussian
asked Nov 26 '18 at 9:32
ArchieArchie
590725
add a comment |
The GaussianMixture() implementation in scikit-learn offers four different types of covariance matrices when fitting the model. One of those is the 'spherical' type, in which each component has its own single variance.
My question, isn't this the same as doing k-means on a dataset?
python scikit-learn gaussian
asked Nov 26 '18 at 9:32
ArchieArchie
590725
add a comment |
The GaussianMixture() implementation in scikit-learn offers four different types of covariance matrices when fitting the model. One of those is the 'spherical' type, in which each component has its own single variance.
My question, isn't this the same as doing k-means on a dataset?
python scikit-learn gaussian
asked Nov 26 '18 at 9:32
ArchieArchie
590725
The GaussianMixture() implementation in scikit-learn offers four different types of covariance matrices when fitting the model. One of those is the 'spherical' type, in which each component has its own single variance.
My question, isn't this the same as doing k-means on a dataset?
python scikit-learn gaussian
python scikit-learn gaussian
asked Nov 26 '18 at 9:32
ArchieArchie
590725
asked Nov 26 '18 at 9:32
ArchieArchie
590725
asked Nov 26 '18 at 9:32
ArchieArchie
590725
asked Nov 26 '18 at 9:32
ArchieArchie
590725
asked Nov 26 '18 at 9:32
ArchieArchie
590725
590725
add a comment |
add a comment |
1 Answer
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K-Means is exactly like a Hard Assignment GMM, where each mixture component has isotropic variance, and they are all equal.
Just being isotropic ('spherical') does not guarantee equivalence to K-Means. the variance should also be the same.
More detailed explanation can be found here.
answered Nov 26 '18 at 9:50
DinariDinari
1,694523
So if the implementation would be that each component share the same single variance value, then the two are equivalent (combined with hard assignments)?
– Archie
Nov 26 '18 at 12:47
1
Yes, if the above conditions exist, then the likelihood some pointxis a member of some gaussian, depends only on the distance from itsmean, exactly like inK-Means, and during the re-evaluating of the Gaussians, you only evaluate its mean, by averaging all the points, which is too, exactly like inK-means(As you don't re-evaluate the variance).
– Dinari
Nov 26 '18 at 12:51
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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oldest
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active
oldest
votes
K-Means is exactly like a Hard Assignment GMM, where each mixture component has isotropic variance, and they are all equal.
Just being isotropic ('spherical') does not guarantee equivalence to K-Means. the variance should also be the same.
More detailed explanation can be found here.
answered Nov 26 '18 at 9:50
DinariDinari
1,694523
So if the implementation would be that each component share the same single variance value, then the two are equivalent (combined with hard assignments)?
– Archie
Nov 26 '18 at 12:47
1
Yes, if the above conditions exist, then the likelihood some pointxis a member of some gaussian, depends only on the distance from itsmean, exactly like inK-Means, and during the re-evaluating of the Gaussians, you only evaluate its mean, by averaging all the points, which is too, exactly like inK-means(As you don't re-evaluate the variance).
– Dinari
Nov 26 '18 at 12:51
add a comment |
K-Means is exactly like a Hard Assignment GMM, where each mixture component has isotropic variance, and they are all equal.
Just being isotropic ('spherical') does not guarantee equivalence to K-Means. the variance should also be the same.
More detailed explanation can be found here.
answered Nov 26 '18 at 9:50
DinariDinari
1,694523
So if the implementation would be that each component share the same single variance value, then the two are equivalent (combined with hard assignments)?
– Archie
Nov 26 '18 at 12:47
1
Yes, if the above conditions exist, then the likelihood some pointxis a member of some gaussian, depends only on the distance from itsmean, exactly like inK-Means, and during the re-evaluating of the Gaussians, you only evaluate its mean, by averaging all the points, which is too, exactly like inK-means(As you don't re-evaluate the variance).
– Dinari
Nov 26 '18 at 12:51
add a comment |
K-Means is exactly like a Hard Assignment GMM, where each mixture component has isotropic variance, and they are all equal.
Just being isotropic ('spherical') does not guarantee equivalence to K-Means. the variance should also be the same.
More detailed explanation can be found here.
answered Nov 26 '18 at 9:50
DinariDinari
1,694523
K-Means is exactly like a Hard Assignment GMM, where each mixture component has isotropic variance, and they are all equal.
Just being isotropic ('spherical') does not guarantee equivalence to K-Means. the variance should also be the same.
More detailed explanation can be found here.
answered Nov 26 '18 at 9:50
DinariDinari
1,694523
answered Nov 26 '18 at 9:50
DinariDinari
1,694523
answered Nov 26 '18 at 9:50
DinariDinari
1,694523
answered Nov 26 '18 at 9:50
DinariDinari
1,694523
1,694523
So if the implementation would be that each component share the same single variance value, then the two are equivalent (combined with hard assignments)?
– Archie
Nov 26 '18 at 12:47
1
Yes, if the above conditions exist, then the likelihood some pointxis a member of some gaussian, depends only on the distance from itsmean, exactly like inK-Means, and during the re-evaluating of the Gaussians, you only evaluate its mean, by averaging all the points, which is too, exactly like inK-means(As you don't re-evaluate the variance).
– Dinari
Nov 26 '18 at 12:51
add a comment |
So if the implementation would be that each component share the same single variance value, then the two are equivalent (combined with hard assignments)?
– Archie
Nov 26 '18 at 12:47
1
Yes, if the above conditions exist, then the likelihood some pointxis a member of some gaussian, depends only on the distance from itsmean, exactly like inK-Means, and during the re-evaluating of the Gaussians, you only evaluate its mean, by averaging all the points, which is too, exactly like inK-means(As you don't re-evaluate the variance).
– Dinari
Nov 26 '18 at 12:51
So if the implementation would be that each component share the same single variance value, then the two are equivalent (combined with hard assignments)?
– Archie
Nov 26 '18 at 12:47
So if the implementation would be that each component share the same single variance value, then the two are equivalent (combined with hard assignments)?
– Archie
Nov 26 '18 at 12:47
1
1
Yes, if the above conditions exist, then the likelihood some point
x is a member of some gaussian, depends only on the distance from its mean, exactly like in K-Means, and during the re-evaluating of the Gaussians, you only evaluate its mean, by averaging all the points, which is too, exactly like in K-means (As you don't re-evaluate the variance).– Dinari
Nov 26 '18 at 12:51
Yes, if the above conditions exist, then the likelihood some point
x is a member of some gaussian, depends only on the distance from its mean, exactly like in K-Means, and during the re-evaluating of the Gaussians, you only evaluate its mean, by averaging all the points, which is too, exactly like in K-means (As you don't re-evaluate the variance).– Dinari
Nov 26 '18 at 12:51
add a comment |
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