Unsupervised Learning: Difference between revisions
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&= \sum_{j=1}{k} \sum_{i\mid z(i)=j} \nabla \Vert x^{(i)} - \mu_{j} \Vert ^2\\ | &= \sum_{j=1}{k} \sum_{i\mid z(i)=j} \nabla \Vert x^{(i)} - \mu_{j} \Vert ^2\\ | ||
&= \sum_{j=1}{k} \sum_{i\mid z(i)=j} 2(x^{(i)} - \mu_{j})\\ | &= \sum_{j=1}{k} \sum_{i\mid z(i)=j} 2(x^{(i)} - \mu_{j})\\ | ||
\implies \mu_{j} &= (\sum_{i\mid z(i)=j} x^(i))/(\sum_{i\mid z(i)=j} 1) \quad \forall j | \implies \mu_{j} &= (\sum_{i\mid z(i)=j} x^{(i)})/(\sum_{i\mid z(i)=j} 1) \quad \forall j | ||
\end{align} | \end{align} | ||
</math> | </math> |
Revision as of 17:46, 12 November 2019
Basics of Unsupervised Learning
Clustering
Given points \(\displaystyle \{x^{(1)},...,x^{(m)}\}\) we wish to group them into clusters 1,...,k.
Usually we are given the number of clusters, k.
The problem is to assign labels \(\displaystyle \{z^{(1)},...,z^{(m)}\}\) where \(\displaystyle z^{(i)} \in \{1,...,k\}\)
K-means
Here we wish to cluster them to minimize the distance to the mean of their cluster.
In our formulation we have k clusters.
The mean of each cluster \(\displaystyle \mu_i\) is called the centroid.</math>
Optimization
Let \(\displaystyle \mathbf{\mu}\) denote the centroids and let \(\displaystyle \mathbf{z}\) denote the cluster labels for our data.
Our loss function is \(\displaystyle L(\mu, \mathbf{z}) = \sum_{i} \Vert x^{(i)} - \mu_{z^{(i)}} \Vert ^2\).
- Using coordinate-block descent
\(\displaystyle L(\mu, \mathbf{z}) = \sum_{i} \Vert x^{(i)} - \mu_{z^{(i)}} \Vert ^2\).
If we fix \(\displaystyle \mu\) then we have only need to minimize \(\displaystyle z^{(i)}\).
Since each term of the sum is independent, we simply choose the closest centroid for each point
If we fix \(\displaystyle \mathbf{z}\) then we need to minimize \(\displaystyle L(\mu, \mathbf{z})\) wrt \(\displaystyle \mu\).
Taking the gradient and setting it to 0 we get:
\(\displaystyle
\begin{align}
\nabla L(\mu, \mathbf{z}) &= \nabla \sum_{i} \Vert x^{(i)} - \mu_{z^{(i)}} \Vert ^2\\
&= \nabla \sum_{j=1}{k} \sum_{i\mid z(i)=j} \Vert x^{(i)} - \mu_{z^{(i)}} \Vert ^2\\
&= \nabla \sum_{j=1}{k} \sum_{i\mid z(i)=j} \Vert x^{(i)} - \mu_{j} \Vert ^2\\
&= \sum_{j=1}{k} \sum_{i\mid z(i)=j} \nabla \Vert x^{(i)} - \mu_{j} \Vert ^2\\
&= \sum_{j=1}{k} \sum_{i\mid z(i)=j} \nabla \Vert x^{(i)} - \mu_{j} \Vert ^2\\
&= \sum_{j=1}{k} \sum_{i\mid z(i)=j} 2(x^{(i)} - \mu_{j})\\
\implies \mu_{j} &= (\sum_{i\mid z(i)=j} x^{(i)})/(\sum_{i\mid z(i)=j} 1) \quad \forall j
\end{align}
\)
Algorithm
Soft K-means
We will develop a model for how our data is generated:
Given \(\displaystyle k\) clusters, the probability of a point being from cluster k is \(\displaystyle \phi_k = P(z^{(i)} = k)\)