Lecture 4

Unsupervised LearningClustering & Dimensionality Reduction

273A Intro Machine Learning

What is Unsupervised Learning?• In supervised learning we were given attributes & targets (e.g. class labels). In unsupervised learning we are only given attributes.

• Our task is to discover structure in the data.

• Example I: the data may be structured in clusters:

• Example II: the data may live on a lower dimensional manifold:

Is this a good clustering?

Why Discover Structure ?

• Data compression: If you have a good model you can encode the data more cheaply.

• Example: To encode the data I have to encode the x and y position of each data-case. However, I could also encode the offset and angle of the line plus the deviations from the line. Small numbers can be encoded more cheaply than large numbers with the same precision.

• This idea is the basis for model selection: The complexity of your model (e.g. the number of parameters) should be such that you can encode the data-set with the fewest number of bits (up to a certain precision).

Homework: Argue why a larger dataset will require a more complex model to achieve maximal compression.

Why Discover Structure ?

• Often, the result of an unsupervised learning algorithm is a new representation for the same data. This new representation should be more meaningful and could be used for further processing (e.g. classification).

• Example I: Clustering. The new representation is now given by the label of a cluster to which the data-point belongs. This tells us how similar data-cases are.

• Example II: Dimensionality Reduction. Instead of a 100 dimensional vector of real numbers, the data are now represented by a 2 dimensional vector which can be drawn in the plane.

• The new representation is smaller and hence more convenient computationally.

• Example I: A text corpus has about 1M documents. Each document is represented as a 20,000 dimensional count vector for each word in the vocabulary. Dimensionality reduction turns this into a (say) 50 dimensional vector for each doc. However: in the new representation documents which are on the same topic, but do not necessarily share keywords have moved closer together!

54700013500000010004001...

theaon...

Clustering: K-means• We iterate two operations:

1. Update the assignment of data-cases to clusters2. Update the location of the cluster.

• Denote the assignment of data-case “i” to cluster “c”. • Denote the position of cluster “c” in a d-dimensional space.

• Denote the location of data-case i

• Then iterate until convergence:

1. For each data-case, compute distances to each cluster and the closest one:

2. For each cluster location, compute the mean location of all data-cases assigned to it:

[1,2,3,..., ]iz K

dc

dix

argmin|| ||i i cc

z x i

1

c

c ii Sc

x cN

Set of data-cases assigned to cluster cNr. of data-cases in cluster c

K-means

• Cost function:

• Each step in k-means decreases this cost function.

• Often initialization is very important since there are very many local minima in C. Relatively good initialization: place cluster locations on K randomly chosen data-cases.

• How to choose K? Add complexity term: and minimize also over K Or X-validation Or Bayesian methods

2

1

|| ||zi

N

ii

C x

Homework: Derive the k-means algorithm by showing that:step 1 minimizes C over z, keeping the cluster locations fixed.step 2 minimizes C over cluster locations, keeping z fixed.

1[# ] log( )

2C C parameters N

Vector Quantization

• K-means divides the space up in a Voronoi tesselation.• Every point on a tile is summarized by the code-book vector “+”. This clearly allows for data compression !

Mixtures of Gaussians

• K-means assigns each data-case to exactly 1 cluster. But what if clusters are overlapping? Maybe we are uncertain as to which cluster it really belongs.

• The mixtures of Gaussians algorithm assigns data-cases to cluster with a certain probability.

MoG Clustering

1

/ 2

1 1[ ; , ] exp[ ( ) ( )]

22 det( )T

dN x x x

Covariance determines the shape of these contours

• Idea: fit these Gaussian densities to the data, one per cluster.

EM Algorithm: E-step

' ' '' 1

[ ; , ]

[ ; , ]

c i c cic K

ic c cc

N xr

N x

• “r” is the probability that data-case “i” belongs to cluster “c”.

• is the a priori probability of being assigned to cluster “c”.

• Note that if the Gaussian has high probability on data-case “i” (i.e. the bell-shape is on top of the data-case) then it claims high responsibility for this data-case.

• The denominator is just to normalize all responsibilities to 1:

c

1

1K

icc

r i

Homework: Imagine there are only two identical Gaussians and they both have theirmeans equal to Xi (the location of data-case “i”). Compute the responsibilities for data-case “i”. What happens if one Gaussian has much larger variance than the other?

EM Algorithm: M-Step

1c ic i

ic

r xN

c ici

N r

1( )( )Tc ic i c i c

ic

r x xN

cc

NN

total responsibility claimed by cluster “c”

expected fraction of data-cases assigned to this cluster

weighted sample mean where every data-case is weighted according to the probability that it belongs to that cluster.

weighted sample covariance

Homework: show that k-means is a special case of the E and M steps.

EM-MoG

• EM comes from “expectation maximization”. We won’t go through the derivation.

• If we are forced to decide, we should assign a data-case to the cluster which claims highest responsibility.

• For a new data-case, we should compute responsibilities as in the E-step and pick the cluster with the largest responsibility.

• E and M steps should be iterated until convergence (which is guaranteed).

• Every step increases the following objective function (which is the total log-probability of the data under the model we are learning):

1 1

log [ ; , ]N K

c i c ci c

L N x

Dimensionality Reduction• Instead of organized in clusters, the data may be approximately lying on a (perhaps curved) manifold.

• Most information in the data would be retained if we project the data on this low dimensional manifold.

• Advantages: visualization, extracting meaning attributes, computational efficiency

Principal Components Analysis

• We search for those directions in space that have the highest variance.

• We then project the data onto the subspace of highest variance.

• This structure is encoded in the sample co-variance of the data:

1

1

1

1( )( )

N

ii

NT

i ii

xN

C x xN

PCA• We want to find the eigenvectors and eigenvalues of this covariance:

TC U U

12

d

0

0

1u

2u

du

eigenvalue = variancein direction eigenvector

( in matlab [U,L]=eig(C) )

1u

2u

Orthogonal, unit-length eigenvectors.

PCA properties

1

1 1

( )

( ) ( )

dT

i i ii

d dT T

j i i i j i i i j j ji i

C uu

Cu uu u u u u u

check this

(U eigevectors)

T TU U UU I (u orthonormal U rotation)

1:T

i iky U x1u

2u

3u1: 1: 1:

Tk k kC U U

12

0

0 3

1:3U

1:3

(rank-k approximation)

(projection)

1: 1: 1: 1: 1: 1: 1:1 1

1 1N NT T T T T T

y i i i ik k k k k k ki i

C U x x U U x x U U U U UN N

Homework: What projection z has covariance C=I in k dimensions ?

PCA properties

1:kC is the optimal rank-k approximation of C in Frobenius norm. I.e. it minimizes the cost-function:

12 2

1 1 1

( )d d k

Tij il lj

i j l

C A A with A U

Note that there are infinite solutions that minimize this norm. If A is a solution, then is also a solution.

The solution provided by PCA is unique because U is orthogonal and orderedby largest eigenvalue.

Solution is also nested: if I solve for a rank-k+1 approximation, I will find that the first k eigenvectors are those found by an rank-k approximation (etc.)

TAR with RR I

Homework

Imagine I have 1000 20x20 images of faces.

Each pixel is an attribute Xi and can take continuous values in the interval [0,1].

Let’s say I am interested in finding the four “eigen-faces” that span most of the variance in the data.

Provide pseudo-code of how to find these four eigen-faces.