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**Algorithms and applications**

Clustering (Chap 7) Algorithms and applications

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**Introduction Clustering is an important data mining task.**

Clustering makes it possible to “almost automatically” summarize large amounts of high-dimensional data. Clustering (aka segmentation) is applied in many domains: retail, web, image analysis, bioinformatics, psychology, brain science, medical diagnosis etc.

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**Key Idea Given: (i) a data set D and (ii) similarity function s**

Goal: break up D into groups such that items which are similar to each other end in the same group AND items which are dis-similar to each other end up in different groups. Example: customers with similar buying patterns end up in the same cluster. People with similar “biological signals” end in the same cluster. Regions with similar “weather” patterns.

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**Clustering Example: Iris Data**

Iris data set is a “very small” data set consisting of attributes of 3 flower types. Flower types: Iris_setosa; Iris_versicolor; Iris-virginica 4 attributes: sepal-length; sepal-width; petal-length; petal-width Thus each flower is “data vector” in a four-dimensional space. The “fifth” dimension is a label. Example: [5.1,3.5,1.4,0.2,iris-setosa] It is hard to visualize in four-dimensions!

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Example Iris Data Set Which is the “best” pair of attributes to distinguish the three flower types ?

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**Attributes and Features**

In almost all cases the labels are not given but we can acquire attributes and construct new features. Notice the terminology: attributes, features, variables, dimensions – they are often used interchangeably. I prefer to distinguish between attributes and features. We acquire attributes and construct features. Thus there are finitely many attributes but infinitely many possible features. For example, Iris data set had four attributes. Lets construct a new feature: sepal-length/sepal-width.

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**Similarity and Distance Functions**

We have already seen the cosine similarity function. sim(x,y) = (x.y)/||x||||y|| We can also define a “distance” function which in some sense is like an “opposite” of similarity. Entities which are highly similar to each other have a small distance between them; while items which are less similar have a large distance between them. Example: dist(x,y) = 1 – sim(x,y) “identical” objects: sim(x,y) = 1 dist(x,y) = 0 sim(x,y) = 0 dist(x,y) = 1

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**Euclidean Distance Take two data vectors d1 and d2 D1 = (x1,x2,…,xn)**

D2 = (y1,y2,…yn) Then Euclidean Distance between D1 and D2 is

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**Euclidean Distance: Examples**

D1=(1,1,1); D2=(1,1,1) dist(D1,D2) = 0 D1=(1,2,1); D2= (2,1,1)

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**Simple Clustering: One-dimensional/one-centroid**

1 2 6 3 Three data points: 1,2 and 6 Want to summarize with one point The average of (1,2,6) = 9/3 = 3 3 is called the centroid

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**High-Dimensional/one-centroid**

Let (2,4), (4,6) and (8,10) be three data points in two dimensions. The average (centroid) is:

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**Different interpretation of centroid**

Let x1, x2, x3 be data points. Define a function f as: Now, it turns out that the minimum of the function f occurs at the centroid of the three points. This interpretation can be easily be generalized across dimensions and centroids.

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**High-dimension/many centroids**

Let X be a set of d-dimensional data points (data vectors). Find k data points (not necessarily in X, lets call them cluster centers), such that the sum of the square Euclidean distances of each point in X to its nearest cluster center is minimized. This is called the K-means problem. The data points which get assigned to their nearest cluster center end up as the k-clusters.

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K-means algorithm Let C = initial k cluster centroids (often selected randomly) Mark C as unstable While <C is unstable> Assign all data points to their nearest centroid in C. Compute the centroids of the points assigned to each element of C. Update C as the set of new centroids. Mark C as stable or unstable by comparing with previous set of centroids. End While Complexity: O(nkdI) n:num of points; k: num of clusters; d: dimension; I: num of iterations Take away: complexity is linear in n.

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An important question ? Does the K-means algorithm “solve” the K-means problem. Answer: only partially. In fact nobody has yet been able to design an algorithm to “solve” the K-means problem. Part of the “P =\= NP” problem. A million dollar prize if you have a “solution” or show that the “solution” cannot exist.

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**Example: 2 Clusters A(-1,2) B(1,2) c c 4 (0,0) C(-1,-2) D(1,-2) c c 2**

K-means Problem: Solution is (0,2) and (0,-2) and the clusters are {A,B} and {C,D} K-means Algorithm: Suppose the initial centroids are (-1,0) and (1,0) then {A,C} and {B,D} end up as the two clusters.

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**Kmeans algorithm on Iris Data**

Note: The K-means algorithms does not know the labels!

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**Kmeans in Matlab In the tutorial you will be working with Matlab**

Matlab is a very easy to use data processing language For example, the K-means algorithm can be run using one line >> [IDX,C]= kmeans(irisdata,3); The cluster labels (1,2,3) are stored in the variable IDX.

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Summary Clustering is a way of compressing data into information from which we can derive “knowledge” The first step is to map data into “data vectors” and then use a “distance” function. We formalized clustering as a K-means problem and then used the K-means algorithm to obtain a solution. The applications of clustering are almost limitless. Almost every data mining exercise begins with clustering – to get initial understanding of the data.

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