Implementation of Single Pass Algorithm for Clustering – BE(IT) CLP-II Practical
Aim : To implement Single Pass Algorithm for Clustering in Documents and Files .
Objective : To study Clustering in files or Documents using single pass algorithm
Given below is the Single Pass Algorithm for Clustering with source code in Java Language . For this code to work you should have three files for sample input (Text Files ) .
Source Code for Single Pass Algorithm
package com.prac.prac;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
public class singlepass {
public static void main(String[] args) throws IOException{
BufferedReader stdInpt = new BufferedReader(new InputStreamReader(System.in));
System.out.println("Enter the no of Tokens");
int noOfDocuments=Integer.parseInt(stdInpt.readLine());
System.out.println("Enter the no of Documents");
int noOfTokens=Integer.parseInt(stdInpt.readLine());
System.out.println("Enter the threshhold");
float threshhold=Float.parseFloat(stdInpt.readLine());
System.out.println("Enter the Document Token Matrix");
int [][]input= new int [noOfDocuments][noOfTokens];
for(int i=0;i{
for(int j=0;j{
System.out.println("Enter("+i+","+j+")");
input[i][j]=Integer.parseInt(stdInpt.readLine());
}
}
SinglePassAlgorithm(noOfDocuments, noOfTokens, threshhold, input);
}
private static void SinglePassAlgorithm(int noOfDocuments,int noOfTokens,float threshhold,int [][]input)
{
int [][] cluster = new int [noOfDocuments][noOfDocuments+1];
ArrayListclusterRepresentative = new ArrayList ();
cluster [0][0]=1;
cluster [0][1]=0;
int noOfClusters=1;
Float []temp= new Float[noOfTokens];
temp=convertintArrToFloatArr(input[0]);
clusterRepresentative.add(temp);
for(int i=1;i{
float max=-1;
int clusterId=-1;
for(int j=0;j{
float similarity=calculateSimilarity(convertintArrToFloatArr(input[i]),clusterRepresentative.get(j) );
if(similarity>threshhold)
{
if(similarity>max)
{
max=similarity;
clusterId=j;
}
}
}
if(max==-1)
{
cluster[noOfClusters][0]=1;
cluster[noOfClusters][1]=i;
noOfClusters++;
clusterRepresentative.add(convertintArrToFloatArr(input[i]));
}
else
{
cluster[clusterId][0]+=1;
int index=cluster[clusterId][0];
cluster[clusterId][index]=i;
clusterRepresentative.set(clusterId,calculateClusterRepresentative(cluster[clusterId],input, noOfTokens));
}
}
for(int i=0;i{
System.out.print("\n"+i+"\t");
for(int j=1;j<=cluster[i][0];++j)
{
System.out.print(" "+cluster[i][j]);
}
}
}
private static Float[] convertintArrToFloatArr(int[] input)
{
int size=input.length;
Float[] answer = new Float[size];
for(int i=0;i{
answer[i]=(float)input[i];
}
return answer;
}
private static float calculateSimilarity(Float[] a,Float[] b)
{
float answer=0;
for(int i=0;i{
answer+=a[i]*b[i];
}
return answer;
}
private static Float[] calculateClusterRepresentative(int[] cluster,int [][] input,int noOFTokens)
{
Float[] answer= new Float[noOFTokens];
for(int i=0;i{
answer[i]=Float.parseFloat("0");
}
for(int i=1;i<=cluster[0];++i)
{
for(int j=0;j{
answer[j]+=input[cluster[i]][j];
}
}
for(int i=0;i{
answer[i]/=cluster[0];
}
return answer;
}
}
Output of Single Pass Algorithm
Enter the no of Tokens
5
Enter the no of Documents
5
Enter the threshhold
10
Enter the Document Token Matrix
Enter(0,0)
1
Enter(0,1)
3
Enter(0,2)
3
Enter(0,3)
2
Enter(0,4)
2
Enter(1,0)
2
Enter(1,1)
1
Enter(1,2)
0
Enter(1,3)
1
Enter(1,4)
2
Enter(2,0)
0
Enter(2,1)
2
Enter(2,2)
0
Enter(2,3)
0
Enter(2,4)
1
Enter(3,0)
0
Enter(3,1)
3
Enter(3,2)
1
Enter(3,3)
0
Enter(3,4)
5
Enter(4,0)
1
Enter(4,1)
0
Enter(4,2)
1
Enter(4,3)
0
Enter(4,4)
1
0 0 1 3
1 2
2 4
bhamare says
incomplete code plz send full,& correct code
ProjectsGeek says
Let us check this code and we will let you know.