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資源簡介

可以用的經典密度聚類算法實例,

資源截圖

代碼片段和文件信息

function?[class]?=?Dbscan(?Xkeps?)
[mn]=size(X);
class=zeros(1m);
clusterId=1;
X=[class‘X];
n=n+1;

for?i=1:m
??if?X(i1)==0
??X=expandcluster(iXclusterIdepskn);
??if?X(i1)==-1??%?if?is?noisekeep?the?clusterId?not?increasing?
??clusterId=clusterId-1;
??end
??clusterId=clusterId+1;
??end
end
clusterId=clusterId-1;
%?disp(‘clusterNum:‘);
%?disp(clusterId);
class=X(:1)‘;
end

function?[X]=expandcluster(indexXclusterIdepskn)
???Di=dist(X(index(2:n))X(:(2:n)));
???D=find(Di???if?length(D)????X(index1)=-1;%noise
???return
???else
???for?i=1:length(D)
???????if(X(D(i)1)==0)
???????X(D(i)1)=clusterId;
???????end
???end
???end
???D(1)=[];??
??
while?~isempty(D)
???Di=dist(X(D(1)(2:n))X(:(2:n)));
???Result=find(Di???if?length(Result)>=k
???????for?t=1:length(Result)?
?????????if?X(Result(t)1)==0||X(Result(t)1)==-1
????????????if?X(Result(t)1)==0
???????????????D(length(D)+1)=Result(t);
????????????end
????????????X(Result(t)1)=clusterId;
?????????end
???????end
???end
???????D(1)=[];???
end

end


function?[Di]=dist(ix)
[mn]=size(x);
Di=sqrt(sum((((ones(m1)*i)-x).^2)‘));
if?n==1
????Di=abs((ones(m1)*i-x))‘;
end
end



?



?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件????????1251??2014-03-21?21:46??Dbscan.m

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