How to display separate disconnected trees in MATLAB when doing hierarchical clustering and producing dendrograms? - matlab

Question

I am working with MATLAB, and I have an adjacency matrix:
mat =
0 1 0 0 0 0
1 0 0 0 1 0
0 0 0 1 0 0
0 0 1 0 0 1
0 1 0 0 0 0
0 0 0 1 0 0
which is not fully connected. Nodes {1,2,5} are connected, and {3,4,6} are connected (the edges are directed).
I would like to see the separate clusters in a dendrogram on a single plot. Since there is not path from one cluster to the next, I would like to see separate trees with separate roots for each cluster. I am using the commands:
mat=zeros(6,6)
mat(1,2)=1;mat(2,1)=1;mat(5,2)=1;mat(2,5)=1;
mat(6,4)=1;mat(4,6)=1;mat(3,4)=1;mat(4,3)=1;
Y=pdist(mat)
squareform(Y)
Z=linkage(Y)
figure()
dendrogram(Z)
These commands are advised from Hierarchical Clustering. And the result is attached: imageDendrogram. Other than that the labels don't make sense, the whole tree is connected, and I connect figure out how to have several disconnected trees which reflect the disconnected nature of the data. I would like to avoid multiple plots as I wish to work with larger datasets that may have many disjoint clusters.

Answer 1

I see this was asked a while ago, but in case you're still interested, here's something to try:
First extract the values above the diagonal from the adjacency matrix, like so:
>> matY = [];
>> for n = 2:6
for m = n:6
matY = [matY mat(n,m)];
end
end
>> matY
matY =
Columns 1 through 13
0 0 0 1 0 0 1 0 0 0 0 1 0
Columns 14 through 15
0 0
Now you have something that looks like the Y vector pdist would have produced. But the values here are the opposite of what you probably want; the unconnected vertices have a "distance" of zero and the connected ones are one apart. Let's fix that:
>> matY(matY == 0) = 10
matY =
Columns 1 through 13
10 10 10 1 10 10 1 10 10 10 10 1 10
Columns 14 through 15
10 10
Better. Now we can compute a regular cluster tree, which will represent connected vertices as "close together" and non-connected ones as "far apart":
>> linkage(matY)
ans =
3 6 1
1 5 1
2 4 1
7 8 10
9 10 10
>> dendrogram(ans)
The resulting diagram:
Hope this is a decent approximation of what you're looking for.