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clustering, matlab, nominal data

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Radoslav Vandzura
Radoslav Vandzura il 14 Gen 2016
Commentato: Tom Lane il 30 Gen 2016
Hello All. I need an advice. I need recommend method of clustering which is suitable for nominal data in Matlab. Could you help me, please? I appreciate every idea. Thank you in advance.

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Walter Roberson
Walter Roberson il 15 Gen 2016

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Image Analyst
Image Analyst il 15 Gen 2016
Try the Classification Learner app on the Apps tab.
  1 Commento
Tom Lane
Tom Lane il 16 Gen 2016
This could work as a post-processing step to assign new data to classes found from the original data. But classificationLearner would require that you know the clusters (groups) for the original data.

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Tom Lane
Tom Lane il 16 Gen 2016
For hierarchical clustering, consider using Hamming distance. Here's an example that isn't realistic but that illustrates what to do:
x=randi(3,100,4); % noisy coordinates
x(1:50,5:6) = randi(2,50,2); % try to make 1st 50 points closer
x(51:100,5:6) = 2+randi(2,50,2); % next 50 points different
z = linkage(x,'ave','hamming'); % try average linkage clustering
dendrogram(z,100) % show dendrogram with all points
  2 Commenti
Radoslav Vandzura
Radoslav Vandzura il 20 Gen 2016
But I have categorical data and I need to do clustering....My data consist of names of operating system (UNIX, WINDOWS,...), type of virtualisation (virtual system, virtualization host,...)... Can I change these data to number? For example, UNIX-1, WINDOWS-2...??????
I don´t know what do you mean...:( IS the Classification Learner app for Classification not for clustering, isn´t it?
Tom Lane
Tom Lane il 30 Gen 2016
You are right that the clustering functions operate on matrices so you would need to convert your data to numbers. The grp2idx function could be helpful. And yes, the Classification Learner app is aimed at classifying data into known groups. Here is a simple example where you can see the Hamming distance between data represented by a three-category variable and a two-category variable.
>> x = [1 1;2 1;3 1;1 2;2 2;2 3];
>> squareform(pdist(x,'hamming'))
ans =
0 0.5000 0.5000 0.5000 1.0000 1.0000
0.5000 0 0.5000 1.0000 0.5000 0.5000
0.5000 0.5000 0 1.0000 1.0000 1.0000
0.5000 1.0000 1.0000 0 0.5000 1.0000
1.0000 0.5000 1.0000 0.5000 0 0.5000
1.0000 0.5000 1.0000 1.0000 0.5000 0

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