How can I perform K-means clustering on time series data?

Question

How can I do K-means clustering of time series data? I understand how this works when the input data is a set of points, but I don't know how to cluster a time series with 1XM, where M is the data length. In particular, I'm not sure how to update the mean of the cluster for time series data.

I have a set of labelled time series, and I want to use the K-means algorithm to check whether I will get back a similar label or not. My X matrix will be N X M, where N is number of time series and M is data length as mentioned above.

Does anyone know how to do this? For example, how could I modify this k-means MATLAB code so that it would work for time series data? Also, I would like to be able to use different distance metrics besides Euclidean distance.

To better illustrate my doubts, here is the code I modified for time series data:

---

% Check if second input is centroids
if ~isscalar(k) 
    c=k;
    k=size(c,1);
else
    c=X(ceil(rand(k,1)*n),:); % assign centroid randomly at start
end

% allocating variables
g0=ones(n,1); 
gIdx=zeros(n,1);
D=zeros(n,k);

% Main loop converge if previous partition is the same as current
while any(g0~=gIdx)
%     disp(sum(g0~=gIdx))
    g0=gIdx;
    % Loop for each centroid
    for t=1:k
        %  d=zeros(n,1);
        % Loop for each dimension
        for s=1:n
            D(s,t) = sqrt(sum((X(s,:)-c(t,:)).^2)); 
        end
    end
    % Partition data to closest centroids
    [z,gIdx]=min(D,[],2);
    % Update centroids using means of partitions
    for t=1:k

        % Is this how we calculate new mean of the time series?
        c(t,:)=mean(X(gIdx==t,:));

    end
end

Answer 1

Time series are usually high-dimensional. And you need specialized distance function to compare them for similarity. Plus, there might be outliers.

k-means is designed for low-dimensional spaces with a (meaningful) euclidean distance. It is not very robust towards outliers, as it puts squared weight on them.

Doesn't sound like a good idea to me to use k-means on time series data. Try looking into more modern, robust clustering algorithms. Many will allow you to use arbitrary distance functions, including time series distances such as DTW.

Answer 2

It's probably too late for an answer, but:

• k-means can be used to cluster longitudinal data
• Anony-Mousse is right, DWT distance is the way to go for time series

The methods above use R. You'll find more methods by looking, e.g., for "Iterative Incremental Clustering of Time Series".

Answer 3

I have recently come across the kml R package which claims to implement k-means clustering for longitudinal data. I have not tried it out myself.

Also the Time-series clustering - A decade review paper by S. Aghabozorgi, A. S. Shirkhorshidi and T. Ying Wah might be useful to you to seek out alternatives. Another nice paper although somewhat dated is Clustering of time series data-a survey by T. Warren Liao.

Answer 4

If you did really want to use clustering, then dependent on your application you could generate a low dimensional feature vector for each time series. For example, use time series mean, standard deviation, dominant frequency from a Fourier transform etc. This would be suitable for use with k-means, but whether it would give you useful results is dependent on your specific application and the content of your time series.

Answer 5

I don't think k-means is the right way for it either. As @Anony-Mousse suggested you can utilize DTW. In fact, I had the same problem for one of my projects and I wrote my own class for that in Python. The logic is;

1. Create your all cluster combinations. k is for cluster count and n is for number of series. The number of items returned should be n! / k! / (n-k)!. These would be something like potential centers.
2. For each series, calculate distances for each center in each cluster groups and assign it to the minimum one.
3. For each cluster groups, calculate total distance within individual clusters.
4. Choose the minimum.

And, the Python implementation is here if you're interested.