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I have written k means clustering code in c#. I am clustering random 99 text articles of Sports Area which I downloaded from Github for different values of K i.e.3,4,5,6,7. I want to analyze the time period taken for K means Clustering

Following is the time taken for different values of K

K = 3     Time taken = 2 and half hour
K = 4     Time taken = 3 hour 30 minute
K = 5     Time taken = 3 hour 50 minute
K = 6     Time taken = 3 hour 32 minute
K = 7     Time taken = 3 hour 35 minute

As you can see that time taken increases from K= 3 to K =5 and then decreases for K=6 and increases again for K = 7.

My question : Is this result valid? As the value of K increases shouldn't time taken by K means algorithm also increase? Suggestions needed

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    $\begingroup$ Did you check also the number of iterations taken before convergence in each case? $\endgroup$ – Giuseppe Nov 24 '18 at 13:07
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    $\begingroup$ It (the number of iterations needed to converge to a given threshold) is data-dependent and not clearly related to k. Moreover, if k is very large so it approaches the number of objects, less time will be needed. $\endgroup$ – ttnphns Nov 24 '18 at 16:09
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It seems you have only performed one run per k. As K-Means clustering is very sensitive to initialization and often does not produce the same result twice, I consider your sample too small to judge the influence of k on the runtime of k-means.

Further, as Giuseppe points out, the number of iterations are most certainly not equal across your 5 runs.

Finally, as ttnphns points out, the number of iterations - thus the time needed - is data-dependent.

Your result therefore seems valid. The runtime of your algorithm depends on multiple factors (data properties, initialization) and therefore does not neccesarily increase proportionally with k.

Have a look at Run time analysis of the clustering algorithm (k-means) for further information and material on the topic.

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  • $\begingroup$ The no of iterations were different for each value of K...I did not note them but I remember it being different..... $\endgroup$ – Annonymous programmer Nov 25 '18 at 10:54

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