I wonder what is the usefulness of k-means clustering in high dimensional spaces, and why it can be better (or not) than other clustering methods when dealing with high dimensional spaces.
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1$\begingroup$ The answer can be found here: How to tell if data is "clustered" enough for clustering algorithms to produce meaningful results. Although that is only a particular answer to a more general question, this Q may be a duplicate. $\endgroup$– gung - Reinstate MonicaCommented Apr 11, 2014 at 18:01
1 Answer
Is k-means meaningful at all?
See for example my answer here: https://stats.stackexchange.com/a/35760/7828
k-means optimizes variances. Is the unweighted sum of variances meaningful on your data set? Probably not. How can then k-means be meaningful? In high-dimensional data, distance doesn't work. But variance = squared Euclidean distance; so is it meaningful to optimize something of which you know it doesn't work in high-dimensional data?
For the particular problems of high-dimensional data, I recommend the following study:
Zimek, A., Schubert, E. and Kriegel, H.-P. (2012), A survey on unsupervised outlier detection in high-dimensional numerical data. Statistical Analy Data Mining, 5: 363–387. doi: 10.1002/sam.11161
It's main focus is outlier detection, but the observations on the challenges of high-dimensional data apply to a much broader context. They show some simple experiments how high-dimensional data can be a problem. What I like about this study is they also show that high-dimensional data can be easy, too; it's not black and white, but you need to carefully study your data.
Useful is different. Often people use k-means not to actually discover clusters.
But to find representative objects. It's a clever way of semi-random sampling k objects that aren't too similar to be useful.
If you only need a clever way of sampling, k-means may be very useful.
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4$\begingroup$ This answer might be really meaningful if you show
In high-dimensional data, distance doesn't work
- elaborate it, in the specific context of clustering. It is what the OP presumably wants to hear - demonstration or proof. For example, please do show that euclidean distance becomes less meaningful in 1D-2D-3D sequence. $\endgroup$– ttnphnsCommented Apr 11, 2014 at 19:12 -
$\begingroup$ Thank you for the suggestion. I've added a reference that discusses this in detail, and that I found very valueable. $\endgroup$ Commented Apr 12, 2014 at 18:43
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$\begingroup$ @Anony-Mousse thanks for answering this. And thanks to ttnphns too for the suggestion :) $\endgroup$ Commented Feb 20, 2016 at 20:16