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Jul 06 '10 at 20:42

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sbirch
2415711

I've always used k-means and set k=n when I didn't know k ahead of time (+some center merging criteria, which seems to come down to h in eq. 1), but it's pretty expensive. From a cursory glance it looks like this is O(n) to my O(n^2). That said, k-means is a lot simpler--I imagine that's why. There's reasonably complex math (at least, non-intuitive), and you have to choose h. Everyone learns bubble sort, even though nobody uses it in practice (although given, everyone learns quicksort & mergesort too).

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eq. 3 is per-element! Making the algorithm quadratic (in line with ogrisel's answer)

Jul 06 '10 at 20:45

sbirch's gravatar image

sbirch
2415711

I've always used k-means and set k=n when I didn't know k ahead of time (+some center merging criteria, which seems to come down to h in eq. 1), but it's pretty expensive. From This is probably a cursory glance it looks like this is O(n) to my O(n^2). That said, more powerful algorithm, but k-means is a lot simpler--I imagine that's why. There's reasonably complex math (at least, non-intuitive), and you have to choose h. Everyone learns bubble sort, even though nobody uses it in practice (although given, everyone learns quicksort & mergesort too).

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