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| bio | website | google.com |
|---|---|---|
| location | North Pole | |
| age | 14 | |
| visits | member for | 1 year, 5 months |
| seen | May 14 at 21:30 | |
| stats | profile views | 306 |
Prefers to stay anonymous and eat mousse au chocolat.
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Mar 22 |
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Cluster on high dimensional categorical data (Images with keywords) Any combination of keywords is an itemset. There is no notion for disjoint itemsets; this constraint it usually not considered interesting. Usually, one is interested in maximal itemsets, i.e. such that no larger superset of keywords exist that still is frequent. |
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Mar 22 |
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Pros and cons of clustering algorithms? Actually, pre-built decision trees are the cheapest classifiers, resource-wise. You can translate them to native code trivially; that is much of their beauty (plus their understandability). |
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Mar 21 |
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Pros and cons of clustering algorithms? That sounds more as if you want to do classification with respect to the clusters you obtained earlier. Which is very natural to do with decision trees, actually. |
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Mar 21 |
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Cluster on high dimensional categorical data (Images with keywords) For this definition of "clusters", you may want to look at frequent itemset mining instead. It's not cluster analysis, but probably what you are looking for: frequently used keyword combinations implicitely also group transactions (= images). |
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Mar 20 |
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Cluster on high dimensional categorical data (Images with keywords) What would be a good cluster? I mean, you don't want just random partitions! |
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Mar 18 |
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Cycling in k-means algorithm Let me emphasize (since this overlooked often) that the convergence proofs need (squared) euclidean distance, so that the distance function and the mean function optimize the same criterion. If you use a different distance (actually, you should not use a distance, but "least sum of squares") you may lose convergence in k-means. |
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Mar 15 |
answered | Pros and cons of clustering algorithms? |
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Mar 13 |
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difference in training and testing procedure of model edited tags |
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Mar 10 |
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How do I mathematically prove that k-means clustering converges to minimum squared error? Your question is not talking about k-means, but about the Lloyd/Forgy algorithm, which is a heuristic and does not guarantee to find the optimum assignment. |
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Mar 10 |
answered | How do I mathematically prove that k-means clustering converges to minimum squared error? |
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Mar 7 |
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rounding when making a histogram Except that when you have data in e.g. $[0:1]$, you end up having an extra bin, containing $[1;1]$ only, as $1$ will not be included in the previous bins range. |
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Mar 4 |
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Measuring k-means clustering quality on training and test sets Whenever I tried k-means, the returned clusters were useless to me. It's optimizing a statistic - sum of squares - that is meaningless for any real data that I have worked with so far. Other clustering algorithms provided much more useful results. |
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Mar 4 |
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Comparison of k-means clustering output Yes, but uniform data does not have reasonable clusters. The clustering algorithm should reply "no clusters found". |
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Mar 4 |
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Measuring k-means clustering quality on training and test sets I don't think there is a reasonable test to check that the k-means numbers are more than a local minimum. Sorry. My experience is that k-means results are quite crappy. |
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Mar 4 |
answered | Comparison of k-means clustering output |
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Mar 4 |
answered | Measuring k-means clustering quality on training and test sets |
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Mar 4 |
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Appropriate cluster method? Again, that is the theory of dummy variables. The actual clustering result will have a value there, but that may still not reflect useful clusters. Variance is optimized by 0 or 1, so chances are that you end up just doing some randomized feature selection this way. |
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Feb 27 |
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Numerical Instability of calculating inverse covariance matrix With 21 dimensions, you should try to get much more samples. This is barely enough to define a covariance matrix at all, but no way enough to make the results statistically meaningful. So either, drop some columns, or get more samples. |
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Feb 26 |
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Clustering samples from non-stationary multivariate distributions Statistics requires assumptions. You can do EM with other distributions as well, it's just that nobody seems to use that in practise. Either way it is not "based on geometry". And in fact, EM uses covariances. |
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Feb 26 |
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Clustering samples from non-stationary multivariate distributions Then you must have overlooked EM clustering... |
