How can I get top terms for a subset of documents in a Lucene index?

Question

I know its possible to get the top terms within a Lucene Index, but is there a way to get the top terms based on a subset of a Lucene index?

I.e. What are the top terms

Answer 1

Ideally there'd be a utility somewhere to do this, but I'm not aware of one. However, it's not too hard to do this "by hand" in a reasonably efficient way. I'll assume that you already have a Query and/or Filter object that you can use to define the subset of interest.

First, build a list in memory of all of the document IDs in your index subset. You can use IndexSearcher.search(Query, Filter, HitCollector) to do this very quickly; the HitCollector documentation includes an example that seems like it ought to work, or you can use some other container to store your doc IDs.

Next, initialize an empty HashMap (or whatever) to map terms to total frequency, and populate the map by invoking one of the IndexReader.getTermFreqVector methods for every document and field of interest. The three-argument form seems simpler, but either should be just fine. For the three-argument form, you'd make a TermVectorMapper whose map method checks if term is in the map, associates it with frequency if not, or adds frequency to the existing value if so. Be sure to use the same TermVectorMapper object across all of the calls to getTermFreqVector in this pass, rather than instantiating a new one for each document in the loop. You can also speed things up quite a bit by overriding isIgnoringPositions() and isIgnoringOffsets(); your object should return true for both of those. It looks like your TermVectorMapper might also be forced to define a setExpectations method, but that one doesn't need to do anything.

Once you've built your map, just sort the map items by descending frequency and read off however many top terms you like. If you know in advance how many terms you want, you might prefer to do some kind of fancy heap-based algorithm to find the top k items in linear time instead of using an O(n log n) sort. I imagine the plain old sort will be plenty fast in practice. But it's up to you.

If you prefer, you can combine the first two stages by having your HitCollector invoke getTermFreqVector directly. This should certainly produce equally correct results, and intuitively seems like it would be simpler and better, but the docs seem to warn that doing so is likely to be quite a bit slower than the two-pass approach (on same page as the HitCollector example above). Or I could be misinterpreting their warning. If you're feeling ambitious you could try it both ways, compare, and let us know.