Hello people, yes, there were several threads about this topic, but I sadly have to respawn it, I'm sorry.
The first I found was a discussion from May 2005: http://mail-archives.apache.org/mod_mbox/lucene-java-user/200505.mbox/[EMAIL PROTECTED] There the final solution suggestion from Hoss was to try it with a binary search on the TermEnum The second one was also from May 2005, it seems that it was a follow-up: http://mail-archives.apache.org/mod_mbox/lucene-java-user/200505.mbox/[EMAIL PROTECTED] http://markmail.org/message/rp4xfdclsha7h5uq#query:termenum%20lucene%20maximum%20value+page:1+mid:q3doh6tvyf6swl6h+state:results Here, the solution with the FieldCache was discussed, and also another direction with a RangeQuery which results in a TooMAnyClauses that can be avoided by setting the 'allowed clauses count' to a bigger number. I use the solution with the FieldCache, which worked fine for a long time, but when I use it with bigger fields with millions of entries, I have big peaks in memory consumption, which sometimes result in an OutOfMemory Error. I assume that with the range query, I will fall in the same Problem. I now want to try out the solution from Hoss with the binary search over the TermEnum, but it is not clear for me how to perform this. The only methods in TermEnum are public abstract boolean next() public abstract Term term(); public abstract int docFreq(); public abstract void close() public boolean skipTo(Term target) Whereby skipTo "Skips terms to the first beyond the current whose value is greater or equal to 'target'. Returns true iff there is such an entry." How to avoid to perfom the 'big loop with next' until I am at the last entry, like the current implementation of skipTo: do { if (!next()) return false; } while (target.compareTo(term()) > 0); return true; Whereby target() would be the over biggest value we could think about, and I remember the term bevore the method returns false. Because of the tree-like architecture of the index, where the letters are some kind of nodes, e.g. a z ab ar ze zu abi ark zer zul I would assume that there is a fast possibility to determine that 'abi' is the minimum and 'zul' the maximum for that field - by simply walking through the tree 'left - or rightwise' (when I only get the left node, I will walk to the minimum, when I only get the right node through walking, I will get the maximum) But this is a theoretical view. Enables the Lucene implementation walkthroughs like this? At least the RangeQuery implementation I would assume walks throgh the tree. Thanks for all answers! kindly regards Chris
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