xsl-list
[Top] [All Lists]

RE: performance issues saxon

2003-02-18 06:49:14
Ok if  the requirement is to group all the products of the 
same id together, I would follow the following steps. 1. Get 
all the unique nodes 2. Group all the products with the id as 
defined by the id of the unique node.


Is there any other way we could avoid using two nested  loops 
for a requirement like this? Thanks Vasu

Absolutely. Muenchian grouping solves this in O(n log n) time. See
www.jenitennison.com/xslt/grouping for the algorithm.

Actually it's essentially the algorithm you described above. Step 1
requires a serial scan of the document, Step 2 requires an index lookup
using a key. Your mistake was to do step 2 with a (second, nested)
serial scan.

Michael Kay
Software AG
home: Michael(_dot_)H(_dot_)Kay(_at_)ntlworld(_dot_)com
work: Michael(_dot_)Kay(_at_)softwareag(_dot_)com 



 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list



<Prev in Thread] Current Thread [Next in Thread>