Subject: Re: [xsl] Testing by counting or positional predicate From: Dimitre Novatchev <dnovatchev@xxxxxxxxx> Date: Thu, 11 Jan 2001 04:29:44 -0800 (PST) |
Hi Jeni, I'd try something else -- while in all provided expressions you have to compute the union of the two node-sets, why not try to find the set difference b/n $superset and $subset. That is, not($subset[count(. | $superset) > count($superset)]) In case of a "smart" processor it will stop evaluating the above expression immediately when the first node from $subset that does not belong to $superset is found. In the best case just one node may be tried. In the worst case, the time must be similar to the time for uniting the two node-sets. Cheers, Dimitre. Jeni Tennison wrote: David C. and I have both brought this up recently, but hidden away in the depths of other messages and I don't think anyone has commented on it. So, I was wondering: which is more efficient for testing whether a $subset node set is a subset of a $superset node set? count($subset|$superset) = count($superset) or: not(($subset|$superset)[count($superset) + 1]) Does it make any difference if the $subset is just one node? count($node|$node-set) = count($node-set) or: not(($node|$node-set)[count($node-set) + 1]) Does this extend to testing for identity or is the unioned set so small that it doesn't make any odds? count($node1|$node2) = 1 or: not(($node1|$node2)[2]) Thanks, Jeni --- Jeni Tennison http://www.jenitennison.com/ __________________________________________________ Do You Yahoo!? Yahoo! Photos - Share your holiday photos online! http://photos.yahoo.com/ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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