> def hasObject(self, id): > """Test if an object is in the current object.""" > return id in self.objectIds()
Because then it always computes all ids, before testing presence (second list scan). Time is N+pos(id) whereas with the scan below, time is pos(id) (N being the total number of subobjects). Florent > > def hasObject(self, id): > > """Test if an object is in the current object.""" > > for o in self._objects: > > if o['id'] == id: > > return 1 > > return 0 > > > > to ObjectManager. > > > > This would bring it in line with BTreeFolder2 (who already has an > > hasObject method) and we could then always use the most efficient method > > to test if a folder has a given subobject id. > > > > Opinions ? > > > > Florent -- Florent Guillaume, Nuxeo (Paris, France) +33 1 40 33 71 59 http://nuxeo.com mailto:[EMAIL PROTECTED] _______________________________________________ Zope-Dev maillist - [EMAIL PROTECTED] http://mail.zope.org/mailman/listinfo/zope-dev ** No cross posts or HTML encoding! ** (Related lists - http://mail.zope.org/mailman/listinfo/zope-announce http://mail.zope.org/mailman/listinfo/zope )
