Hi herbie,
If your query includes an inequality (such as x>50), then your first sort
order has to be on the same property as that inequality, which means you
can't (directly) fetch the most recent 200 results with x>50. You either
need to change your query to use only equality filters, or you need to fetch
extra results, then sort them in memory and only take the most recent ones.
-Nick Johnson
On Tue, Jun 23, 2009 at 1:44 PM, herbie <4whi...@o2.co.uk> wrote:
>
> Thanks for your help Nick.
>
> No my threshold value 'x' isn't constant. I still havn't got my head
> round this yet! Can you tell me how to get the latest entities
> (assuming I don't want all of them) out of the datastore and filter
> on another property?
>
> For example: Get the latest 200 entities where x > 50. I don't
> care what their 'date's are as long as I get the latest and x > 50.
>
>
> On Jun 23, 1:16 pm, "Nick Johnson (Google)" <nick.john...@google.com>
> wrote:
> > On Tue, Jun 23, 2009 at 12:42 PM, herbie <4whi...@o2.co.uk> wrote:
> >
> > > So will this :
> > > query = Foo.all().filter("property_x >" 50).order("property_x") .order
> > > ("-timestamp")
> > > results = query.fetch(200)
> >
> > > ..get the latest entities where property_x > 50 ? Or will it get the
> > > 200 properties with the largest 'property_x' which are then ordered
> > > by 'timestamp' ? A subtle but important difference.
> >
> > It will get the 200 entities with the smallest property_x greater than 50
> > (since you're filtering >50 and ordering first by property_x). If two
> > entities have the same value for property_x, they will be sorted by
> > timestamp, descending.
> >
> > If you need the latest, and your threshold of 50 is a constant, you can
> add
> > a BooleanProperty to your entity group encoding the condition 'is greater
> > than 50', and filter on that using an equality filter.
> >
> > -Nick Johnson
> >
> >
> >
> >
> >
> > > As I said I need make sure I get the latest entities.
> >
> > > On Jun 22, 11:33 pm, Tony <fatd...@gmail.com> wrote:
> > > > Yes, that is what it means. I forgot about that restriction.
> >
> > > > I see what you mean about changing 'x' values. Perhaps consider
> > > > keeping two counts - a running sum and a running count (of the # of x
> > > > properties). If a user modifies an 'x' value, you can adjust the sum
> > > > up or down accordingly.
> >
> > > > On Jun 22, 5:40 pm, herbie <4whi...@o2.co.uk> wrote:
> >
> > > > > I tried your query below but I get "BadArgumentError: First
> ordering
> > > > > property must be the same as inequality filter property, if
> specified
> > > > > for this query;"
> > > > > Does this mean I have to order on 'x' first, then order on 'date'?
> > > > > Will this still return the latest 200 of all entities with x > 50
> if
> > > > > I call query.fetch(200)?
> >
> > > > > I take your's and Nick's about keeping a 'running average'. But
> in
> > > > > my example the user can change the 'x' value so the average has to
> be
> > > > > recalculated from the latest entities.
> >
> > > > > On Jun 22, 9:46 pm, Tony <fatd...@gmail.com> wrote:
> >
> > > > > > You could accomplish this task like so:
> >
> > > > > > xlist = []
> > > > > > query = Foo.all().filter("property_x >" 50).order("-timestamp")
> > > > > > for q in query:
> > > > > > xlist.append(q.property_x)
> > > > > > avg = sum(xlist) / len(xlist)
> >
> > > > > > What Nick is saying, I think, is that fetching 1000 entities is
> going
> > > > > > to be very resource-intensive, so a better way to do it is to
> > > > > > calculate this data at write-time instead of read-time. For
> example,
> > > > > > every time you add an entity, you could update a separate entity
> that
> > > > > > has a property like "average = db.FloatProperty()" with the
> current
> > > > > > average, and then you could simply fetch that entity and get the
> > > > > > current running average.
> >
> > > > > > On Jun 22, 4:25 pm, herbie <4whi...@o2.co.uk> wrote:
> >
> > > > > > > Ok. Say I have many (>1000) Model entities with two properties
> 'x'
> > > > > > > and 'date'. What is the most efficient query to fetch say
> the
> > > > > > > latest 200 entities where x > 50. I don't care what their
> > > 'date's
> > > > > > > are as long as I get the latest and x > 50
> >
> > > > > > > Thanks again for your help.
> >
> > > > > > > On Jun 22, 4:11 pm, "Nick Johnson (Google)" <
> > > nick.john...@google.com>
> > > > > > > wrote:
> >
> > > > > > > > Consider precalculating this data and storing it against
> another
> > > entity.
> > > > > > > > This will save a lot of work on requests.
> >
> > > > > > > > -Nick Johnson
> >
> > > > > > > > On Mon, Jun 22, 2009 at 3:55 PM, herbie <4whi...@o2.co.uk>
> > > wrote:
> >
> > > > > > > > > No the users won't need to read 1000 entities, but I want
> to
> > > calculate
> > > > > > > > > the average of a property from the latest 1000 entities.
> >
> > > > > > > > > On Jun 22, 3:30 pm, "Nick Johnson (Google)" <
> > > nick.john...@google.com>
> > > > > > > > > wrote:
> > > > > > > > > > Correct. Are you sure you need 1000 entities, though?
> Your
> > > users probably
> > > > > > > > > > won't read through all 1000.
> >
> > > > > > > > > > -Nick Johnson
> >
> > > > > > > > > > On Mon, Jun 22, 2009 at 3:23 PM, herbie <
> 4whi...@o2.co.uk>
> > > wrote:
> >
> > > > > > > > > > > So to be sure to get the latest 1000 entities I should
> add
> > > a datetime
> > > > > > > > > > > property to my entitie model and filter and sort on
> that?
> >
> > > > > > > > > > > On Jun 22, 1:42 pm, herbie <4whi...@o2.co.uk> wrote:
> > > > > > > > > > > > I know that if there are more than 1000 entities that
> > > match a query,
> > > > > > > > > > > > then only 1000 will be return by fetch(). But my
> > > question is which
> > > > > > > > > > > > 1000? The last 1000 added to the datastore? The
> first
> > > 1000 added to
> > > > > > > > > > > > the datastore? Or is it undedined?
> >
> > > > > > > > > > > > Thanks
> > > > > > > > > > > > Ian
> >
> > > > > > > > > > --
> > > > > > > > > > Nick Johnson, App Engine Developer Programs Engineer
> > > > > > > > > > Google Ireland Ltd. :: Registered in Dublin, Ireland,
> > > Registration
> > > > > > > > > Number:
> > > > > > > > > > 368047
> >
> > > > > > > > --
> > > > > > > > Nick Johnson, App Engine Developer Programs Engineer
> > > > > > > > Google Ireland Ltd. :: Registered in Dublin, Ireland,
> > > Registration Number:
> > > > > > > > 368047
> >
> > --
> > Nick Johnson, App Engine Developer Programs Engineer
> > Google Ireland Ltd. :: Registered in Dublin, Ireland, Registration
> Number:
> > 368047
> >
>
--
Nick Johnson, App Engine Developer Programs Engineer
Google Ireland Ltd. :: Registered in Dublin, Ireland, Registration Number:
368047
--~--~---------~--~----~------------~-------~--~----~
You received this message because you are subscribed to the Google Groups
"Google App Engine" group.
To post to this group, send email to google-appengine@googlegroups.com
To unsubscribe from this group, send email to
google-appengine+unsubscr...@googlegroups.com
For more options, visit this group at
http://groups.google.com/group/google-appengine?hl=en
-~----------~----~----~----~------~----~------~--~---
