Hi all,
I confirm that indexing the edges is the only way to have a O(1)-O(LogN)
time. Once indexed you can start from edges and get in/out vertices.


Best Regards,

Luca Garulli
CEO at Orient Technologies LTD
the Company behind OrientDB
http://about.me/luca.garulli


On 12 May 2015 at 12:47, <[email protected]> wrote:

> Hi,
>
> That is not applicable. Any "where clause" would demand the graph to visit
> all edges-vertexes even when looking for the newest 100 (to look up for the
> remote property).
> When there are 1 million+ edges that takes a very long time and is not
> acceptable.
>
> Thanks for playing!
>
> regards,
>  -Stefan
>
>
> On Tuesday, 12 May 2015 07:54:03 UTC, Izzet Pembeci wrote:
>>
>> select out('ACTOR')[79220,79221,79222,79223,79224,79225,79226,79227,79228
>> ,79229,79230] from #140:0
>>
>> Can you transform the above query to something like:
>>
>> select out('ACTOR') from #140:0 WHERE ...
>>
>>
>> If this works in a faster way, one may conclude that implementation of []
>> has some performance problems. I also think that with your reasonable bumps
>> you earned yourself the right to open an issue on this one.
>>
>> iZzeT
>>
>>
>> On Monday, May 11, 2015 at 2:21:27 AM UTC+3, [email protected]
>> wrote:
>>>
>>> Hi,
>>>
>>> I do believe that in this case neither are applicable.
>>> I think the ridbag offers only traversal (not fetching by positions
>>> directly) and the edge has not property to index by (it's on the outgoing
>>> vertex).
>>>
>>> Regards,
>>>   -Stefan
>>>
>>> On Sunday, 10 May 2015 15:25:18 UTC, Ziink A wrote:
>>>>
>>>> I'm still evaluating OrientDB so I might be totally off but I would
>>>> index the edge class (SOME_LABEL).
>>>>
>>>> Also take a look at http://orientdb.com/docs/last/RidBag.html
>>>>
>>>>
>>>> On Thursday, April 30, 2015 at 4:27:40 AM UTC-7,
>>>> [email protected] wrote:
>>>>>
>>>>> Hi,
>>>>>
>>>>> I have very dense graph that contains vertexes with a lot of edges and
>>>>> I need to fetch the X last edges added to the Vertex.
>>>>>
>>>>> Orient SQL allows me to do it like this:
>>>>>
>>>>>    - select out('SOME_LABEL')[80000] from #1:0
>>>>>    - please note that this is single direction (out) and a single
>>>>>    link type / label ('SOME_LABEL')
>>>>>
>>>>> I have several questions regarding this:
>>>>>
>>>>>    - Are the edges in a consistent order?
>>>>>    - Assuming append-only operations and no deletions
>>>>>
>>>>>    - Can anything be done to speed this up?
>>>>>    - I ask because this query is very slow (0.7 sec.)
>>>>>    - Asking for a list "select
>>>>>    
>>>>> out('ACTOR')[79220,79221,79222,79223,79224,79225,79226,79227,79228,79229,79230]
>>>>>    from #140:0" takes almost n*req_time longer
>>>>>
>>>>>    - What happens underneath (is the whole list iterated from top to
>>>>>    get to this)
>>>>>
>>>>>    - Can this be achieved using the Java API?
>>>>>
>>>>> Assistance is highly appreciated.
>>>>>
>>>>> Best regards,
>>>>>  -Stefan
>>>>>
>>>>  --
>
> ---
> You received this message because you are subscribed to the Google Groups
> "OrientDB" group.
> To unsubscribe from this group and stop receiving emails from it, send an
> email to [email protected].
> For more options, visit https://groups.google.com/d/optout.
>

-- 

--- 
You received this message because you are subscribed to the Google Groups 
"OrientDB" group.
To unsubscribe from this group and stop receiving emails from it, send an email 
to [email protected].
For more options, visit https://groups.google.com/d/optout.

Reply via email to