Does the dataset have bidirectional friendship relationships? Or just one
relationship per friendship?
If the former, then removing the duplicate and ignoring the direction in
neo4j makes sense.
For a potentially faster query you could use:
*MATCH* *(*n*{*id *:* *{*id*}})-[:*CONNECTED*]-(f1)-[**:*CONNECTED*]-(f2)*
*WITH DISTINCT f2*
*MATCH (f2)**-[**:*CONNECTED*]-(f3)*
*WITH DISTINCT f3*
*MATCH (f3)**-[**:*CONNECTED*]-(f4)*
*WITH DISTINCT f4*
*MATCH (f4)**-[**:*CONNECTED*]-(*friend_of_friend*)*
*RETURN* *distinct* friend_of_friend*.*id
You can also try the new query optimizer that will be available soon by
prefixing your query with "cypher 2.1.experimental"
I can check the dataset.
On Sat, Nov 8, 2014 at 6:22 PM, Kamilos <[email protected]> wrote:
>
> hi, guys,
>>
> thx a lot for your help
> but unfortunately I don't have time right now to do the benchamark
> completely
>
> What I found out is that with the querry "friend of friends" at level 5,
> Neo4J is faster than mysql.
>
> I used the standford dataset from facebook to populate my db :
> http://snap.stanford.edu/data/egonets-Facebook.html
>
> The queries are
>
> Neo4j
>
> *MATCH* *(*n*{*id *:* *{*id*}})-[:*CONNECTED***1..5*]-(*friend_of_friend
> *)*
>
> *RETURN* *distinct* friend_of_friend*.*id
>
> Mysql
>
> *SELECT* *distinct* uid2
>
> *FROM* edge
>
> *WHERE* uid1 *IN* *(*
>
> *SELECT* uid2
>
> *FROM* edge
>
> *WHERE* uid1 *in* *(* *SELECT* uid2
>
> *FROM* edge
>
> *WHERE* uid1 *in* *(* *SELECT* uid2
>
> *FROM* edge
>
> *WHERE* uid1 *in* *(* *SELECT*
> uid2
>
> *FROM* edge
>
> *WHERE* uid1
> *= *?
>
> *)*
>
> *)*
>
> *)*
>
> *);*
>
>
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