我有点难过。
在我的数据库中，我有这样的关系：
(u:User)-[r1:LISTENS_TO]->(a:Artist)<-[r2:LISTENS_TO]-(u2:User)
我想执行一个查询，对于给定的用户，我找到该用户和其他所有用户之间的常见艺术家。
为了了解我的数据库大小，我有大约600位用户，47,546位艺术家，以及184,211位用户和艺术家之间的关系。
我尝试的第一个查询如下：
START me=node(553314), other=node:userLocations("withinDistance:[38.89037,-77.03196,80.467]")
OPTIONAL MATCH
pMutualArtists=(me:User)-[ar1:LISTENS_TO]->(a:Artist)<-[ar2:LISTENS_TO]-(other:User)
WHERE
other:User
WITH other, COUNT(DISTINCT pMutualArtists) AS mutualArtists
ORDER BY mutualArtists DESC
LIMIT 10
RETURN other.username, mutualArtists
这需要大约20秒才能返回。此查询的配置文件如下：
+----------------------+-------+--------+------------------------+------------------------------------------------------------------------------------------------+
| Operator | Rows | DbHits | Identifiers | Other |
+----------------------+-------+--------+------------------------+------------------------------------------------------------------------------------------------+
| ColumnFilter(0) | 10 | 0 | | keep columns other.username, mutualArtists |
| Extract | 10 | 20 | | other.username |
| ColumnFilter(1) | 10 | 0 | | keep columns other, mutualArtists |
| Top | 10 | 0 | | { AUTOINT0}; Cached( INTERNAL_AGGREGATEb6facb18-1c5d-45a6-83bf-a75c25ba6baf of type Integer) |
| EagerAggregation | 563 | 0 | | other |
| OptionalMatch | 52806 | 0 | | |
| Eager(0) | 563 | 0 | | |
| NodeByIndexQuery(1) | 563 | 564 | other, other | Literal(withinDistance:[38.89037,-77.03196,80.467]); userLocations |
| NodeById(1) | 1 | 1 | me, me | Literal(List(553314)) |
| Eager(1) | 82 | 0 | | |
| ExtractPath | 82 | 0 | pMutualArtists | |
| Filter(0) | 82 | 82 | | (hasLabel(a:Artist(1)) AND NOT(ar1 == ar2)) |
| SimplePatternMatcher | 82 | 82 | a, me, ar2, ar1, other | |
| Filter(1) | 1 | 3 | | ((hasLabel(me:User(3)) AND hasLabel(other:User(3))) AND hasLabel(other:User(3))) |
| NodeByIndexQuery(1) | 563 | 564 | other, other | Literal(withinDistance:[38.89037,-77.03196,80.467]); userLocations |
| NodeById(1) | 1 | 1 | me, me | Literal(List(553314)) |
+----------------------+-------+--------+------------------------+------------------------------------------------------------------------------------------------+
我很沮丧。它似乎不需要20秒。
我后来回到了问题，并尝试从头开始调试。
我开始打破查询，我注意到我得到了更快的结果。没有Neo4J Spatial查询，我得到的结果大约是1.5秒。
我最后添加了一些东西，最后得到了以下查询：
START u=node(553314), u2=node:userLocations("withinDistance:[38.89037,-77.03196,80.467]")
OPTIONAL MATCH
pMutualArtists=(u:User)-[ar1:LISTENS_TO]->(a:Artist)<-[ar2:LISTENS_TO]-(u2:User)
WHERE
u2:User
WITH u2, COUNT(DISTINCT pMutualArtists) AS mutualArtists
ORDER BY mutualArtists DESC
LIMIT 10
RETURN u2.username, mutualArtists
此查询在4240毫秒内返回。改进了5倍！此查询的配置文件如下：
+----------------------+-------+--------+--------------------+------------------------------------------------------------------------------------------------+
| Operator | Rows | DbHits | Identifiers | Other |
+----------------------+-------+--------+--------------------+------------------------------------------------------------------------------------------------+
| ColumnFilter(0) | 10 | 0 | | keep columns u2.username, mutualArtists |
| Extract | 10 | 20 | | u2.username |
| ColumnFilter(1) | 10 | 0 | | keep columns u2, mutualArtists |
| Top | 10 | 0 | | { AUTOINT0}; Cached( INTERNAL_AGGREGATEbdf86ac1-8677-4d45-967f-c2dd594aba49 of type Integer) |
| EagerAggregation | 563 | 0 | | u2 |
| OptionalMatch | 52806 | 0 | | |
| Eager(0) | 563 | 0 | | |
| NodeByIndexQuery(1) | 563 | 564 | u2, u2 | Literal(withinDistance:[38.89037,-77.03196,80.467]); userLocations |
| NodeById(1) | 1 | 1 | u, u | Literal(List(553314)) |
| Eager(1) | 82 | 0 | | |
| ExtractPath | 82 | 0 | pMutualArtists | |
| Filter(0) | 82 | 82 | | (hasLabel(a:Artist(1)) AND NOT(ar1 == ar2)) |
| SimplePatternMatcher | 82 | 82 | a, u2, u, ar2, ar1 | |
| Filter(1) | 1 | 3 | | ((hasLabel(u:User(3)) AND hasLabel(u2:User(3))) AND hasLabel(u2:User(3))) |
| NodeByIndexQuery(1) | 563 | 564 | u2, u2 | Literal(withinDistance:[38.89037,-77.03196,80.467]); userLocations |
| NodeById(1) | 1 | 1 | u, u | Literal(List(553314)) |
+----------------------+-------+--------+--------------------+------------------------------------------------------------------------------------------------+
并且，为了证明我连续运行它们并得到了非常不同的结果：
neo4j-sh (?)$ START u=node(553314), u2=node:userLocations("withinDistance:[38.89037,-77.03196,80.467]")
>
> OPTIONAL MATCH
> pMutualArtists=(u:User)-[ar1:LISTENS_TO]->(a:Artist)<-[ar2:LISTENS_TO]-(u2:User)
> WHERE
> u2:User
>
> WITH u2, COUNT(DISTINCT pMutualArtists) AS mutualArtists
> ORDER BY mutualArtists DESC
> LIMIT 10
> RETURN u2.username, mutualArtists
> ;
+------------------------------+
| u2.username | mutualArtists |
+------------------------------+
| "573904765" | 644 |
| "28600291" | 601 |
| "1092510304" | 558 |
| "1367963461" | 521 |
| "1508790199" | 455 |
| "1335360028" | 447 |
| "18200866" | 444 |
| "1229430376" | 435 |
| "748318333" | 434 |
| "5612902" | 431 |
+------------------------------+
10 rows
4240 ms
neo4j-sh (?)$ START me=node(553314), other=node:userLocations("withinDistance:[38.89037,-77.03196,80.467]")
>
> OPTIONAL MATCH
> pMutualArtists=(me:User)-[ar1:LISTENS_TO]->(a:Artist)<-[ar2:LISTENS_TO]-(other:User)
> WHERE
> other:User
>
> WITH other, COUNT(DISTINCT pMutualArtists) AS mutualArtists
> ORDER BY mutualArtists DESC
> LIMIT 10
> RETURN other.username, mutualArtists;
+--------------------------------+
| other.username | mutualArtists |
+--------------------------------+
| "573904765" | 644 |
| "28600291" | 601 |
| "1092510304" | 558 |
| "1367963461" | 521 |
| "1508790199" | 455 |
| "1335360028" | 447 |
| "18200866" | 444 |
| "1229430376" | 435 |
| "748318333" | 434 |
| "5612902" | 431 |
+--------------------------------+
10 rows
20418 ms
除非我疯了，否则这两个查询之间的唯一区别就是节点的名称（我已经改变了＃34;我＆＃34;到了＆＃34;你＆＃34;和＆＃34 ;其他＆＃34;到＆＃34; u2＆＃34;）。
为什么这会导致5倍的提升？！？！
如果有人对此有任何见解，我将永远感激。
谢谢，
- 亚当
编辑8.1.14
根据@ ulkas的建议，我尝试简化查询。
结果是：
START u=node(553314), u2=node:userLocations("withinDistance:[38.89037,-77.03196,80.467]")
OPTIONAL MATCH pMutualArtists=(u:User)-[ar1:LISTENS_TO]->(a:Artist)<-[ar2:LISTENS_TO]-(u2:User)
RETURN u2.username, COUNT(DISTINCT pMutualArtists) as mutualArtists
ORDER BY mutualArtists DESC
LIMIT 10
~4秒
START me=node(553314), other=node:userLocations("withinDistance:[38.89037,-77.03196,80.467]")
OPTIONAL MATCH pMutualArtists=(me:User)-[ar1:LISTENS_TO]->(a:Artist)<-[ar2:LISTENS_TO]-(other:User)
RETURN other.username, COUNT(DISTINCT pMutualArtists) as mutualArtists
ORDER BY mutualArtists DESC
LIMIT 10
~20秒
太奇怪了。看起来好像字面上的＆＃34;其他＆＃34;和＆＃34;我＆＃34;导致查询时间大幅跳跃。我很困惑。
谢谢， - 亚当
答案 0 :(得分：1)
这听起来像是在看到缓存的效果。第一次访问时，不会填充缓存。由于节点/关系已在缓存中可用，因此后续查询命中同一图表的速度会快得多。
答案 1 :(得分：1)
在OPTIONAL MATCH之后使用WHERE other:User没有任何意义，因为结束节点other（u2）必须匹配。尝试在没有optional match和where且没有最后with的情况下执行查询，只需
START me=node(553314), other=node:userLocations("withinDistance[38.89037,-77.03196,80.467]")
MATCH
pMutualArtists=(me:User)-[ar1:LISTENS_TO]->(a:Artist)<-[ar2:LISTENS_TO]-(other:User)
RETURN other.username, count(DISTINCT pMutualArtists) as mutualArtists
ORDER BY mutualArtists DESC
LIMIT 10