5

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)

我想对给定用户执行查询，在哪里找到该用户和其他所有用户之间的共同艺术家。

为了大致了解我的数据库的大小，我有大约 600 个用户、47,546 位艺术家以及用户和艺术家之间的 184,211 个关系。

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。这似乎不需要 20 秒。

后来我又回到了这个问题上，并尝试从一开始就对其进行调试。

我开始分解查询，我注意到我得到了更快的结果。如果没有 Neo4J Spatial 查询，我在大约 1.5 秒内得到结果。

我终于添加了一些东西，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。5倍的提升！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)) |
+----------------------+-------+--------+--------------------+------------------------------------------------------------------------------------------------+

而且，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

除非我发疯了，否则这两个查询之间的唯一区别是节点的名称（我已将“me”更改为“u”，将“other”更改为“u2”）。

为什么这会导致 5 倍的改进？？！？！

如果有人对此有任何见解，我将永远感激不尽。

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。似乎从字面上看，“其他”和“我”的命名节点导致查询时间急剧增加。我很困惑。

谢谢，-亚当

4

2 回答 2

1

听起来您正在看到缓存的效果。在第一次访问时，缓存不会被填充。由于节点/关系已经在缓存中可用，后续查询同一个图会更快。

于 2014-08-01T06:22:02.443 回答
1

使用OPTIONAL MATCH以下WHERE other:User没有意义，因为结束节点other( u2) 必须匹配。尝试在没有和没有最后一个的情况下执行查询optional match，where简单with地说

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
于 2014-08-01T10:31:08.333 回答