blocks|key|837688|text|嗯，这真的是关于Lucene指数的表现。如果您大部分时间只需要这个查询，那么为什么不更新一个整数，在某个节点上使用总计数，并可能将其与索引插入一起更新，以便更好地使用每晚以上的查询运行更新？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|837689|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Mmh,
this is really about the performance of that Lucene index. If you just need this single query most of the time, why not update an integer with the total count on some node somewhere, and maybe update that together with the index insertions, for good measure run an update with the query above every night on it?

blocks|key|1691962|text|最后，用Gremlin而不是Cypher，我找到了解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1691963|g.getRawGraph().index().forNodes('NAME_OF_USERS_INDEX').query(
++++new+org.neo4j.index.lucene.QueryContext('*')
).size()|code-block|syntax|javascript|1691964|此方法使用lucene索引获取“接近”行。|1691965|再次感谢大家。|1691966|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Finaly, using Gremlin instead of Cypher, I found the solution. 

<pre><code>g.getRawGraph().index().forNodes('NAME_OF_USERS_INDEX').query(
 new org.neo4j.index.lucene.QueryContext('*')
).size()
</code></pre>

This method uses the lucene index to get "aproximate" rows.

Thanks again to all.

blocks|key|1214248|text|如果希望获得最佳性能，请不要将实体类别建模为节点上的属性。相反，你要这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1214249|公司1-：IS_ENTITY->公司实体|1214250|或者如果您正在使用2.0|1214251|公司1:公司|1214252|第二个功能还允许您在一个单独的后台线程中自动更新索引，顺便说一句，imo是2.0最好的新特性之一|1214253|第一种方法还应该证明效率更高，因为与从节点读取属性相比，创建一个“跃点”通常花费更少的时间。但是，它确实要求您为实体创建单独的索引。|1214254|您的查询如下所示：|1214255|v2.0|offset|length|style|BOLD|1214256|MATCH+company:COMPANY
RETURN+count(company)|code-block|syntax|javascript|1214257|v1.9|1214258|START+entity=node:entityindex(value='company')
MATCH+company-[:IS_ENTITIY]->entity
RETURN+count(company)|1214259|entityMap^0|0|0|0|0|0|0|0|0|4|0|0|0|4|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|15|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|16|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|17|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|18|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|19|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|1A|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|1B|8|@$P|1C|Q|1D|R|S]]|9|@]|A|$]]|$1|T|3|U|5|V|7|1E|8|@]|9|@]|A|$W|X]]|$1|Y|3|Z|5|6|7|1F|8|@$P|1G|Q|1H|R|S]]|9|@]|A|$]]|$1|10|3|11|5|V|7|1I|8|@]|9|@]|A|$W|X]]|$1|12|3|-4|5|6|7|1J|8|@]|9|@]|A|$]]]|13|$]]

If you want the best performance, don't model your entity categories as properties on the node. In stead, do it like this :

company1-[:IS_ENTITY]->companyentity

Or if you are using 2.0

company1:COMPANY

The second would also allow you automatically update your index in a separate background thread by the way, imo one of the best new features of 2.0

The first method should also proof more efficient, since making a "hop" in general takes less time than reading a property from a node. It does however require you to create a separate index for the entities.

Your queries would look like this :

v2.0

<pre><code>MATCH company:COMPANY
RETURN count(company)
</code></pre>

v1.9

<pre><code>START entity=node:entityindex(value='company')
MATCH company-[:IS_ENTITIY]-&gt;entity
RETURN count(company)
</code></pre>

We're developing an application based on <a href="/questions/tagged/neo4j" class="post-tag" title="show questions tagged 'neo4j'" rel="tag">neo4j</a> and <a href="/questions/tagged/php" class="post-tag" title="show questions tagged 'php'" rel="tag">php</a> with about 200k nodes, which every node has a property like <code>type='user'</code> or <code>type='company'</code> to denote a specific entity of our application. We need to get the count of all nodes of a specific type in the graph.

We created an index for every entity like <code>users</code>, <code>companies</code> which holds the nodes of that property. So inside <code>users</code> index resides 130K nodes, and the rest on <code>companies</code>.

With Cypher we quering like this.

<pre><code>START u=node:users('id:*')
RETURN count(u)
</code></pre>

And the results are

<pre><code>Returned 1 row.Query took 4080ms
</code></pre>

The Server is configured as default with a little tweaks, but 4 sec is too for our needs. Think that the database will grow in 1 month 20K, so we need this query performs very very much.

Is there any other way to do this, maybe with Gremlin, or with some other server plugin?
I'll cache those results, but I want to know if is possible to tweak this.

Thanks a lot and sorry for my poor english.

neo4j count nodes performance on 200K nodes and 450K relations

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我们正在开发一个基于和的应用程序，其中包含大约200 k节点，每个节点都有一个属性(如type='user'或type='company' )来表示应用程序的特定实体。我们需要得到图中特定类型的所有节点的计数。我们为每个实体创建了一个索引，比如保存该属性节点的users，companies。因此，在users索引中驻留...

问neo4j计数节点在200 K节点上的性能及450 K关系
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