我投了妮可和萨姆的票。
但我认为还有更简单的：

MATCH (a)
RETURN id(a), labels(a) as stereotype, 
size((a)-->()) as out, size((a)<--()) as in

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你可以用你想要的任何属性来改变id(a)。

更新NEO4J 5

MATCH (a)
RETURN 
  COUNT { (a)-->() } AS out,
  COUNT { (a)<--() } AS in

型

我会这样做。假设你已经加载了电影数据集：

MATCH (a)-[r]-(b)
RETURN COALESCE(a.name, a.title), LABELS(a) AS Label,
       SUM(CASE WHEN STARTNODE(r) = a THEN 1 ELSE 0 END) AS outgoing,
       SUM(CASE WHEN STARTNODE(r) = b THEN 1 ELSE 0 END) AS incoming
ORDER BY outgoing DESC

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你可以在这里看到它的结果：http://console.neo4j.org/r/6ebvy8

这个查询应该对你有用。逻辑非常简单。

MATCH (a)
OPTIONAL MATCH (a)-->(b)
WITH a, LENGTH(COLLECT(b)) AS out
OPTIONAL MATCH (c)-->(a)
RETURN a.name, labels(a) AS Stereotype, out, LENGTH(COLLECT(c)) AS in 
ORDER BY out DESC;

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Here is a console，用于演示查询。
查询不能正常工作的原因之一是它使用了COUNT()函数，该函数返回行数，而不是集合的长度（或大小）。

MATCH outg=(a)-[og]->(b), incom=(c)-[ic]->(a) RETURN id(a), labels(a) AS Stereotype,count(og) AS out, count(ic) AS in ORDER BY out DESC

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