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Neo4j 计数不可预测

[英]Neo4j count is not predictable

Using the Preferential attachment example in the guides, I have a node C with degree 3 and node E with degree 1使用指南中的优先附件示例,我有一个度数为 3 的节点 C 和度数为 1 的节点 E

UNWIND [["A", "C"], ["A", "B"], ["B", "D"],
    ["B", "C"], ["B", "E"], ["C", "D"]] AS pair
MERGE (n1:Node {name: pair[0]})
MERGE (n2:Node {name: pair[1]})
MERGE (n1)-[:FRIENDS]-(n2)

and when I try a simple degree query for node EI get the correct answer.当我尝试对节点 EI 进行简单度数查询时,得到正确答案。 But when I add another node, the answer changes to 3但是当我添加另一个节点时,答案变为 3

MATCH (e:Node {name: 'E'})--(othere)
RETURN e,othere, count(othere) 

returns 1 for count(othere)计数返回 1(其他)

MATCH (c:Node {name: 'C'})--(otherc)
MATCH (e:Node {name: 'E'})--(othere)
RETURN e,othere, count(othere)

returns 3 for count(othere).计数(其他)返回 3。 Why should this be?为什么会这样?

In your last query's RETURN clause, the COUNT aggregating function counts the number of result rows that have the same e and othere values.在最后一个查询的RETURN子句中, 聚合 functionCOUNT计算具有相同eothere的结果行数 With your sample data, there are 3 such result rows.使用您的样本数据,有 3 个这样的结果行。

Here is one way to get a correct count of the number of relationships between e and othere :这是正确计算eothere之间关系数量的一种方法:

MATCH (c:Node {name: 'C'})--(otherc)
MATCH (e:Node {name: 'E'})-[r]-(othere)
RETURN e, othere, COUNT(DISTINCT r)

[DISCUSSION] [讨论]

In general, there can be any number of relationships between any 2 nodes.通常,任何 2 个节点之间可以有任意数量的关系。 So, to make this a more general discussion, suppose the "E" node has 2 relationships to the "B" node (and no other relationships).因此,为了进行更一般的讨论,假设“E”节点与“B”节点有 2 个关系(并且没有其他关系)。

  • My query would return a correct COUNT (degree) of 2. Remember, the degree of a node is its number of relationships .我的查询将返回正确的COUNT (degree) 2。请记住,节点的度数是它的关系数

  • A similar-looking query that returned COUNT(DISTINCT othere) instead of COUNT(DISTINCT r) would return 1, which is incorrect.返回COUNT(DISTINCT othere)而不是COUNT(DISTINCT r)的类似查询将返回 1,这是不正确的。

  • Your second query's return clause ( RETURN e, othere, COUNT(othere) ) clause would return a COUNT of 6 (because there'd be 6 result rows with othere ).您的第二个查询的 return 子句( RETURN e, othere, COUNT(othere) )子句将返回COUNT 6(因为othere会有 6 个结果行)。

I hope this helps to make clear why I used COUNT(DISTINCT r) .我希望这有助于弄清楚我为什么使用COUNT(DISTINCT r) You should also read the aggregating function documentation carefully.您还应该仔细阅读汇总的 function文档。

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