如何使用py2neo计算与一种节点的关系（一种类型）

Question

Using py2neo 4.x, neo4j 3.5.3, python 3.7.x

What I have: a node from the graph a

graph = Graph(
    host="alpha.graph.domain.co",
    auth=('neo4j', 'theActualPassword')
)
# grab the graph
a = Node("Type", url="https://en.wikipedia.org/wiki/Vivendi")
# create a local node with attributes I should be able to MERGE on
graph.merge(a,"Type","url")
# do said merge
graph.pull(a)
# pull any attributes (in my case Labels) that exist on the node in neo4j...
# ...but not on my local node
# better ways to do this also would be nice in the comments
relMatch = RelationshipMatcher(graph)

What I want: the count of how many "CREATED" relationships are connected to a (in this case, 7)

What I've tried:

x = relMatch.get(20943820943) using one of the IDs of the relationships to see what's what. It returns None , which the docs say means

If no such Relationship is found, py:const:None is returned instead. Contrast with matcher[1234] which raises a KeyError if no entity is found.

which leaves me thinking I'm coming at it all wrong.

also: relMatch.match(a,"CREATED") which raises

raise ValueError("Nodes must be supplied as a Sequence or a Set")

telling me that I'm definitely not reading the docs right.

Not necessarily using this class, which probably isn't what I think it is, how do I get a count of how many ["CREATED"] are pointed at a ?

Answer 1

Below works, and is easier to write than the previous implementation, but I don't think it's the right way to do it.

result = graph.run(
    "MATCH(a) < -[r:CREATED]-(b) WHERE ID(a)=" + str(a.identity) + " RETURN count(r)"
).evaluate()
print(result)

Answer 2

With a RelationshipMatcher , you can simply take a count using len . So if I read your code correctly, you'll need something like:

count = len(RelationshipMatcher(graph).match((None, a), "CREATED"))

Or even easier:

count = len(graph.match((None, a), "CREATED"))

(since graph.match is a shortcut to RelationshipMatcher(graph) )

The (None, a) tuple specifies an ordered pair of nodes (relationships in one direction only) where the start node is any node and the end node is a . Using len simply evaluates the match and returns a count.