如何通过密码查询在neo4j中获取不同的标签及其计数？

Question

I was needing to check what different labels present in graph database neo4j.

How to get different labels and their count in neo4j by cypher query?

Answer 1

I finally found a solution to the multiple label problem that is less complicated:

MATCH (a) WITH DISTINCT LABELS(a) AS temp, COUNT(a) AS tempCnt
UNWIND temp AS label
RETURN label, SUM(tempCnt) AS cnt

Answer 2

通过这个密码查询，我们可以得到 neo4j 中存在的不同标签及其计数。

MATCH (n) RETURN DISTINCT LABELS(n), COUNT(n)

Answer 3

It's surprisingly complex to get a count per label since nodes can have multiple labels and labels (n) returns a collection of strings representing those labels. On a graph consisting of three nodes and two labels, as {:A} , {:B} and {:A:B} , labels (n) returns three distinct string collections. Instead of counting two nodes with :A and two nodes with :B , the result would be one for each of the three label combinations. See console . To aggregate per label, not per label collection, you'd have to group by the values within collection, which is cumbersome.

I have an ugly way to do it, maybe someone can suggest a better one: first find out max number of labels any node has.

MATCH (n)
RETURN max(length(labels(n)))

Then chain that many queries with UNION , counting nodes by the label at position i in the collection, where i starts at 0 and increments to the max-1. If the nodes have at most 3 labels,

MATCH (n)
RETURN labels (n)[0] as name, count (n) as cnt
UNION MATCH (n)
RETURN labels (n)[1] as name, count (n) as cnt
UNION MATCH (n)
RETURN labels (n)[2] as name, count (n) as cnt

This aggregates the label counts correctly, but it returns a null count for every case where the index is out of the collection bounds. For the first return (the [0] index) this signifies nodes that don't have a label. For the other lines, the null count similarly signifies nodes with less labels than queried for, but this information is irrelevant, so can be ignored

MATCH (n)
RETURN labels (n)[0] as name, count (n) as cnt
UNION MATCH (n)
WITH labels (n)[1] as name, count (n) as cnt
WHERE name IS NOT NULL
RETURN name, cnt
UNION MATCH (n)
WITH labels (n)[2] as name, count (n) as cnt
WHERE name IS NOT NULL
RETURN name, cnt

I'm sure this could be done more gracefully, but that's as far as I got.

Answer 4

You can leverage apoc library meta graph as described here https://stackoverflow.com/a/52489029 --> run the code below, it works even if there is more labels attached to a one node.

CALL apoc.meta.stats() YIELD labels
RETURN labels