Gremlin查询，仅返回具有边的顶点

Question

有没有更有效的方法来返回前100个具有边且没有特定属性的顶点的id？

g.V()
 .filter(hasNot("SOME_PROPERTY").bothE())
 .limit(100)
 .id()

Answer 1

I don't think you can write that in a much more optimal fashion. That traversal will only be as fast as the the underlying graphs ability to optimize the absence of a property which typically isn't that fast. It's generally treated as a global operation that has to iterate every vertex in the graph (or until it finds 100 matches) and I don't think that any graph allows indices that can help in this sort of case.

If this traversal is meant to be a real-time traversal (OLTP) then you should probably consider defaulting that "SOME_PROPERTY" so that it can be indexed in some way to detect negative values or if it is more of an administrative traversal (e.g. detecting bad data to clean up) (OLAP) then you should probably execute that traversal with Gremlin Spark.