根据顶点属性在Tinkerpop3中创建边？

Question

I'm working to migrate information from another database to a Tinkerpop database, using TinkerPop3 Java. I have a large number of ids stored in a single vertex, which I need to transform into an edge between two vertices.

Let's say I run the following code to create the Modern toy graph, and add a property "friend" of josh, with the corresponding name of his friend "marko".

graph = TinkerFactory.createModern()
g = graph.traversal()
g.V().has("name","josh").property("friend","marko")

What I want is a query that will programmatically find all vertices with the property "friend", find the corresponding vertex whose name matches, and create an edge labeled "friends" between them. Ideally this would be a single query, as I have to scan a large number of vertices.

The following query will find the vertices that have friends and the names of those friends, but I don't know how to find the vertex with the matching name without breaking the search into multiple queries.

g.V().has("friend").as("a").values("friend").as("b").select("a","b")

Thanks!!

Answer 1

I found the following worked:

g.V().has("friend").as("x").
V().has("name").where(eq("x")).by("name").by("friend").
as("y").
addE("friends").from("x").to("y")

inspired from the answer found here: https://groups.google.com/forum/#!msg/gremlin-users/l-Xtknn3Loo/0SljkNFiDQAJ;context-place=forum/gremlin-users

Explanation: The query is finding a series of nodes with friends, labelling them "x". Then it looks at all nodes with names, and labels them as "y" if they match - that is, if their name matches the friend property of "x". (That's what the two .by() statements are doing). Then it creates an edge between them.