Question

我有一个像这样的递归数据结构：

@Canonical
static class Person {
    String name
    Set<Person> knows
}

我有一个代表这种结构的Tinkerpop图：

(Jon) -- knows --> (Billy) -- knows --> (Jane) -- ... -->
                           \- knows --> (Suzy) -- ... -->

将任意深度的Tinkerpop图表映射到数据结构的最有效方法是什么？

我可以想象使用PathStep，但似乎应该有一个更好的方法，而且我还没有很好地了解TP3。

def 'build recursive person object'() {
    when:
    def g = TinkerGraph.open()
    def jon = g.addVertex(T.label, 'person', 'name', 'jon')
    def bill = g.addVertex(T.label, 'person', 'name', 'bill')
    def jane = g.addVertex(T.label, 'person', 'name', 'jane')

    jon.addEdge('knows', bill)
    bill.addEdge('knows', jane)

    Multimap<String, Person> relationships = HashMultimap.create()

    g.V().has('name', 'jon')
        .as('a')
        .jump('b', { traverser -> !traverser.get().out('knows').hasNext() })
        .outE('knows')
        .inV()
        .jump('a')
        .as('b')
        .path(
            Function.<Vertex>identity(),
            { Edge e ->
                relationships.put(e.outV().next().value('name') as String,
                        new Person(e.inV().next().value('name') as String))
            } as Function
        )
        .next()

    Person root = new Person('jon')
    Stack<Person> people = new Stack<>()
    Set<String> alreadySeen = new HashSet<>()
    people.push(root)
    alreadySeen.add('jon')

    while(!people.isEmpty()) {
        def person = people.pop()
        person.knows = relationships.get(person.name)
        for(Person related : person.knows) {
            if(alreadySeen.add(related.name))
               people.push(related)
        }
    }

    then:
    root.name == 'jon'
    root.knows*.name == ['bill']
    root.knows[0].knows*.name == ['jane']
}

Answer 1

我会利用Gremlin的tree结构和一些递归方法调用：

import com.google.common.collect.Iterators;
import com.tinkerpop.gremlin.process.T;
import com.tinkerpop.gremlin.process.Traverser;
import com.tinkerpop.gremlin.process.graph.util.Tree;
import com.tinkerpop.gremlin.structure.Vertex;
import com.tinkerpop.gremlin.tinkergraph.structure.TinkerGraph;

import java.util.HashSet;
import java.util.Iterator;
import java.util.Set;
import java.util.function.Function;

/**
 * @author Daniel Kuppitz (daniel at thinkaurelius.com)
 */
public class App {

    static class Person {

        private String name;
        private Set<Person> knows;

        public static Person fromTree(final String name, final Tree node) {
            final Person person = new Person(name);
            final Tree self = (Tree) node.get(name);
            final Set<String> children = (Set<String>) self.keySet();
            children.forEach(n -> person.knows.add(Person.fromTree(n, self)));
            return person;
        }

        public Person(final String name) {
            this.name = name;
            this.knows = new HashSet<>();
        }

        public String name() {
            return name;
        }

        public Set<Person> knows() {
            return knows;
        }
    }

    public static void main(final String[] args) throws Exception {

        final TinkerGraph g = TinkerGraph.open();

        Vertex jon = g.addVertex(T.label, "person", "name", "jon");
        Vertex bill = g.addVertex(T.label, "person", "name", "bill");
        Vertex jane = g.addVertex(T.label, "person", "name", "jane");
        Vertex suzy = g.addVertex(T.label, "person", "name", "suzy");

        jon.addEdge("knows", bill);
        bill.addEdge("knows", jane);
        bill.addEdge("knows", suzy);

        final Tree tree = (Tree) g.V().has("name", "jon").as("x").out("knows").until("x", (Traverser<Vertex> t) ->
                t.get().outE("knows").hasNext(), t -> true).tree((Function<Vertex, String>) v -> v.value("name")).cap().next();

        if (!tree.isEmpty()) {
            final Person p1 = Person.fromTree("jon", tree);
            assert p1.name().equals("jon");
            assert p1.knows().size() == 1;

            final Person p2 = p1.knows().iterator().next();
            assert p2.name().equals("bill");
            assert p2.knows().size() == 2;

            final Iterator<Person> pitty = p2.knows().iterator();
            Iterators.all(pitty, p -> (p.name().equals("jane") || p.name().equals("suzy")) && p1.knows().isEmpty());
        }
    }
}

这种方式归结为一句话：Person.fromTree("jon", tree)

如何将Tinkerpop3图形映射到递归数据结构？

1 个答案: