https://groups.google.com/forum/#!topic/gremlin-users/UZMD1qp5mfg使用脚本通过示例讨论了一个组:
gremlin> g.V.outE.groupBy{it.inV.next().name}{it.weight}{it.sum().doubleValue()}.cap.orderMap(T.decr)
我想用Java尝试这个,但是:
首先,我只有:
// g.V.outE.groupBy{it.inV.next().name}{it.weight}{it.sum().doubleValue()}.cap.orderMap(T.decr)
GraphTraversalSource g = TinkerFactory.createModern().traversal();
Map<Object, Object> map = g.V().outE().group().by().next();
if (debug) {
System.out.println(map.values().size());
for (Entry<Object, Object> entry : map.entrySet()) {
System.out
.println(String.format("%s=%s", entry.getKey(), entry.getValue()));
}
}
给出调试输出:
6
e[7][1-knows->2]=[e[7][1-knows->2]]
e[8][1-knows->4]=[e[8][1-knows->4]]
e[9][1-created->3]=[e[9][1-created->3]]
e[10][4-created->5]=[e[10][4-created->5]]
e[11][4-created->3]=[e[11][4-created->3]]
e[12][6-created->3]=[e[12][6-created->3]]
然后我尝试了一下这样的事情:
import static org.apache.tinkerpop.gremlin.process.traversal.dsl.graph.__.*;
/**
* show the given map entries
*
* @param map
*/
public void showMap(String title, Map<Object, Object> map) {
System.out.println(title + ":" + map.values().size());
for (Entry<Object, Object> entry : map.entrySet()) {
System.out
.println(String.format("\t%s=%s", entry.getKey(), entry.getValue()));
}
}
public void showObject(String title, Object object) {
System.out.println(title+":"+object.toString());
}
@Test
/**
* https://groups.google.com/forum/#!topic/gremlin-users/UZMD1qp5mfg
* https://stackoverflow.com/questions/55771036/it-keyword-in-gremlin-java
*/
public void testGroupBy() {
// gremlin:
// g.V.outE.groupBy{it.inV.next().name}{it.weight}{it.sum().doubleValue()}.cap.orderMap(T.decr)
GraphTraversalSource g = TinkerFactory.createModern().traversal();
debug=true;
if (debug) {
g.V().outE().group().by().forEachRemaining(m -> showMap("by()", m));
g.V().outE().group().by(inV().id())
.forEachRemaining(m -> showMap("by(inV().id())", m));
g.V().outE().group("edges").by(inV().id()).cap("edges")
.forEachRemaining(o -> showObject("cap", o));
}
结果:
by():6
e[7][1-knows->2]=[e[7][1-knows->2]]
e[8][1-knows->4]=[e[8][1-knows->4]]
e[9][1-created->3]=[e[9][1-created->3]]
e[10][4-created->5]=[e[10][4-created->5]]
e[11][4-created->3]=[e[11][4-created->3]]
e[12][6-created->3]=[e[12][6-created->3]]
by(inV().id()):4
2=[e[7][1-knows->2]]
3=[e[9][1-created->3], e[11][4-created->3], e[12][6-created->3]]
4=[e[8][1-knows->4]]
5=[e[10][4-created->5]]
cap:{2=[e[7][1-knows->2]], 3=[e[9][1-created->3], e[11][4-created->3], e[12][6-created->3]], 4=[e[8][1-knows->4]], 5=[e[10][4-created->5]]}
所以看起来“it”可以简单地省略,而对于cap步骤,必须命名“group”。 gremlin-groovy到Java翻译的其他部分我还是不明白。
上面的脚本如何完全翻译成Java?
在TinkerPop 3中,它很简单:
g.V().outE().
group().
by(inV().values("name")).
by(values("weight").sum()).
order(local).
by(values, desc)
或者以完整的java语法:
import org.apache.tinkerpop.gremlin.structure.Column;
import org.apache.tinkerpop.gremlin.process.traversal.Order;
import org.apache.tinkerpop.gremlin.process.traversal.Scope;
import org.apache.tinkerpop.gremlin.process.traversal.dsl.graph.GraphTraversalSource;
import static org.apache.tinkerpop.gremlin.process.traversal.dsl.graph.__.*;
import org.apache.tinkerpop.gremlin.process.traversal.dsl.graph.__;
g.V().outE().
group().
by(inV().values("name")).
by(values("weight").sum()).
order(Scope.local).
by(Column.values, Order.desc)
结果:
总和:{ripple = 1.0,josh = 1.0,lop = 1.0,vadas = 0.5}
UPDATE
要回答关于使用两个顶点的评论中的问题,它将是这样的:
g.E().
group().
by(bothV().values("name").fold()).
by(values("weight").sum()).
order(Scope.local).
by(Column.values, Order.desc)