可以按_from和_to分组，计算每组有多少条边，然后过滤出唯一的组合：

FOR ec IN edge_collection
  COLLECT from = ec._from, to = ec._to WITH COUNT INTO count
  FILTER count > 1
  RETURN { from, to, count }

或者，如果您也要返回边缘关键点：

FOR ec IN edge_collection
  COLLECT from = ec._from, to = ec._to INTO edges = ec._key
  LET count = LENGTH(edges)
  FILTER count > 1
  RETURN { from, to, count, edges }

或者，使用聚合代替后期计算：

FOR ec IN edge_collection
  COLLECT from = ec._from, to = ec._to AGGREGATE count = LENGTH(1) INTO edges = ec._key
  FILTER count > 1
  RETURN { from, to, count, edges }

要返回完整的边缘，请使用INTO edges = ec。您也可以只使用INTO edges，但这样每个边缘都将嵌套在对象{"ec": … }中。

经过进一步的研究，我发现了一种使用COLLECT语句的明显更快的方法。
此外，完全披露，这是我建立了this other answer。

LET duplicates = (
    FOR ec IN edge_collection
        COLLECT from = ec._from, to = ec._to WITH COUNT INTO count
        FILTER count > 1
        RETURN {
            from: from,
            to: to,
            count: count
        }
)

FOR d IN duplicates
    FOR ec IN edge_collection
        FILTER d.from == ec._from AND d.to == ec._to
        RETURN ec

编辑：
基于@CodeManX的答案，我的最终目标是能够删除这些重复的值，或者更确切地说是REMOVE这些重复的值。如果有人出于类似的目的偶然发现了这一点，这可能会有所帮助：

LET duplicates = (
    FOR ec IN edge_collection
        COLLECT from = ec._from, to = ec._to AGGREGATE count = LENGTH(1) INTO edges = ec._key
        FILTER count > 1
        RETURN { from, to, count, edges }
)

FOR d IN duplicates
    LET key_to_delete = FIRST(d.edges)
    REMOVE { _key: key_to_delete } IN edge_collection