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      listUnion()

      函数 listUnion() 可以计算两个列表的全部元素并去重,以列表的形式返回,即返回两个列表的合集(允许有重复元素)。

      Arguments:

      • 第一个列表 <list>
      • 第二个列表 <list>

      Returns:

      • 和集 <list>

      一般用法

      示例:异源别名直接计算

      uncollect [[1,2,2],[2,4,5]] as a
      uncollect [[2,4,7],[4,5,7]] as b
      return table(toString(a), toString(b), toString(listUnion(a, b)))
      

      | toString(a) | toString(b) | toString(listUnion(a, b)) |
      |-------------|-------------|---------------------------|
      | [1,2,2]     | [2,4,7]     | [1,2,4,7]                 |
      | [2,4,5]     | [4,5,7]     | [2,4,5,7]                 |
      

      示例:异源别名笛卡尔积后计算

      uncollect [[1,2,2],[2,4,5]] as a
      uncollect [[2,4,7],[4,5,7]] as b
      with listUnion(a, b) as c
      return table(toString(a), toString(b), toString(c))
      

      | toString(a) | toString(b) | toString(c) |
      |-------------|-------------|-------------|
      | [1,2,2]     | [2,4,7]     | [1,2,4,7]   |
      | [1,2,2]     | [4,5,7]     | [1,2,4,5,7] |
      | [2,4,5]     | [2,4,7]     | [2,4,5,7]   |
      | [2,4,5]     | [4,5,7]     | [2,4,5,7]   |
      

      示例图集:(以下示例将在本图基础上运行)

      在空图集中依次运行以下各行代码以创建该图数据:

      create().node_schema("student").node_schema("course")
      create().node_property(@*, "name").node_property(@student, "age", int32).node_property(@course, "credit", int32)
      insert().into(@student).nodes([{_id:"S001", _uuid:1, name:"Jason", age:25}, {_id:"S002", _uuid:2, name:"Lina", age:23}, {_id:"S003", _uuid:3, name:"Eric", age:24}, {_id:"S004", _uuid:4, name:"Emma", age:26}, {_id:"S005", _uuid:5, name:"Pepe", age:24}])
      insert().into(@course).nodes([{_id:"C001", _uuid:6, name:"French", credit:4}, {_id:"C002", _uuid:7, name:"Math", credit:5}])
      insert().into(@default).edges([{_uuid:1, _from_uuid:1, _to_uuid:6}, {_uuid:2, _from_uuid:2, _to_uuid:6}, {_uuid:3, _from_uuid:3, _to_uuid:6}, {_uuid:4, _from_uuid:2, _to_uuid:7}, {_uuid:5, _from_uuid:3, _to_uuid:7}, {_uuid:6, _from_uuid:4, _to_uuid:7}, {_uuid:7, _from_uuid:5, _to_uuid:7}])
      

      示例:查找选择了法语或数学课的学生

      khop().src({name == "French"}).depth(1) as n1
      with collect(n1) as l1
      khop().src({name == "Math"}).depth(1) as n2
      with collect(n2) as l2
      return listUnion(l1, l2)
      

      [
        {"id":"S001","uuid":"1","schema":"student","values":{"name":"Jason","age":"25"}},
        {"id":"S002","uuid":"2","schema":"student","values":{"name":"Lina","age":"23"}},
        {"id":"S003","uuid":"3","schema":"student","values":{"name":"Eric","age":"24"}},
        {"id":"S004","uuid":"4","schema":"student","values":{"name":"Emma","age":"26"}},
        {"id":"S005","uuid":"5","schema":"student","values":{"name":"Pepe","age":"24"}}
      ]
      
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