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是否可以从 ArangoDB / Foxx 服务中的 Traverser 步骤中获取结果？

我希望在电网图上实现欧姆电阻计算。在这里，我沿着图的边缘遍历 BFS（广度优先）到它的叶子。- 没问题。使用过滤器、扩展器和访客就可以了！

我的兴趣是在分支节点/步骤处对每条线执行1/x计算，以解决“回到图的根/顶部”时的并联电阻。

所以，我的问题是：“是否有可能从每一步中获得返回值？” - 如果是，我在哪里可以指定计算返回值？

感谢您的任何提示！

更新 2017-12-21：进一步解释

这是我的案例演示图的图像：演示图 的图像 图像解释： * “黑色”值是用于计算的原始欧姆值（例如加权图中边缘的权重）。* “红色”值是每一步计算的欧姆值（子结果）。

图表的 CSV 数据：

  • 节点：
"_key","stationId","voltage","voltageKey","pathId","elementId","type","typeKey","name","status","oStatus","isSource","sourceColor ","isStrandSource","isStrandEnd","strandSourceColor"
"NE-2_J_1","2","20","J","","1","conn","","root","1","1","","", "","",""
"NE-2_J_2","2","20","J","","2","conn","","Abzweig 1","1","1","","" ,"","",""
"NE-2_J_3","2","20","J","","3","conn","","Abzweig 2","1","1","","" ,"","",""
"NE-2_J_4","2","20","J","","4","conn","","Abzweig 3","1","1","","" ,"","",""
"NE-2_J_5","2","21","J","","5","conn","","Abzweig 4","1","1","","" ,"","",""
"NE-2_J_6","2","22","J","","6","conn","","Abzweig 5","1","1","","" ,"","",""
"NE-2_J_7","2","23","J","","7","conn","","Abzweig 6","1","1","","" ,"","",""
"NE-2_J_8","2","24","J","","8","conn","","Abzweig 7","1","1","","" ,"","",""
  • 边缘：
"_key","_from","_to","elementId","type","typeKey","voltage","voltageKey","name",
"状态","oStatus","xOhm"
"NE-J19","NE-2_J_1","NE-2_J_2","19","实心","","20","J","","1","1","1 "
"NE-J20","NE-2_J_2","NE-2_J_3","20","实心","","20","J","","1","1","2 "
"NE-J21","NE-2_J_2","NE-2_J_4","21","实心","","20","J","","1","1","3 "
"NE-J22","NE-2_J_3","NE-2_J_5","22","实心","","20","J","","1","1","4 "
"NE-J23","NE-2_J_3","NE-2_J_6","23","实心","","20","J","","1","1","5 "
"NE-J24","NE-2_J_4","NE-2_J_7","24","实心","","20","J","","1","1","6 "
"NE-J25","NE-2_J_4","NE-2_J_8","25","实心","","20","J","","1","1","7 "

计算说明：

  • 计算是“自下而上”执行的，所以我的意图是在遍历器“返回”底层图形的结果之后处理数学。

  • 第 1 步：计算“NE-2_J_3”处“左下”分支（“NE-2_J_5”和“NE-2_J6”）的最终（平行）电阻：

    • 数学是1 / ( 1/4 Ohm + 1/5 Ohm ) = 2,22 Ohm

  • 第 2 步：计算“NE-2_J_4”处“右下”分支（“NE-2_J_7”和“NE-2_J_8”）的最终（平行）电阻：

    • 数学是1 / ( 1/6 Ohm + 1/7 Ohm ) = 3,23 Ohm

  • 第 3 步：在计算第 5 步之前，计算“左下”分支的结果（内联）电阻（“NE-2_J_3 原始”值和“NE-2_J_3 计算”）：

    • 数学是2 Ohm + 2,22 Ohm = 4,22 Ohm

  • 第 4 步：在计算第 5 步之前，计算“右下”分支的结果（内联）电阻（“NE-2_J_4 原始”值和“NE-2_J_4 计算”）：

    • 数学是3 Ohm + 3,23 Ohm = 6,23 Ohm

  • 第 5 步：计算“中心”分支（第 3 步和第 4 步）在“NE-2_J_2”处产生的（平行）电阻：

    • 数学是1 / ( 1/4,22 Ohm + 1/6,23 Ohm ) = 2,52 Ohm

  • 第 6 步：为“NE-2_J_1”处的最终结果计算“中心”分支的结果（内联）电阻（“NE-2_J_2 原始”值和“NE-2_J_2 计算”）：

    • 数学是1 Ohm + 2,52 Ohm = 3,52 Ohm
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