apoc вернул путь превращения в сгруппированные данные

Question

это скриншот примера графика:

образец данных графика генерации:

CREATE (r:Red {key:'someid'})
CREATE (t1:Teal)
CREATE (t2:Teal)
CREATE (t3:Teal)
CREATE (t4:Teal)
CREATE (t5:Teal)
CREATE (t6:Teal)
CREATE (r2:Red)
CREATE (b1:Blue)
CREATE (b2:Blue)
CREATE (b3:Blue)
CREATE (b4:Blue)
CREATE (f:FINAL_NODE)
CREATE (r)<-[:TO_ACTIVE_DRAFT]-(t1)
CREATE (r)<-[:TO_ACTIVE_DRAFT]-(t2)
CREATE (r)<-[:TO_ACTIVE_DRAFT]-(t3)
CREATE (r)<-[:TO_ACTIVE_DRAFT]-(t4)
CREATE (r)<-[:TO_ACTIVE_DRAFT]-(t5)
CREATE (r)<-[:TO_ACTIVE_DRAFT]-(t6)
CREATE (t1)<-[:TO_ATL]-(r2)
CREATE (t2)<-[:TO_ATL]-(r2)
CREATE (t3)<-[:TO_ATL]-(r2)
CREATE (t4)<-[:TO_ATL]-(r2)
CREATE (t5)<-[:TO_ATL]-(r2)
CREATE (t6)<-[:TO_ATL]-(r2)
CREATE (r2)<-[:TO_ACTIVE_DRAFT]-(b1)
CREATE (r2)<-[:TO_ACTIVE_DRAFT]-(b2)
CREATE (r2)<-[:TO_ACTIVE_DRAFT]-(b3)
CREATE (r2)<-[:TO_ACTIVE_DRAFT]-(b4)
CREATE (b1)-[:TO_FINAL]->(f)
CREATE (b2)-[:TO_FINAL]->(f)
CREATE (b3)-[:TO_FINAL]->(f)
CREATE (b4)-[:TO_FINAL]->(f)

если мой запрос выглядит так:

MATCH (f:Red {key:'someid'}) 
CALL apoc.path.expand(f, '<TO_ACTIVE_DRAFT|<TO_ATL|TO_FINAL>', '>FINAL_NODE',-1,-1) yield path as p

RETURN p

при использовании браузера и просмотре табличного представления похоже, что данные возвращаются в виде отдельных путей в массиве. Таким образом, массив в массиве.

[
  [
     {red},
     {relationship},
     {teal},
     {relationship},
     {red},
     {relationship},
     {blue},
     {relationship},
     {green}
  ],
  [
     {red},
     {relationship},
     {teal},
     {relationship},
     {red},
     {relationship},
     {blue},
     {relationship},
     {green}
  ],
  etc
]

{} - это свойства каждого узла и отношения.

Если я использую

RETURN Nodes (p)

он просто дает мне список всех элементов без глубины, и если я пытаюсь использовать размер или длину с путем, в этом случае все они равны 5 и не дают мне отдельные узлы.

Что я действительно хотел бы, так это иметь возможность воссоздать уровни графика, так что в идеале данные должны быть возвращены так:

[
  [
    {red}
  ],[
    {teal},
    {teal},
    {teal},
    {teal},
    {teal},
    {teal}
  ],[
    {red}
  ],[
    {blue},
    {blue},
    {blue},
    {blue},
    {blue},
    {blue},
    {blue}
  ],[
    {green}
  ]
]

Возможно ли это с помощью функции расширения, или мне нужно выяснить, как сделать то же самое со стандартным шаблоном с помощью шифра - т. Е. MATCH (r) - [: *] - (final)?

// пара уточнений: Там может быть произвольное количество маршрутов и узлов. Так что в этом примере это красный, чирок, красный, синий, зеленый. Это может быть красный, чирок, красный, чирок, красный, синий, зеленый. Может быть просто красным, синим, зеленым.

Я использую C # и драйвер neo4j, если это имеет значение, для каких данных я могу получить возвращенный пример ответа, который я получаю от простого возврата пути в C #

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[Дополнительное уточнение]

Более конкретный пример: если вы перемещаете животных, у вас есть источник рождения, а затем у вас есть несколько движений, пока вы не доберетесь до рынка. таким образом, график выглядит как начальный скриншот, где красные узлы - это определенное место, куда животное было перемещено, а бирюзовый и синий - разные типы графиков. Тем не менее, единственная информация, которую я точно знаю, это то, где она находится в данный момент (так что за красный узел), и я хочу добраться до источника рождения (зеленый узел). Количество движений и местоположений, в которых он находился, являются переменными. Возможно, он был перемещен один раз или мог быть перемещен x раз. Я склонен знать, что пути ограничены 3 помеченными отношениями. Поэтому, хотя мне нравится синтаксис во многих процедурах apoc, если я не могу управлять этим путем после его получения, то, возможно, мне просто нужно сделать что-то вроде

MATCH (f:Red {key:'someid'})
OPTIONAL MATCH (f)<-[:TO_ACTIVE_DRAFT|TO_ATL*]-()-[:SOURCE_FOR_INC]->(g:Green)
RETURN g

, но для меня приведенное выше вернет только f или g, а не путь, поэтому должно быть что-то вроде этого:

MATCH (f:Red {key:'someid'})
p = OPTIONAL MATCH (f)<-[:TO_ACTIVE_DRAFT|TO_ATL*]-()-[:SOURCE_FOR_INC]->(g:Green)
RETURN p

, который, как я полагаю, даст мне путь, но будет таким же сценарием, как и в случае с apoc, где я не могу преобразовать данные в группы ...

Answer 1

Должна быть возможность придумать код Cypher, но это потребует некоторых усилий и, вероятно, будет довольно сложным.

Вероятно, было бы лучше, если бы вы взяли результат:

MATCH p=(f:Red {key:'someid'}<-[:TO_ACTIVE_DRAFT|TO_ATL*]-()-[:SOURCE_FOR_INC]->(g:Green)
RETURN p

и сгенерируйте нужные данные на каком-нибудь внешнем языке программирования. В большинстве случаев пути - это то, чего хотят люди, поэтому Сайфер направлен на это.

Answer 2

Этот запрос (с некоторыми предполагаемыми метками узлов), который не использует процедуру APOC, может работать для вас:

MATCH (f:Node {key:'someid'})
  <-[:TO_ACTIVE_DRAFT]-(t:Teal)
  <-[:ATL_TO_DRAFT]-(r:Red)
  <-[:TO_ACTIVE_DRAFT]-(b:Blue)
  -[:SOURCE_FOR_INC]->(g:Green)
RETURN COLLECT(DISTINCT f), COLLECT(DISTINCT t), COLLECT(DISTINCT r), COLLECT(DISTINCT b), COLLECT(DISTINCT g)