我使用以下代碼在 Neo4j 中創(chuàng)建了節(jié)點，from py2neo import Graph, Node, Relationshipg = Graph(password='neo4j')tx = g.begin()node1 = Node('Node', name='Node-1')node2 = Node('Node', name='Node-2')node3 = Node('Node', name='Node-3')node4 = Node('Node', name='Node-4')node5 = Node('Node', name='Node-5')node6 = Node('Node', name='Node-6')node7 = Node('Node', name='Node-7')tx.create(node1)tx.create(node2)tx.create(node3)tx.create(node4)tx.create(node5)tx.create(node6)tx.create(node7)rel12 = Relationship(node1, '0.2', node2, weight=0.2)rel13 = Relationship(node1, '0.2', node3, weight=0.2)rel14 = Relationship(node1, '0.6', node4, weight=0.6)rel45 = Relationship(node4, '0.5', node5, weight=0.5)rel46 = Relationship(node4, '0.3', node6, weight=0.3)rel47 = Relationship(node4, '0.2', node7, weight=0.2)tx.create(rel12)tx.create(rel13)tx.create(rel14)tx.create(rel45)tx.create(rel46)tx.create(rel47)tx.commit()這是界面中的圖形Neo4j，我想按名稱選擇一個節(jié)點，然后隨機走到另一個節(jié)點。但是隨機選擇應(yīng)該是這樣的，import randomrandom.choices(['Node-2', 'Node-3', 'Node-4'], weights=(0.2, 0.2, 0.6))我可以用下面的代碼選擇節(jié)點，但我不知道如何隨機走到另一個節(jié)點。from py2neo import Graphfrom py2neo.matching import NodeMatcherg = Graph(password='neo4j')nodes = NodeMatcher(g)node1 = nodes.match('Node', name='Node-1').first()如果以node-1為起點，可以走的路，Node-1 -> Node-2Node-1 -> Node-3Node-1 -> Node-4 -> Node-5Node-1 -> Node-4 -> Node-6Node-1 -> Node-4 -> Node-7任何想法？提前致謝。
1 回答
臨摹微笑
TA貢獻1982條經(jīng)驗 獲得超2個贊
Py2neo 支持進行 Cypher 查詢，這里有一個很好的 hello-world 教程，介紹如何做到這一點。
因此，我將提供一個帶注釋的 Cypher 查詢，希望對您有用。
但首先，請注意幾點：
您不應(yīng)該為服務(wù)于相同目的的關(guān)系提供幾乎無限數(shù)量的類型（如“0.2”、“0.5”等），因為當(dāng)您想要按類型搜索特定關(guān)系時，這是非常無益的（這是其中之一你會想做的最常見的事情）并且會導(dǎo)致大量的關(guān)系類型。所以我在我的回答中假設(shè)利益關(guān)系實際上都有類型
TO。我的查詢使用一個臨時
Temp節(jié)點來存儲查詢的臨時狀態(tài)，因為它遍歷隨機路徑中的關(guān)系。該Temp節(jié)點將在查詢結(jié)束時被刪除。
查詢?nèi)缦拢?/p>
// Get the (assumed-unique) starting Node `n`?
MATCH (n:Node)
WHERE n.name = 'Node-1'
// Create (if necessary) the unique `Temp` node, and initialize
// it with the native ID of the starting node and an empty `pathRels` list
MERGE (temp:Temp)
SET temp = {id: ID(n), pathRels: []}
WITH temp
// apoc.periodic.commit() repeatedly executes the query passed to it
// until it returns 0 or NULL.
// The query passed here iteratively extends the list of relationships
// in `temp.pathRels`. In each iteration, if the current `temp.id`
// node has any outgoing `TO` relationships, the query:
// - appends to `temp.pathRels` a randomly-selected relationship, taking
//? ?into account the relationship weights (which MUST sum to 1.0),
// - sets `temp.id` to the ID of the end node of that selected relationship,
// - and returns 1.
// But if the current `temp.id` node has no outgoing `TO` relationships, then
// the query returns 0.
CALL apoc.periodic.commit(
? "
? ? MATCH (a:Node)
? ? WHERE ID(a) = $temp.id
? ? WITH a, [(a)-[rel:TO]->() | rel] AS rels
? ? LIMIT 1 // apoc.periodic.commit requires a LIMIT clause. `LIMIT 1` should be harmless here.
? ? CALL apoc.do.when(
? ? ? SIZE(rels) > 0,
? ? ? '
? ? ? ?WITH temp, a, REDUCE(s={x: rand()}, r IN rels | CASE
? ? ? ? ?WHEN s.x IS NULL THEN s
? ? ? ? ?WHEN s.x < r.weight THEN {x: NULL, pathRel: r}
? ? ? ? ?ELSE {x: s.x - r.weight} END
? ? ? ?).pathRel AS pathRel
? ? ? ?SET temp.id = ID(ENDNODE(pathRel)), temp.pathRels = temp.pathRels + pathRel
? ? ? ?RETURN 1 AS result
? ? ? ',
? ? ? '
? ? ? ?RETURN 0 AS result
? ? ? ',
? ? ? {temp: $temp, a: a, rels: rels}
? ? ) YIELD value
? ? RETURN value.result
? ",
? {temp: temp}
) YIELD batchErrors
// Use the `temp.pathRels` list to generate the `weightedRandomPath`
// (or you could just return `pathRels` as-is).
// Then delete the `Temp` node, since it is no longer needed.
// Finally, return `weightedRandomPath`, and also the `batchErrors` returned by
// apoc.periodic.commit() (in case it had any errors).?
WITH temp, apoc.path.create(STARTNODE(temp.pathRels[0]), temp.pathRels) AS weightedRandomPath, batchErrors
DELETE temp
RETURN weightedRandomPath, batchErrors
添加回答
舉報
0/150
提交
取消