我有两个查询，每个查询都返回一个由三列组成的表。两个表的前两列相似，第三列的名称不同。参见以下内容：
MATCH (n:PI)-[r:SIMILAR_STUDY]->(p:PI) where ID(n) = 3162 RETURN ID(n) , ID(p),r.score as score1 ORDER BY r.score DESC LIMIT 5
ID(n) ID(p) score1
3162 4978 0.6666666666666666
3162 4054 0.6363636363636364
3162 4194 0.6
3162 4980 0.5555555555555556
3162 3661 0.5555555555555556
和：
MATCH (n:PI)-[r:SIMILAR_SITE]->(p:PI) where ID(n) = 3162 RETURN ID(n) ,ID(p) ,r.score as score2 ORDER BY r.score DESC LIMIT 5
ID(n) ID(p) score2
3162 4980 0.6
3162 3183 0.5
3162 4612 0.42857142857142855
3162 4595 0.4
3162 6926 0.4
如何将这些结果组合到一个具有4列ID(n)、ID(p)、score1、score2的唯一表中，以及如何为分数的平均值添加一列？以下是预期结果：
ID(n) ID(p) score1 score2 avg
3162 4980 0.5555555555555556 0.6 0.57
3162 3183 0 0.5 0.25
第二次编辑：
两个都用一个查询怎么样？类似于：
MATCH (n:PI)-[:SIMILAR_STUDY|SIMILAR_SITE]->(t:PI)
WHERE ID(n) = 3162
WITH distinct t, n
OPTIONAL MATCH (t)<-[r_ste:SIMILAR_STUDY]-(n)
WITH n, t, COALESCE(r_ste.score, 0) as r1
OPTIONAL MATCH (t)<-[r_se:SIMILAR_SITE]-(n)
WITH ID(n) as n, ID(t) as t, r1, COALESCE(r_se.score, 0) as r2
WITH n, t, r1, r2, ((r1 + r2)/2) as avg
RETURN n, t, r1, r2, avg
OPTIONAL MATCH允许我们处理没有匹配的情况，COALESCE允许我们用0替换null。
要查看示例，可以使用：
MERGE (a:PI{ID: 3162})
MERGE (b:PI{ID: 4978})
MERGE (c:PI{ID: 4054})
MERGE (d:PI{ID: 4194})
MERGE (e:PI{ID: 4980})
MERGE (f:PI{ID: 3661})
MERGE (g:PI{ID: 3183})
MERGE (h:PI{ID: 4612})
MERGE (i:PI{ID: 4595})
MERGE (j:PI{ID: 6926})
MERGE (a)-[:SIMILAR_STUDY{score: 0.6666666}]-(b)
MERGE (a)-[:SIMILAR_STUDY{score: 0.63636}]-(c)
MERGE (a)-[:SIMILAR_STUDY{score: 0.6}]-(d)
MERGE (a)-[:SIMILAR_STUDY{score: 0.55555}]-(e)
MERGE (a)-[:SIMILAR_STUDY{score: 0.55555}]-(f)
MERGE (a)-[:SIMILAR_SITE{score: 0.6}]-(e)
MERGE (a)-[:SIMILAR_SITE{score: 0.5}]-(g)
MERGE (a)-[:SIMILAR_SITE{score: 0.428}]-(h)
MERGE (a)-[:SIMILAR_SITE{score: 0.4}]-(i)
MERGE (a)-[:SIMILAR_SITE{score: 0.4}]-(j)
并运行：
MATCH (n:PI)-[:SIMILAR_STUDY|SIMILAR_SITE]->(t:PI)
WHERE n.ID = 3162
WITH distinct t, n
OPTIONAL MATCH (t)<-[r_ste:SIMILAR_STUDY]-(n)
WITH n, t, COALESCE(r_ste.score, 0) as r1
OPTIONAL MATCH (t)<-[r_se:SIMILAR_SITE]-(n)
WITH n.ID as n, t.ID as t, r1, COALESCE(r_se.score, 0) as r2
WITH n, t, r1, r2, ((r1 + r2)/2) as avg
RETURN n, t, r1, r2, avg
其中ID是节点上的键。通过我添加的示例数据，我得到：
╒════╤════╤═════════╤═════╤══════════════════╕
│"n" │"t" │"r1" │"r2" │"avg" │
╞════╪════╪═════════╪═════╪══════════════════╡
│3162│4595│0 │0.4 │0.2 │
├────┼────┼─────────┼─────┼──────────────────┤
│3162│4054│0.63636 │0 │0.31818 │
├────┼────┼─────────┼─────┼──────────────────┤
│3162│4978│0.6666666│0 │0.3333333 │
├────┼────┼─────────┼─────┼──────────────────┤
│3162│4980│0.55555 │0.6 │0.5777749999999999│
├────┼────┼─────────┼─────┼──────────────────┤
│3162│3183│0 │0.5 │0.25 │
├────┼────┼─────────┼─────┼──────────────────┤
│3162│4612│0 │0.428│0.214 │
├────┼────┼─────────┼─────┼──────────────────┤
│3162│6926│0 │0.4 │0.2 │
├────┼────┼─────────┼─────┼──────────────────┤
│3162│3661│0.55555 │0 │0.277775 │
├────┼────┼─────────┼─────┼──────────────────┤
│3162│4194│0.6 │0 │0.3 │
└────┴────┴─────────┴─────┴──────────────────┘