mayer
2021 年1 月 18 日 11:29
1
需求是：取我关注的关注，过滤掉我已关注的人，然后需要知道我关注的谁关注了他。
所以会用到 minus 取差集，取差集之后，该如何获取dst的src信息呢？
下面是测试的数据：
获取我的二度关注 和 获取我的一度关注，执行结果如下，可以看到都包含一个10000001的点
(user@nebula) [friend]> go 2 step from 33630944 OVER follow yield distinct follow._dst as dst|order by dst asc|limit 5
============
| dst |
============
| 10000001 |
------------
| 10000005 |
------------
| 10011385 |
------------
| 10036270 |
------------
| 10079598 |
------------
Got 5 rows (Time spent: 14.338/63.063 ms)
Mon Jan 18 09:55:31 2021
(user@nebula) [friend]> go from 33630944 over follow yield distinct follow._dst as dst|order by dst asc|limit 5
============
| dst |
============
| 10000001 |
------------
| 10019476 |
------------
| 10044807 |
------------
| 10058555 |
------------
| 10062883 |
------------
Got 5 rows (Time spent: 3.153/46.851 ms)
Mon Jan 18 09:55:53 2021
使用 minus 取差集可以过滤掉10000001这个点，如下
(user@nebula) [friend]> (go 2 STEPS from 33630944 OVER follow where follow._dst!=33630944 yield distinct follow._dst as dst minus go from 33630944 OVER follow yield follow._dst as dst) | order by dst asc | limit 5
============
| dst |
============
| 10000005 |
------------
| 10011385 |
------------
| 10036270 |
------------
| 10079598 |
------------
| 10079979 |
------------
Got 5 rows (Time spent: 84.311/128.351 ms)
Mon Jan 18 09:58:42 2021
但是如果在查询结果上加上 follow._src 信息，minus会不起作用，是因为会把 dst 和 src加在一起来取了差集
(user@nebula) [friend]> (go 2 STEPS from 33630944 OVER follow where follow._dst!=33630944 yield distinct follow._dst as dst,follow._src as src minus go from 33630944 OVER follow yield follow._dst as dst,follow._src as src) | order by dst asc | limit 5
=======================
| dst | src |
=======================
| 10000001 | 62100520 |
-----------------------
| 10000001 | 76764760 |
-----------------------
| 10000001 | 47633975 |
-----------------------
| 10000005 | 47633975 |
-----------------------
| 10011385 | 76764760 |
-----------------------
Got 5 rows (Time spent: 83.475/128.034 ms)
所以想请教下：使用minus取差集后，该如何获取dst的src信息？
做社交应用中推荐用户的需求，不论什么规则，基本都需要用minus做过滤，比如过滤掉我的好友，过滤掉我关注的人等等，最后的结果dst也要知道src，所以如果现在1.x版本不支持，希望新发布的2.0版本能支持下。
不能利用取差集以后，利用dst，作为源，查反向边获取src么？
mayer
2021 年1 月 18 日 14:21
3
利用dst作为源来查反向边，一个dst有N个反向边，怎么能确定是从哪个src查过来的啊
嗯，我重新看了你的需求，应该类似之前提到的 "我的朋友的朋友不是我的朋友问题”，这里你好像也有提到：怎么样获取这些二度关注来自谁的关注呢？ follow._src 好像不能与distinct 和 MINUS同时使用 - #19 by yee
我跟同事确认了，目前我们确实不支持直接查出来，我会把你这个需求在内部反馈，有消息会在这里跟进
mayer
2021 年3 月 15 日 13:36
7
你好，请问这个需求现在进展怎么样呀，新的版本支持了么，：）
wey
2021 年10 月 20 日 06:22
9
非常抱歉 Mayer，才看到，您看看Match 能满足您的需求不？
MATCH path = (origin:player {name:"LaMarcus Aldridge"})-[:follow]-(f:player)-[:follow]-(fof:player) \
WHERE NOT(fof == origin or fof == f) \
RETURN DISTINCT fof.name AS new_Friends_of_Friends, f.name AS new_FoF_followed_by
两列分别是新两跳好友和被哪个好友follow
(root@nebula) [basketballplayer]> MATCH path = (origin:player {name:"LaMarcus Aldridge"})-[:follow]-(f:player)-[:follow]-(fof:player) \
-> WHERE NOT(fof == origin or fof == f) \
-> RETURN DISTINCT fof.name AS new_Friends_of_Friends, f.name AS new_FoF_followed_by
+------------------------+---------------------+
| new_Friends_of_Friends | new_FoF_followed_by |
+------------------------+---------------------+
| "Tim Duncan" | "Tony Parker" |
+------------------------+---------------------+
| "Manu Ginobili" | "Tony Parker" |
+------------------------+---------------------+
| "Tracy McGrady" | "Rudy Gay" |
+------------------------+---------------------+
| "Dejounte Murray" | "Tony Parker" |
+------------------------+---------------------+
| "Marco Belinelli" | "Tony Parker" |
+------------------------+---------------------+
| "Boris Diaw" | "Tony Parker" |
+------------------------+---------------------+
Got 6 rows (time spent 316915/329628 us)
Wed, 20 Oct 2021 14:20:32 CST
Sxod
2022 年4 月 21 日 11:42
10
你好，快一年了，这个获得dst的src需求现在怎么样了呢
wey
2022 年4 月 22 日 02:00
11
现在的 master，或者马上的 3.1.0 支持 WHERE 里的 路径 PATTERN 表达，可以这么写了
match (n:player)-[:follow]->(x:player)-[:follow]->(m:player) \
where id(n)=="player100" and not (n:player)-[:follow]->(m:player) \
return m,x
比如:
(root@nebula) [basketballplayer]>
match (n:player)-[:follow]->(x:player)-[:follow]->(m:player) \
where id(n)=="player100” \
and not (n:player)-[:follow]->(m:player) \
return m as new_fof, collect(x) as common_friend
+----------------------------------------------------+------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+
| new_fof | common_friend |
+----------------------------------------------------+------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+
| ("player100" :player{age: 42, name: "Tim Duncan"}) | [("player102" :player{age: 33, name: "LaMarcus Aldridge"}), ("player125" :player{age: 41, name: "Manu Ginobili"}), ("player101" :player{age: 36, name: "Tony Parker"}), ("player101" :player{age: 36, name: "Tony Parker"})] |
+----------------------------------------------------+------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+
Got 1 rows (time spent 10929/91577 us)
Fri, 22 Apr 2022 10:06:02 CST
PS. 应该再加一个条件 and id(m) != id(n)，过滤掉自己，这里的结果因为数据集的原因，加上 and id(m) != id(n) 其实就没有 new friend of friend 了，这就是一个例子哈，欢迎试试、探索哈。
update：这个101的情况下有数据了。
(root@nebula) [basketballplayer]> match (n:player)-[:follow]->(x:player)-[:follow]->(m:player) \
where id(n)=="player101" and \
not (n:player)-[:follow]->(m:player) and \
id(m) != id(n) \
return m as new_fof,collect(x) as common_friend
+--------------------------------------------------+------------------------------------------------------+
| new_fof | common_friend |
+--------------------------------------------------+------------------------------------------------------+
| ("player103" :player{age: 32, name: "Rudy Gay"}) | [("player100" :player{age: 42, name: "Tim Duncan"})] |
+--------------------------------------------------+------------------------------------------------------+
Got 1 rows (time spent 11080/110073 us)
1 个赞
Sxod
2022 年4 月 22 日 07:33
12
数据量比较大时，match的效率太慢了，有go语法实现该目标的例子吗
