I want to store hierarchical data in the PostgreSQL database.
I found ltree extension but it is used to store tree-structured data i.e there can be only one parent node.
Is there any way I can tweak it to store multiple parent nodes?
Yes, it is possible to do so using the ltree extension, by using two tables: one for your nodes and another for your paths. Bustawin has a great writeup on how to work with DAGs using this extension with this strategy.
For a node to have multiple parents, it is necessary that we define more than one ltree path for such node, one that leads from each parent. Say we have a DAG that forms a diamond between nodes A, B, C, and D:
A -> B
B -> D
A -> C
C -> D
Using the extension, you should have a node D related to two paths, A.B.D and A.C.D. Both queries should return node D
SELECT * FROM nodes
JOIN paths ON paths.node_id = nodes.id
WHERE paths.ltree_path ~ "A.B.*{1}"
SELECT * FROM nodes
JOIN paths ON paths.node_id = nodes.id
WHERE paths.ltree_path ~ "A.C.*{1}"
Related
I'm trying to create a new node with label C and relationships from a-->c and b-->c, but if and only if the whole pattern a-->c,b-->c does exist.
a and b already exist (merged before the rest of the query).
The below query is a portion of the query I want to write to accomplish this.
However, it creates a random empty node devoid of properties and labels and attaches the relationship to that node instead. This shouldn't be possible and is certainty not what I want. How do I stop that from happening?
merge (a: A {id: 1})
merge (b: B {id:1})
with *
call {with a, b
match (a)-[:is_required]->(dummy:C), (a)-[:is_required]->(b)
with count(*) as cnt
where cnt = 0
merge (temp: Temporary {id: 12948125})
merge (a)-[:is_required]->(temp)
return temp
}
return *
Thanks
I think there are a couple of problems here:
There are restrictions on how you can use variables introduced with WITH in a sub-query. This article helps to explain them https://neo4j.com/developer/kb/conditional-cypher-execution/
I think you may be expecting the WHERE to introduce conditional flow like IF does in other languages. WHERE is a filter (maybe FILTER would have been a better choice of keyword than WHERE). In this case you are filtering out 'cnt's where they are 0, but then never reference cnt again, so the merge (temp: Temporary {id: 12948125}) and merge (a)-[:is_required]->(temp) always get executed. The trouble is, due to the above restrictions on using variables inside sub-queries, the (a) node you are trying to reference doesn't exist, it's not the one in the outer query. Neo4j then just creates an empty node, with no properties or labels and links it to the :Temporary node - this is completely valid and why you are getting empty nodes.
This query should result in what you intend:
merge (a: A {id: 1})
merge (b: B {id:1})
with *
// Check if a is connected to b or :C (can't use a again otherwise we'd overwrite it)
optional match(x:A {id: 1}) where exists((a)-[:is_required]->(:C)) or exists((a)-[:is_required]->(b))
with *, count(x) as cnt
// use a case to 'fool' foreach into creating the extra :Temporary node required if a is not related to b or :C
foreach ( i in case when cnt = 0 then [1] else [] end |
merge (temp: Temporary {id: 12948125})
merge (a)-[:is_required]->(temp)
)
with *
// Fetch the :Temporary node if it was created
optional match (a)-[:is_required]->(t:Temporary)
return *
There are apoc procedures you could use to perform conditional query execution (they are mentioned in the linked article). You could also play around with looking for a path from (a) and check its length, rather than introduce a new MATCH and the variable x then checking for the existance of related nodes.
If anyone is having the same problem, the answer is that the Neo4j browser is display nonexistent nodes. The query executes fine…
I am using Postgres 13 and have created a table with columns A, B and C. The table is partitioned by A with 2 possible values. Partition 1 contains 100 possible values each for B and C, whereas partition 2 has 100 completely different values for B, and 1 different value for C. I have set the statistics for both columns to maximum so that this definitely doesn't cause any issue
If I group by B and C on either partition, Postgres estimates the number of groups correctly. However if I run the query against the base table where I really want it, it estimates what I assume is no functional dependency between A, B and C, i.e. (p1B + p1C) * (p2B + p2C) for 200 * 101 as opposed to the reality of p1B * p1C + p2B * p2C for 10000 + 100.
I guess I was half expecting it to sum the underlying partitions rather than use the full count of 200 B's and 101 C's that the base table can see. Moreover, if I also add A into the group by then the estimate erroneously doubles further still, as it then thinks that this set will also be duplicated for each value of A.
This all made me think that I need an extended statistic to tell it that A influences either B or C or both. However if I set one on the base partition and analyze, the value in pg_statistic_ext_data->stxdndistinct is null. Whereas if I set it on the partitions themselves, this does appear to work, though isn't particularly useful because the estimation is already correct at this level. How do I go about having Postgres estimate against the base table correctly without having to run the query against all of the partitions and unioning them together?
You can define extended statistics on a partitioned table, but PostgreSQL doesn't collect any data in that case. You'll have to create extended statistics on all partitions individually.
You can confirm that by querying the collected data after an ANALYZE:
SELECT s.stxrelid::regclass AS table_name,
s.stxname AS statistics_name,
d.stxdndistinct AS ndistinct,
d.stxddependencies AS dependencies
FROM pg_statistic_ext AS s
JOIN pg_statistic_ext_data AS d
ON d.stxoid = s.oid;
There is certainly room for improvement here; perhaps don't allow defining extended statistics on a partitioned table in the first place.
I found that I just had to turn enable_partitionwise_aggregate on to get this to estimate correctly
I want to select the first common boss for two employees in draph.
My model is simple:
name: string
boss_of: uids
Lets assume the following data where each arrow denotes the boss_of edge:
A -> B
A -> C
B -> D
C -> E
E -> F
E -> G
So, given F And D the query should return A, for F and G the result is obviously E.
I tried using allofterms but found no solution as there may be a different number of nodes
between the co-workers and their common boss. Is it possible at all to formulate such a query?
I am trying to explore dgraph (or graph databases at all), so maybe I am just overseeing something.
You can use K-Shortest Path Queries
The middle one in the response is the closest common entity.
My use case is:
I have to return whole graph in result but the condition is
If there are more than 1 relationship in between two particular nodes in the same direction then I have to just merge it into 1 relationship. For ex: Lets say there are two nodes 'm' and 'n' and there are 3 relations in between these nodes say r1, r2, r3 (in the same direction) then when I get the result after firing cypher query I should get only 1 relation in between 'n' and 'm'.
I need to perform some operations on top of it like the resultant relation that we got from merging all the relations should contain the properties and their values that I want to retain. Actually I will retain all the properties of any one of the relations that are merging depending upon the timestamp field that is one of the properties in relation.
Note : I have same properties throughout all my relations (The number of properties and name of properties are same across all relations. Values may differ for sure)
Any help would be appreciated. Thanks in advance.
You mean something like this?
Delete all except the first
MATCH (a)-[r]->(b)
WITH a,b,type(r) as type, collect(r) as rels
FOREACH (r in rels[1..] | DELETE r)
Ordering by timestamp first
MATCH (a)-[r]->(b)
WITH a,r,b
ORDER BY r.timestamp DESC
WITH a,b,type(r) as type, collect(r) as rels
FOREACH (r in rels[1..] | DELETE r)
If you want to do all those operations virtually just on query results you'd do them in your programming language of choice.
I have a table called "Tag" which consists of an Id, Name and Description column.
Now lets say I have the tables Character (C), Movie (M), Series (S) etc..
And I want to be able to tag entries in C, M, S with multiple tags and one tag may be used for multiple entries.
So I could realize it like this:
T -> TC <- C
T -> TM <- M
T -> TS <- S
Where TC, TM, TS are the intermediate tables.
I was wondering if I could combine TC, TM, TS into one table with a type column added and still use foreign keys.
As of yet I haven't found a way to do it.
Or is this something I shouldn't be doing?
As the comments above suggested you can't combine multiple table into a single one. If you want to have a single view of the "tag relationships" you can pull the needed information into a View. This way, you only need to write a longer query once and are able to use like a single table. Keep in mind that you can't insert data into a view (there are possibilities to do so, but they are a little advanced)