What is the shortest or fastest SQL select query or SQL procedure to crawl a social graph. Imagine we have this table:
UId FriendId
1 2
2 1
2 4
1 3
5 7
7 5
7 8
5 9
9 7
We have two subset of people here, i'm talking about a sql query or procedure which if we pass:
Uid = 4 return the result set rows with uid : {1, 2, 3}
or if
Uid = 9 return the result set rows with uid : {5, 7, 8}
Sorry for my poor english.
So you want get all friends of someone, including n-th degree friends? I don't think it is possible without recursion.
How you can do that is explained here:
https://inviqa.com/blog/graphs-database-sql-meets-social-network
If you are storing your values in an adjacency list, the easiest way I've found to crawl it is to translate it into a graphing language and query that. For example, if you were working in PHP, you could use the Image_GraphViz package. Or, if you want to use AJAX, you might consider cytoscapeweb. Both work well.
In either case, you'd SELECT * FROM mytable and feed all the records into the graph package as nodes. This means outputting them in dot or GraphML (or other graphing language). Then you can easily query them.
If you don't wish to translate the dataset, consider storing it as nested sets. Nested sets, though a bit of a pain to maintain, are much better than adjacency lists for the kind of queries you are looking to do.
If you are storing your values in an adjacency list, and you want n-th degree you can simply recursively INNER JOIN the UID's. For example:
Select t1.uid, t2.uid, t3.uid FROM t1 INNER JOIN t2 ON t1.uid=t2.uid INNER JOIN t3 ON t2.uid=t3.uid
This query is like a DFS with a fixed depth.
Related
Suppose I have a table of relationships like in a directed graph. For some pairs of ids there are both 1->2 and 2->1 relations, for others there are not. Some nodes are only present in one column.
a b
1 2
2 1
1 3
4 1
5 2
Now I want to work with it as undirected graph. For example, grouping, filtering using both columns present. For example filter node 5 and count neighbors of the rest
node neighbor_count
1 3
2 1
3 1
4 1
Is it possible to compose queries in such a way that first column a is used and then column b is used in the same manner?
I know it is achievable by doubling the table:
select a,count(distinct(b))
from
(select * from grap
union all
select b as a, a as b from grap)
where (not a in (5,6,7)) and (not b in (5,6,7))
group by a;
However, the real tables are quite large (10^9 - 1^10 of pairs). Would union require additional disk usage? A single scan through the base is already quite slow for me. Are there better ways to do this?
(Currently database is sqlite, but the less platform specific the answer the better)
The union all is generated only for the duration of the query. Does it use more disk space? Not permanently.
If the processing of the query requires saving the data out to disk, then it will use more temporary storage for intermediate results.
I would suggests, though, that if you want an undirected graph with this representation, then add in the addition pairs that are not already in the table. This will use more disk space. But you won't have to play games with queries.
For my application I have a table with these three columns: user, item, value
Here's some sample data:
user item value
---------------------
1 1 50
1 2 45
1 23 35
2 1 88
2 23 44
3 2 12
3 1 27
3 5 76
3 23 44
What I need to do is, for a given user, perform simple arithmetic against everyone else's values.
Let's say I want to compare user 1 against everyone else. The calculation looks something like this:
first_user second_user result
1 2 SUM(ABS(50-88) + ABS(35-44))
1 3 SUM(ABS(50-27) + ABS(45-12) + ABS(35-44))
This is currently the bottleneck in my program. For example, many of my queries are starting to take 500+ milliseconds, with this algorithm taking around 95% of the time.
I have many rows in my database and it is O(n^2) (it has to compare all of user 1's values against everyone else's matching values)
I believe I have only two options for how to make this more efficient. First, I could cache the results. But the resulting table would be huge because of the NxN space required, and the values need to be relatively fresh.
The second way is to make the algorithm much quicker. I searched for "postgres SIMD" because I think SIMD sounds like the perfect solution to optimize this. I found a couple related links like this and this, but I'm not sure if they apply here. Also, they seem to both be around 5 years old and relatively unmaintained.
Does Postgres have support for this sort of feature? Where you can "vectorize" a column or possibly import or enable some extension or feature to allow you to quickly perform these sorts of basic arithmetic operations against many rows?
I'm not sure where you get O(n^2) for this. You need to look up the rows for user 1 and then read the data for everyone else. Assuming there are few items and many users, this would be essentially O(n), where "n" is the number of rows in the table.
The query could be phrased as:
select t1.user, t.user, sum(abs(t.value - t1.value))
from t left join
t t1
on t1.item = t.item and
t1.user <> t.user and
t1.user = 1
group by t1.user, t.user;
For this query, you want an index on t(item, user, value).
I have two tables. One is for Task and second is dependency table for the tasks.
I want a query to give me all the tasks (recursively) based on a particular id.
I have two tables. One is for Task
ID TASK
1 Abc
2 Def
3 Ghi
4 Jkl
5 Mno
6 Pqr
The second one is for getting dependent tasks
ID DEPENDENT_ON
2 1
3 1
4 2
4 6
5 2
6 5
Is it possible to write a sql query to get a list of all the tasks (recursive) which are dependent on a particular task.
Example.
I want to check all tasks dependent on ID=1.
Expected output (which is 2 and 3):
2.Def
3.Ghi
Furthermore query should also give output of these two dependent tasks and so on.
Final output should be:
2.Def -- level one
3.Ghi -- level one
4.Jkl -- Dependent on task 2
5.Mno -- Dependent on task 2
6.Pqr -- Dependent on task 5
Formatting is not important. Just output is required
I need to join two tables and then do a recursive search.
You must OUTER JOIN the second table (which you didn't name, so I have called it TASK_TREE) through DEPENDENT_ON to the parent ID. Outer join because task 1 is the top of the tree and depends on no task. Then use Oracle's hierarchical query syntax to walk the tree:
select t.id, t.task, tt.dependent_on, level
from tasks t
left outer join task_tree tt on tt.id = t.id
connect by prior t.id = tt.dependent_on
start with t.id = 1
/
I have included the level so you can see how the tree unfurls. The Oracle SQL documentation covers hierarchical queries in depth. Find out more. If you don't want to use Oracle's proprietary hierarchical syntax, from 11gR2 Oracle supported recursive WITH clause. Find out more.
Incidentally, your posted data contains a error. Task 4 depends on both 2 and 6. Hierarchies must have child nodes which depend on a single parent node. Otherwise you'll get all sorts of weird results.
This is my NEWSPAPER table.
National News A 1
Sports D 1
Editorials A 12
Business E 1
Weather C 2
Television B 7
Births F 7
Classified F 8
Modern Life B 1
Comics C 4
Movies B 4
Bridge B 2
Obituaries F 6
Doctor Is In F 6
When i run this query
select feature,section,page from NEWSPAPER
where section = 'F'
order by page;
It gives this output
Doctor Is In F 6
Obituaries F 6
Births F 7
Classified F 8
But in Kevin Loney's Oracle 10g Complete Reference the output is like this
Obituaries F 6
Doctor Is In F 6
Births F 7
Classified F 8
Please help me understand how is it happening?
If you need reliable, reproducible ordering to occur when two values in your ORDER BY clause's first column are the same, you should always provide another, secondary column to also order on. While you might be able to assume that they will sort themselves based on order entered (almost always the case to my knowledge, but be aware that the SQL standard does not specify any form of default ordering) or index, you never should (unless it is specifically documented as such for the engine you are using--and even then I'd personally never rely on that).
Your query, if you wanted alphabetical sorting by feature within each page, should be:
SELECT feature,section,page FROM NEWSPAPER
WHERE section = 'F'
ORDER BY page, feature;
In relational databases, tables are sets and are unordered. The order by clause is used primarily for output purposes (and a few other cases such as a subquery containing rownum).
This is a good place to start. The SQL standard does not specify what has to happen when the keys on an order by are the same. And this is for good reason. Different techniques can be used for sorting. Some might be stable (preserving original order). Some methods might not be.
Focus on whether the same rows are in the sets, not their ordering. By the way, I would consider this an unfortunate example. The book should not have ambiguous sorts in its examples.
When you use the SELECT statement to query data from a table, the order which rows appear in the result set may not be what you expected.
In some cases, the rows that appear in the result set are in the order that they are stored in the table physically. However, in case the query optimizer uses an index to process the query, the rows will appear as they are stored in the index key order. For this reason, the order of rows in the result set is undetermined or unpredictable.
The query optimizer is a built-in software component in the database
system that determines the most efficient way for an SQL statement to
query the requested data.
I have the following tables, the groups table which contains hierarchically ordered groups and group_member which stores which groups a user belongs to.
groups
---------
id
parent_id
name
group_member
---------
id
group_id
user_id
ID PARENT_ID NAME
---------------------------
1 NULL Cerebra
2 1 CATS
3 2 CATS 2.0
4 1 Cerepedia
5 4 Cerepedia 2.0
6 1 CMS
ID GROUP_ID USER_ID
---------------------------
1 1 3
2 1 4
3 1 5
4 2 7
5 2 6
6 4 6
7 5 12
8 4 9
9 1 10
I want to retrieve the visible groups for a given user. That it is to say groups a user belongs to and children of these groups. For example, with the above data:
USER VISIBLE_GROUPS
9 4, 5
3 1,2,4,5,6
12 5
I am getting these values using recursion and several database queries. But I would like to know if it is possible to do this with a single SQL query to improve my app performance. I am using MySQL.
Two things come to mind:
1 - You can repeatedly outer-join the table to itself to recursively walk up your tree, as in:
SELECT *
FROM
MY_GROUPS MG1
,MY_GROUPS MG2
,MY_GROUPS MG3
,MY_GROUPS MG4
,MY_GROUPS MG5
,MY_GROUP_MEMBERS MGM
WHERE MG1.PARENT_ID = MG2.UNIQID (+)
AND MG1.UNIQID = MGM.GROUP_ID (+)
AND MG2.PARENT_ID = MG3.UNIQID (+)
AND MG3.PARENT_ID = MG4.UNIQID (+)
AND MG4.PARENT_ID = MG5.UNIQID (+)
AND MGM.USER_ID = 9
That's gonna give you results like this:
UNIQID PARENT_ID NAME UNIQID_1 PARENT_ID_1 NAME_1 UNIQID_2 PARENT_ID_2 NAME_2 UNIQID_3 PARENT_ID_3 NAME_3 UNIQID_4 PARENT_ID_4 NAME_4 UNIQID_5 GROUP_ID USER_ID
4 2 Cerepedia 2 1 CATS 1 null Cerebra null null null null null null 8 4 9
The limit here is that you must add a new join for each "level" you want to walk up the tree. If your tree has less than, say, 20 levels, then you could probably get away with it by creating a view that showed 20 levels from every user.
2 - The only other approach that I know of is to create a recursive database function, and call that from code. You'll still have some lookup overhead that way (i.e., your # of queries will still be equal to the # of levels you are walking on the tree), but overall it should be faster since it's all taking place within the database.
I'm not sure about MySql, but in Oracle, such a function would be similar to this one (you'll have to change the table and field names; I'm just copying something I did in the past):
CREATE OR REPLACE FUNCTION GoUpLevel(WO_ID INTEGER, UPLEVEL INTEGER) RETURN INTEGER
IS
BEGIN
DECLARE
iResult INTEGER;
iParent INTEGER;
BEGIN
IF UPLEVEL <= 0 THEN
iResult := WO_ID;
ELSE
SELECT PARENT_ID
INTO iParent
FROM WOTREE
WHERE ID = WO_ID;
iResult := GoUpLevel(iParent,UPLEVEL-1); --recursive
END;
RETURN iResult;
EXCEPTION WHEN NO_DATA_FOUND THEN
RETURN NULL;
END;
END GoUpLevel;
/
Joe Cleko's books "SQL for Smarties" and "Trees and Hierarchies in SQL for Smarties" describe methods that avoid recursion entirely, by using nested sets. That complicates the updating, but makes other queries (that would normally need recursion) comparatively straightforward. There are some examples in this article written by Joe back in 1996.
I don't think that this can be accomplished without using recursion. You can accomplish it with with a single stored procedure using mySQL, but recursion is not allowed in stored procedures by default. This article has information about how to enable recursion. I'm not certain about how much impact this would have on performance verses the multiple query approach. mySQL may do some optimization of stored procedures, but otherwise I would expect the performance to be similar.
Didn't know if you had a Users table, so I get the list via the User_ID's stored in the Group_Member table...
SELECT GroupUsers.User_ID,
(
SELECT
STUFF((SELECT ',' +
Cast(Group_ID As Varchar(10))
FROM Group_Member Member (nolock)
WHERE Member.User_ID=GroupUsers.User_ID
FOR XML PATH('')),1,1,'')
) As Groups
FROM (SELECT User_ID FROM Group_Member GROUP BY User_ID) GroupUsers
That returns:
User_ID Groups
3 1
4 1
5 1
6 2,4
7 2
9 4
10 1
12 5
Which seems right according to the data in your table. But doesn't match up with your expected value list (e.g. User 9 is only in one group in your table data but you show it in the results as belonging to two)
EDIT: Dang. Just noticed that you're using MySQL. My solution was for SQL Server. Sorry.
-- Kevin Fairchild
There was already similar question raised.
Here is my answer (a bit edited):
I am not sure I understand correctly your question, but this could work My take on trees in SQL.
Linked post described method of storing tree in database -- PostgreSQL in that case -- but the method is clear enough, so it can be adopted easily for any database.
With this method you can easy update all the nodes depend on modified node K with about N simple SELECTs queries where N is distance of K from root node.
Good Luck!
I don't remember which SO question I found the link under, but this article on sitepoint.com (second page) shows another way of storing hierarchical trees in a table that makes it easy to find all child nodes, or the path to the top, things like that. Good explanation with example code.
PS. Newish to StackOverflow, is the above ok as an answer, or should it really have been a comment on the question since it's just a pointer to a different solution (not exactly answering the question itself)?
There's no way to do this in the SQL standard, but you can usually find vendor-specific extensions, e.g., CONNECT BY in Oracle.
UPDATE: As the comments point out, this was added in SQL 99.