I have a database in Influxdb 1.7 with several columns and no tag key.
I want to receive only fields where the value is greater than 0 (zero)
(I'm using grafana to show)
like:
table1:
| time | A | B | C | D | E | F |....| AD | AE | ...
|167344...| 0 | 0 | 2 | 0 | 12 | 0 |....| 0 | 0 |
|167356... | 0 | 0 | 2 | 0 | 0 | 0 |....| 0 | 12 |
|167376... | 0 | 0 | 2 | 0 | 1 | 0 |....| 0 | 0 |
|167384... | 1 | 4 | 0 | 0 | 0 | 0 |....| 0 | 1 |
I would like to receive:
| time | A | B | AE |
|167384... | 1 | 4 | 1 |
I've tried:
select last("*") from "table1" where * > 0
select last(*) from "table1" where * > 0
select last(*) as bbb from "table1" where bbb > 0
and other similar queries, but doen't work
I have a Cypher query that shows the following output:
+----------------
| usid | count |
+----------------
| "000" | 1 |
| "000" | 0 |
| "000" | 0 |
| "001" | 1 |
| "001" | 1 |
| "001" | 0 |
| "002" | 2 |
| "002" | 2 |
| "002" | 0 |
| "003" | 4 |
| "003" | 2 |
| "003" | 2 |
| "004" | 4 |
| "004" | 4 |
| "004" | 4 |
+----------------
How can I get the below result with the condition SUM(count) <= 9.
+----------------
| usid | count |
+----------------
| "000" | 1 |
| "001" | 2 |
| "002" | 4 |
| "003" | 8 |
+----------------
Note: I have used the below query to get the 1st table data.
MATCH (us:USER)
WITH us
WHERE us.count <= 4
RETURN us.id as usid, us.count as count;
I don't know how you get your original data, so I will just use a WITH clause and assume the data is there:
// original data
WITH usid, count
// aggregate and filter
WITH usid, sum(count) as new_count
WHERE new_count <= 9
RETURN usid, new_count
Based on the updated question, the new query would look like:
MATCH (us:USER)
WHERE us.count <= 4
WITH us.id as usid, sum(us.count) as count
WHERE new_count <= 9
RETURN usid, count
˙˙˙
I have a tree structure like node(1)->node(2)->node(3). I have name as an property used to retrieve a node.
Given a node say node(3), i wanna retrieve node(1).
Query tried :
MATCH (p:Node)-[:HAS*]->(c:Node) WHERE c.name = "node 3" RETURN p LIMIT 5
But, not able to get node 1.
Your query will not only return "node 1", but it should at least include one path containing it. It's possible to filter the paths to only get the one traversing all the way to the root, however:
MATCH (c:Node {name: "node 3"})<-[:HAS*0..]-(p:Node)
// The root does not have any incoming relationship
WHERE NOT (p)<-[:HAS]-()
RETURN p
Note the use of the 0 length, which matches all cases, including the one where the start node is the root.
Fun fact: even if you have an index on Node:name, it won't be used (unless you're using Neo4j 3.1, where it seems to be fixed since 3.1 Beta2 at least) and you have to explicitly specify it.
MATCH (c:Node {name: "node 3"})<-[:HAS*0..]-(p:Node)
USING INDEX c:Node(name)
WHERE NOT (p)<-[:HAS]-()
RETURN p
Using PROFILE on the first query (with a numerical id property instead of name):
+-----------------------+----------------+------+---------+-------------------------+----------------------+
| Operator | Estimated Rows | Rows | DB Hits | Variables | Other |
+-----------------------+----------------+------+---------+-------------------------+----------------------+
| +ProduceResults | 0 | 1 | 0 | p | p |
| | +----------------+------+---------+-------------------------+----------------------+
| +AntiSemiApply | 0 | 1 | 0 | anon[23], c -- p | |
| |\ +----------------+------+---------+-------------------------+----------------------+
| | +Expand(All) | 1 | 0 | 3 | anon[58], anon[67] -- p | (p)<-[:HAS]-() |
| | | +----------------+------+---------+-------------------------+----------------------+
| | +Argument | 1 | 3 | 0 | p | |
| | +----------------+------+---------+-------------------------+----------------------+
| +Filter | 1 | 3 | 3 | anon[23], c, p | p:Node |
| | +----------------+------+---------+-------------------------+----------------------+
| +VarLengthExpand(All) | 1 | 3 | 5 | anon[23], p -- c | (c)<-[:HAS*]-(p) |
| | +----------------+------+---------+-------------------------+----------------------+
| +Filter | 1 | 1 | 3 | c | c.id == { AUTOINT0} |
| | +----------------+------+---------+-------------------------+----------------------+
| +NodeByLabelScan | 3 | 3 | 4 | c | :Node |
+-----------------------+----------------+------+---------+-------------------------+----------------------+
Total database accesses: 18
and on the second one:
+-----------------------+----------------+------+---------+-------------------------+------------------+
| Operator | Estimated Rows | Rows | DB Hits | Variables | Other |
+-----------------------+----------------+------+---------+-------------------------+------------------+
| +ProduceResults | 0 | 1 | 0 | p | p |
| | +----------------+------+---------+-------------------------+------------------+
| +AntiSemiApply | 0 | 1 | 0 | anon[23], c -- p | |
| |\ +----------------+------+---------+-------------------------+------------------+
| | +Expand(All) | 1 | 0 | 3 | anon[81], anon[90] -- p | (p)<-[:HAS]-() |
| | | +----------------+------+---------+-------------------------+------------------+
| | +Argument | 1 | 3 | 0 | p | |
| | +----------------+------+---------+-------------------------+------------------+
| +Filter | 1 | 3 | 3 | anon[23], c, p | p:Node |
| | +----------------+------+---------+-------------------------+------------------+
| +VarLengthExpand(All) | 1 | 3 | 5 | anon[23], p -- c | (c)<-[:HAS*]-(p) |
| | +----------------+------+---------+-------------------------+------------------+
| +NodeUniqueIndexSeek | 1 | 1 | 2 | c | :Node(id) |
+-----------------------+----------------+------+---------+-------------------------+------------------+
Total database accesses: 13
I have following simple relationship between (:User) nodes.
(:User)-[:FOLLOWS {timestamp}]->(:User)
If I paginate followers ordered by FOLLOWS.timestamp I'm running into performance problems when someone has millions of followers.
MATCH (u:User {Id:{id}})<-[f:FOLLOWS]-(follower)
WHERE f.timestamp <= {timestamp}
RETURN follower
ORDER BY f.timestamp DESC
LIMIT 100
What is suggested approach for paginating big sets of data when ordering is required?
UPDATE
follower timestamp
---------------------------------------
id(1000000) 1455967905
id(999999) 1455967875
id(999998) 1455967234
id(999997) 1455967123
id(999996) 1455965321
id(999995) 1455964123
id(999994) 1455963645
id(999993) 1455963512
id(999992) 1455961343
....
id(2) 1455909382
id(1) 1455908432
I want to slice this list down using timestamp which set on :FOLLOWS relationship. If I want to return batches of 4 followers I take current timestamp first and return 4 most recent, then 1455967123 and 4 most recent and so on. In order to do this the whole list should be order by timestamp which results in performance issues on millions of records.
If you're looking for the most recent followers, i.e. where the timestamp is greater than a given time, it only has to traverse the most recent ones.
You can do it with (2) in 20ms
If you are really looking for the oldest (first) followers it makes sense to skip ahead and don't look at the timestamp of every of the million followers (which takes about 1s on my system, see (3)). If you skip ahead the time goes down to 230ms, see (1)
In general we can see that on my laptop it does 2M db-operations per core and second.
(1) Look at first / oldest followers
PROFILE
> MATCH (u)<-[f:FOLLOWS]-(follower) WHERE id(u) = 0
> // skip ahead
> WITH f,follower SKIP 999000
> // do the actual check
> WITH f,follower WHERE f.ts < 500
> RETURN f, follower
> ORDER BY f.ts
> LIMIT 10;
+---------------------------------+
| f | follower |
+---------------------------------+
| :FOLLOWS[0]{ts:1} | Node[1]{} |
...
+---------------------------------+
10 rows
243 ms
Compiler CYPHER 2.3 Planner COST Runtime INTERPRETED
+-----------------+----------------+---------+---------+-------------------------------------------------------------------------+---------------------------------------+
| Operator | Estimated Rows | Rows | DB Hits | Identifiers | Other |
+-----------------+----------------+---------+---------+-------------------------------------------------------------------------+---------------------------------------+
| +ProduceResults | 1 | 10 | 0 | f, follower | f, follower |
| | +----------------+---------+---------+-------------------------------------------------------------------------+---------------------------------------+
| +Projection | 1 | 10 | 0 | anon[142], anon[155], anon[158], anon[178], f, follower, f, follower, u | anon[155]; anon[158] |
| | +----------------+---------+---------+-------------------------------------------------------------------------+---------------------------------------+
| +Top | 1 | 10 | 0 | anon[142], anon[155], anon[158], anon[178], f, follower, u | Literal(10); |
| | +----------------+---------+---------+-------------------------------------------------------------------------+---------------------------------------+
| +Projection | 1 | 499 | 499 | anon[142], anon[155], anon[158], anon[178], f, follower, u | anon[155]; anon[158]; anon[155].ts |
| | +----------------+---------+---------+-------------------------------------------------------------------------+---------------------------------------+
| +Projection | 1 | 499 | 0 | anon[142], anon[155], anon[158], f, follower, u | f; follower |
| | +----------------+---------+---------+-------------------------------------------------------------------------+---------------------------------------+
| +Filter | 1 | 499 | 0 | anon[142], f, follower, u | anon[142] |
| | +----------------+---------+---------+-------------------------------------------------------------------------+---------------------------------------+
| +Projection | 1 | 1000 | 1000 | anon[142], f, follower, u | f; follower; f.ts < { AUTOINT2} |
| | +----------------+---------+---------+-------------------------------------------------------------------------+---------------------------------------+
| +Skip | 1 | 1000 | 0 | f, follower, u | { AUTOINT1} |
| | +----------------+---------+---------+-------------------------------------------------------------------------+---------------------------------------+
| +Expand(All) | 1 | 1000000 | 1000001 | f, follower, u | (u)<-[ f#12:FOLLOWS]-( follower#24) |
| | +----------------+---------+---------+-------------------------------------------------------------------------+---------------------------------------+
| +NodeByIdSeek | 1 | 1 | 1 | u | |
+-----------------+----------------+---------+---------+-------------------------------------------------------------------------+---------------------------------------+
Total database accesses: 1001501
(2) Look at most recent followers
PROFILE
> MATCH (u)<-[f:FOLLOWS]-(follower) WHERE id(u) = 0
> AND f.ts > 999500
> RETURN f, follower
> LIMIT 10;
+----------------------------------------------+
| f | follower |
+----------------------------------------------+
| :FOLLOWS[999839]{ts:999840} | Node[999840]{} |
...
+----------------------------------------------+
10 rows
23 ms
Compiler CYPHER 2.3 Planner COST Runtime INTERPRETED
+-----------------+----------------+-------+---------+----------------+---------------------------------------------------------------+
| Operator | Estimated Rows | Rows | DB Hits | Identifiers | Other |
+-----------------+----------------+-------+---------+----------------+---------------------------------------------------------------+
| +ProduceResults | 1 | 10 | 0 | f, follower | f, follower |
| | +----------------+-------+---------+----------------+---------------------------------------------------------------+
| +Limit | 1 | 10 | 0 | f, follower, u | Literal(10) |
| | +----------------+-------+---------+----------------+---------------------------------------------------------------+
| +Filter | 1 | 10 | 16394 | f, follower, u | AndedPropertyComparablePredicates(f,f.ts,f.ts > { AUTOINT1}) |
| | +----------------+-------+---------+----------------+---------------------------------------------------------------+
| +Expand(All) | 1 | 16394 | 16395 | f, follower, u | (u)<-[f:FOLLOWS]-(follower) |
| | +----------------+-------+---------+----------------+---------------------------------------------------------------+
| +NodeByIdSeek | 1 | 1 | 1 | u | |
+-----------------+----------------+-------+---------+----------------+---------------------------------------------------------------+
Total database accesses: 32790
(3) Find oldest followers without skipping ahead
PROFILE
> MATCH (u)<-[f:FOLLOWS]-(follower) WHERE id(u) = 0
> AND f.ts < 500
> RETURN f, follower
> LIMIT 10;
+-------------------------------------+
| f | follower |
+-------------------------------------+
...
| :FOLLOWS[491]{ts:492} | Node[492]{} |
+-------------------------------------+
10 rows
1008 ms
Compiler CYPHER 2.3 Planner COST Runtime INTERPRETED
+-----------------+----------------+--------+---------+----------------+---------------------------------------------------------------+
| Operator | Estimated Rows | Rows | DB Hits | Identifiers | Other |
+-----------------+----------------+--------+---------+----------------+---------------------------------------------------------------+
| +ProduceResults | 1 | 10 | 0 | f, follower | f, follower |
| | +----------------+--------+---------+----------------+---------------------------------------------------------------+
| +Limit | 1 | 10 | 0 | f, follower, u | Literal(10) |
| | +----------------+--------+---------+----------------+---------------------------------------------------------------+
| +Filter | 1 | 10 | 999498 | f, follower, u | AndedPropertyComparablePredicates(f,f.ts,f.ts < { AUTOINT1}) |
| | +----------------+--------+---------+----------------+---------------------------------------------------------------+
| +Expand(All) | 1 | 999498 | 999499 | f, follower, u | (u)<-[f:FOLLOWS]-(follower) |
| | +----------------+--------+---------+----------------+---------------------------------------------------------------+
| +NodeByIdSeek | 1 | 1 | 1 | u | |
+-----------------+----------------+--------+---------+----------------+---------------------------------------------------------------+
Total database accesses: 1998998
Can someone explain how these concept works?
I have 1 question. But I don't know have any ideas on constructing the truth table.
f(A,B,C) = AB + A’C
The answer given was ABC + ABC' + A'BC + A'B'C
And i have no idea how it get there. :-(
1. Create a column for each of the inputs, each intermediate functions, and the final function:
A B C | AB | A' | A'C | AB + A'C
--------------------------------
| | | |
| | | |
| | | |
| | | |
| | | |
| | | |
| | | |
| | | |
2. Enumerate all input possibilities, and start filling in the intermediate function values and then the final function value:
A B C | AB | A' | A'C | AB + A'C
--------------------------------
0 0 0 | 0 | 1 | 0 | 0
0 0 1 | | | |
0 1 0 | | | |
0 1 1 | | | |
1 0 0 | | | |
1 0 1 | | | |
1 1 0 | | | |
1 1 1 | | | |
3. Now, you finish the truth table.
Update per OP's edit of question:
The "answer given" can be reduced as follows using Boolean Algebra:
ABC + ABC' + A'BC + A'B'C
AB(C + C') + A'C(B + B')
AB + A'C
...which is the same as the given f(A,B,C). Not sure why ABC + ABC' + A'BC + A'B'C would be considered to be the "answer," but this does show equivalence between the two formulae.