There is a table of the following structure:
CREATE TABLE history
(
pk serial NOT NULL,
"from" integer NOT NULL,
"to" integer NOT NULL,
entity_key text NOT NULL,
data text NOT NULL,
CONSTRAINT history_pkey PRIMARY KEY (pk)
);
The pk is a primary key, from and to define a position in the sequence and the sequence itself for a given entity identified by entity_key. So the entity has one sequence of 2 rows in case if the first row has the from = 1; to = 2 and the second one has from = 2; to = 3. So the point here is that the to of the previous row matches the from of the next one.
The order to determine "next"/"previous" row is defined by pk which grows monotonously (since it's a SERIAL).
The sequence does not have to start with 1 and the to - from does not necessary 1 always. So it can be from = 1; to = 10. What matters is that the "next" row in the sequence matches the to exactly.
Sample dataset:
pk | from | to | entity_key | data
----+--------+------+--------------+-------
1 | 1 | 2 | 42 | foo
2 | 2 | 3 | 42 | bar
3 | 3 | 4 | 42 | baz
4 | 10 | 11 | 42 | another foo
5 | 11 | 12 | 42 | another baz
6 | 1 | 2 | 111 | one one one
7 | 2 | 3 | 111 | one one one two
8 | 3 | 4 | 111 | one one one three
And what I cannot realize is how to partition by "sequences" here so that I could apply window functions to the group that represents a single "sequence".
Let's say I want to use the row_number() function and would like to get the following result:
pk | row_number | entity_key
----+-------------+------------
1 | 1 | 42
2 | 2 | 42
3 | 3 | 42
4 | 1 | 42
5 | 2 | 42
6 | 1 | 111
7 | 2 | 111
8 | 3 | 111
For convenience I created an SQLFiddle with initial seed: http://sqlfiddle.com/#!15/e7c1c
PS: It's not the "give me the codez" question, I made my own research and I just out of ideas how to partition.
It's obvious that I need to LEFT JOIN with the next.from = curr.to, but then it's still not clear how to reset the partition on next.from IS NULL.
PS: It will be a 100 points bounty for the most elegant query that provides the requested result
PPS: the desired solution should be an SQL query not pgsql due to some other limitations that are out of scope of this question.
I don’t know if it counts as “elegant,” but I think this will do what you want:
with Lagged as (
select
pk,
case when lag("to",1) over (order by pk) is distinct from "from" then 1 else 0 end as starts,
entity_key
from history
), LaggedGroups as (
select
pk,
sum(starts) over (order by pk) as groups,
entity_key
from Lagged
)
select
pk,
row_number() over (
partition by groups
order by pk
) as "row_number",
entity_key
from LaggedGroups
Just for fun & completeness: a recursive solution to reconstruct the (doubly) linked lists of records. [ this will not be the fastest solution ]
NOTE: I commented out the ascending pk condition(s) since they are not needed for the connection logic.
WITH RECURSIVE zzz AS (
SELECT h0.pk
, h0."to" AS next
, h0.entity_key AS ek
, 1::integer AS rnk
FROM history h0
WHERE NOT EXISTS (
SELECT * FROM history nx
WHERE nx.entity_key = h0.entity_key
AND nx."to" = h0."from"
-- AND nx.pk > h0.pk
)
UNION ALL
SELECT h1.pk
, h1."to" AS next
, h1.entity_key AS ek
, 1+zzz.rnk AS rnk
FROM zzz
JOIN history h1
ON h1.entity_key = zzz.ek
AND h1."from" = zzz.next
-- AND h1.pk > zzz.pk
)
SELECT * FROM zzz
ORDER BY ek,pk
;
You can use generate_series() to generate all the rows between the two values. Then you can use the difference of row numbers on that:
select pk, "from", "to",
row_number() over (partition by entity_key, min(grp) order by pk) as row_number
from (select h.*,
(row_number() over (partition by entity_key order by ind) -
ind) as grp
from (select h.*, generate_series("from", "to" - 1) as ind
from history h
) h
) h
group by pk, "from", "to", entity_key
Because you specify that the difference is between 1 and 10, this might actually not have such bad performance.
Unfortunately, your SQL Fiddle isn't working right now, so I can't test it.
Well,
this not exactly one SQL query but:
select a.pk as PK, a.entity_key as ENTITY_KEY, b.pk as BPK, 0 as Seq into #tmp
from history a left join history b on a."to" = b."from" and a.pk = b.pk-1
declare #seq int
select #seq = 1
update #tmp set Seq = case when (BPK is null) then #seq-1 else #seq end,
#seq = case when (BPK is null) then #seq+1 else #seq end
select pk, entity_key, ROW_NUMBER() over (PARTITION by entity_key, seq order by pk asc)
from #tmp order by pk
This is in SQL Server 2008
Related
I'd like to get a count of all of the Ids that have have the same value (Drops) as other Ids. For instance, the illustration below shows you that ID 1 and 3 have A drops so the query would count them. Similarly, ID 7 & 18 have B drops so that's another two IDs that the query would count totalling in 4 Ids that share the same values so that's what my query would return.
+------+-------+
| ID | Drops |
+------+-------+
| 1 | A |
| 2 | C |
| 3 | A |
| 7 | B |
| 18 | B |
+------+-------+
I've tried the several approaches but the following query was my last attempt.
With cte1 (Id1, D1) as
(
select Id, Drops
from Posts
),
cte2 (Id2, D2) as
(
select Id, Drops
from Posts
)
Select count(distinct c1.Id1) newcnt, c1.D1
from cte1 c1
left outer join cte2 c2 on c1.D1 = c2.D2
group by c1.D1
The result if written out in full would be a single value output but the records that the query should be choosing should look as follows:
+------+-------+
| ID | Drops |
+------+-------+
| 1 | A |
| 3 | A |
| 7 | B |
| 18 | B |
+------+-------+
Any advice would be great. Thanks
You can use a CTE to generate a list of Drops values that have more than one corresponding ID value, and then JOIN that to Posts to find all rows which have a Drops value that has more than one Post:
WITH CTE AS (
SELECT Drops
FROM Posts
GROUP BY Drops
HAVING COUNT(*) > 1
)
SELECT P.*
FROM Posts P
JOIN CTE ON P.Drops = CTE.Drops
Output:
ID Drops
1 A
3 A
7 B
18 B
If desired you can then count those posts in total (or grouped by Drops value):
WITH CTE AS (
SELECT Drops
FROM Posts
GROUP BY Drops
HAVING COUNT(*) > 1
)
SELECT COUNT(*) AS newcnt
FROM Posts P
JOIN CTE ON P.Drops = CTE.Drops
Output
newcnt
4
Demo on SQLFiddle
You may use dense_rank() to resolve your problem. if drops has the same ID then dense_rank() will provide the same rank.
Here is the demo.
with cte as
(
select
drops,
count(distinct rnk) as newCnt
from
( select
*,
dense_rank() over (partition by drops order by id) as rnk
from myTable
) t
group by
drops
having count(distinct rnk) > 1
)
select
sum(newCnt) as newCnt
from cte
Output:
|newcnt |
|------ |
| 4 |
First group the count of the ids for your drops and then sum the values greater than 1.
select sum(countdrops) as total from
(select drops , count(id) as countdrops from yourtable group by drops) as temp
where countdrops > 1;
after some transformation I have a result from a cross join (from table a and b) where I want to do some analysis on. The table for this looks like this:
+-----+------+------+------+------+-----+------+------+------+------+
| id | 10_1 | 10_2 | 11_1 | 11_2 | id | 10_1 | 10_2 | 11_1 | 11_2 |
+-----+------+------+------+------+-----+------+------+------+------+
| 111 | 1 | 0 | 1 | 0 | 222 | 1 | 0 | 1 | 0 |
| 111 | 1 | 0 | 1 | 0 | 333 | 0 | 0 | 0 | 0 |
| 111 | 1 | 0 | 1 | 0 | 444 | 1 | 0 | 1 | 1 |
| 112 | 0 | 1 | 1 | 0 | 222 | 1 | 0 | 1 | 0 |
+-----+------+------+------+------+-----+------+------+------+------+
The ids in the first column are different from the ids in the sixth column.
In a row are always two different IDs that are matched with each other. The other columns always have either 0 or 1 as a value.
I am now trying to find out how many values(meaning both have "1" in 10_1, 10_2 etc) two IDs have on average in common, but I don't really know how to do so.
I was trying something like this as a start:
SELECT SUM(CASE WHEN a.10_1 = 1 AND b.10_1 = 1 then 1 end)
But this would obviously only count how often two ids have 10_1 in common. I could make something like this for example for different columns:
SELECT SUM(CASE WHEN (a.10_1 = 1 AND b.10_1 = 1)
OR (a.10_2 = 1 AND b.10_1 = 1) OR [...] then 1 end)
To count in general how often two IDs have one thing in common, but this would of course also count if they have two or more things in common. Plus, I would also like to know how often two IDS have two things, three things etc in common.
One "problem" in my case is also that I have like ~30 columns I want to look at, so I can hardly write down for each case every possible combination.
Does anyone know how I can approach my problem in a better way?
Thanks in advance.
Edit:
A possible result could look like this:
+-----------+---------+
| in_common | count |
+-----------+---------+
| 0 | 100 |
| 1 | 500 |
| 2 | 1500 |
| 3 | 5000 |
| 4 | 3000 |
+-----------+---------+
With the codes as column names, you're going to have to write some code that explicitly references each column name. To keep that to a minimum, you could write those references in a single union statement that normalizes the data, such as:
select id, '10_1' where "10_1" = 1
union
select id, '10_2' where "10_2" = 1
union
select id, '11_1' where "11_1" = 1
union
select id, '11_2' where "11_2" = 1;
This needs to be modified to include whatever additional columns you need to link up different IDs. For the purpose of this illustration, I assume the following data model
create table p (
id integer not null primary key,
sex character(1) not null,
age integer not null
);
create table t1 (
id integer not null,
code character varying(4) not null,
constraint pk_t1 primary key (id, code)
);
Though your data evidently does not currently resemble this structure, normalizing your data into a form like this would allow you to apply the following solution to summarize your data in the desired form.
select
in_common,
count(*) as count
from (
select
count(*) as in_common
from (
select
a.id as a_id, a.code,
b.id as b_id, b.code
from
(select p.*, t1.code
from p left join t1 on p.id=t1.id
) as a
inner join (select p.*, t1.code
from p left join t1 on p.id=t1.id
) as b on b.sex <> a.sex and b.age between a.age-10 and a.age+10
where
a.id < b.id
and a.code = b.code
) as c
group by
a_id, b_id
) as summ
group by
in_common;
The proposed solution requires first to take one step back from the cross-join table, as the identical column names are super annoying. Instead, we take the ids from the two tables and put them in a temporary table. The following query gets the result wanted in the question. It assumes table_a and table_b from the question are the same and called tbl, but this assumption is not needed and tbl can be replaced by table_a and table_b in the two sub-SELECT queries. It looks complicated and uses the JSON trick to flatten the columns, but it works here:
WITH idtable AS (
SELECT a.id as id_1, b.id as id_2 FROM
-- put cross join of table a and table b here
)
SELECT in_common,
count(*)
FROM
(SELECT idtable.*,
sum(CASE
WHEN meltedR.value::text=meltedL.value::text THEN 1
ELSE 0
END) AS in_common
FROM idtable
JOIN
(SELECT tbl.id,
b.*
FROM tbl, -- change here to table_a
json_each(row_to_json(tbl)) b -- and here too
WHERE KEY<>'id' ) meltedL ON (idtable.id_1 = meltedL.id)
JOIN
(SELECT tbl.id,
b.*
FROM tbl, -- change here to table_b
json_each(row_to_json(tbl)) b -- and here too
WHERE KEY<>'id' ) meltedR ON (idtable.id_2 = meltedR.id
AND meltedL.key = meltedR.key)
GROUP BY idtable.id_1,
idtable.id_2) tt
GROUP BY in_common ORDER BY in_common;
The output here looks like this:
in_common | count
-----------+-------
2 | 2
3 | 1
4 | 1
(3 rows)
+---+------------+
| V | output |
+---+------------+
| y | 1 |
| y | 2 |
| y | 3 |
| N | 0 |
| y | 1 |
| y | 2 |
| N | 0 |
| N | 1 |
+---+------------+
Let me assume that you have a column (say, id) that has the ordering information. Then, you want to identify groups of "Y"s and "N"s that appear together and then enumerate them.
You can do this using a difference of row numbers trick:
select t.v,
row_number() over (partition by v, seqnum_id - seqnum_vid order by id) as output
from (select t.*,
row_number() over (order by id) as seqnum_id,
row_number() over (partition v by order by id) as seqnum_vid
from t
) t;
Explaining how this works is usually tricky. I recommend that you run the subquery to see what the sequence numbers look like and why the difference is constant for the groups you want to identify.
Your sample output is some how a little bit complex,
I preferred to use a SQL recursive query for the solution of your problem
Of course I assume that the id column is starting from 1 and goes continuosly without any gap. In a more complex case, the row_number() function should be added besides id field and join should be setup on rownumbers
I hope it helps,
--create table bool(id int identity(1,1), bool char(1))
--insert into bool values ('Y'),('N'),('Y'),('Y'),('Y'),('N'),('Y'),('N'),('N'),('Y'),('Y'),('Y'),('Y'),('Y'),('N'),('Y'),('Y')
;with cte as (
select id, bool curr, bool pre, 1 output from bool where id = 1
union all
select
bool.id, bool.bool curr, cte.curr,
case when bool.bool = cte.curr then cte.output + 1 else case when bool.bool = 'Y' then 1 else 0 end end
from cte
inner join bool on bool.id = cte.id + 1
)
select * from cte
Output is as follows
I want to create a query that will count the number of times the following condition is met.
I have a table that consists of multiple records with a matching foreign key. I want to check only for each of the foreign key groups if the highest value of another column of that key occurs more than once. If it does that will up the count.
Data
--------------------------
ID | Foreign Key | Value
--------------------------
1 | 1 | 1
2 | 1 | 2
3 | 1 | 2
4 | 2 | 0
5 | 2 | 2
6 | 2 | 1
7 | 3 | 0
8 | 3 | 1
9 | 3 | 1
The query I want should return the number 2. This is because the maximum value in group 1(Foreign Key) occurs twice, the value is 2. In group 2 the maximum value is 2 but only occurs once so this will not up the count. Then in group 3 the maximum value is 1 which occurs twice which will up the count. The count therefore ends up as two.
All credit goes to the comment from #Bob, but here is the sql that solved this problem.
SELECT Count(1)
FROM (SELECT DISTINCT foreign_key
FROM (SELECT foreign_key,
Count(1)
FROM data
WHERE ( foreign_key, value ) IN (SELECT foreign_key,
Max(value)
FROM data
GROUP BY foreign_key)
GROUP BY foreign_key
HAVING Count(1) > 1) AS data) AS data;
This is one approach:
select max(num_at_max)
from (select t.*, count(val) over(partition by fk) as num_at_max
from tbl t
join (select max(max_val_by_grp) as max_val_all_grps
from (select fk, max(val) as max_val_by_grp
from tbl
group by fk) x) x
on t.val = x.max_val_all_grps) x
I have table with data something like this:
ID | RowNumber | Data
------------------------------
1 | 1 | Data
2 | 2 | Data
3 | 3 | Data
4 | 1 | Data
5 | 2 | Data
6 | 1 | Data
7 | 2 | Data
8 | 3 | Data
9 | 4 | Data
I want to group each set of RowNumbers So that my result is something like this:
ID | RowNumber | Group | Data
--------------------------------------
1 | 1 | a | Data
2 | 2 | a | Data
3 | 3 | a | Data
4 | 1 | b | Data
5 | 2 | b | Data
6 | 1 | c | Data
7 | 2 | c | Data
8 | 3 | c | Data
9 | 4 | c | Data
The only way I know where each group starts and stops is when the RowNumber starts over. How can I accomplish this? It also needs to be fairly efficient since the table I need to do this on has 52 Million Rows.
Additional Info
ID is truly sequential, but RowNumber may not be. I think RowNumber will always begin with 1 but for example the RowNumbers for group1 could be "1,1,2,2,3,4" and for group2 they could be "1,2,4,6", etc.
For the clarified requirements in the comments
The rownumbers for group1 could be "1,1,2,2,3,4" and for group2 they
could be "1,2,4,6" ... a higher number followed by a lower would be a
new group.
A SQL Server 2012 solution could be as follows.
Use LAG to access the previous row and set a flag to 1 if that row is the start of a new group or 0 otherwise.
Calculate a running sum of these flags to use as the grouping value.
Code
WITH T1 AS
(
SELECT *,
LAG(RowNumber) OVER (ORDER BY ID) AS PrevRowNumber
FROM YourTable
), T2 AS
(
SELECT *,
IIF(PrevRowNumber IS NULL OR PrevRowNumber > RowNumber, 1, 0) AS NewGroup
FROM T1
)
SELECT ID,
RowNumber,
Data,
SUM(NewGroup) OVER (ORDER BY ID
ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS Grp
FROM T2
SQL Fiddle
Assuming ID is the clustered index the plan for this has one scan against YourTable and avoids any sort operations.
If the ids are truly sequential, you can do:
select t.*,
(id - rowNumber) as grp
from t
Also you can use recursive CTE
;WITH cte AS
(
SELECT ID, RowNumber, Data, 1 AS [Group]
FROM dbo.test1
WHERE ID = 1
UNION ALL
SELECT t.ID, t.RowNumber, t.Data,
CASE WHEN t.RowNumber != 1 THEN c.[Group] ELSE c.[Group] + 1 END
FROM dbo.test1 t JOIN cte c ON t.ID = c.ID + 1
)
SELECT *
FROM cte
Demo on SQLFiddle
How about:
select ID, RowNumber, Data, dense_rank() over (order by grp) as Grp
from (
select *, (select min(ID) from [Your Table] where ID > t.ID and RowNumber = 1) as grp
from [Your Table] t
) t
order by ID
This should work on SQL 2005. You could also use rank() instead if you don't care about consecutive numbers.