Given the following auction data, how would you find the percent difference between a persons most recent and previous bid for a product using Oracle SQL?
The duplicate sequence (SEQ) for person A and B is representative of data I am working with.
An example of your SQL would be very appreciated.
TXN_TIME | SEQ | PERSON | PRODUCT | TRANSACTION | BID |
2017-11-22 15:41:10:0 | 20 | A | 1 | BID | 12 |
2017-11-22 15:35:10:0 | 10C | A | 1 | CXLBID | NULL |
2017-11-22 15:34:25:0 | 10 | A | 1 | BID | 10 |
2017-11-22 15:35:40:0 | 6 | A | 2 | BID | 4 |
2017-11-22 15:34:50:0 | 1C | A | 2 | CXLBID | NULL |
2017-11-22 15:34:20:0 | 1 | A | 2 | BID | 5 |
2017-11-22 15:35:45:0 | 6 | B | 2 | BID | 2 |
2017-11-22 15:34:55:0 | 1C | B | 2 | CXLBID | NULL |
2017-11-22 15:34:25:0 | 1 | B | 2 | BID | 1 |
We could try to use LEAD/LAG analytic functions if they be available. But one approach here would be to use a CTE to identify just the most recent, and immediately prior, bid for each person, and then compare these two values.
WITH cte AS (
SELECT PERSON, BID,
ROW_NUMBER() OVER (PARTITION BY PERSON ORDER BY TXN_TIME DESC) rn
FROM yourTable
WHERE TRANSACTION = 'BID'
)
SELECT
t1.PERSON,
100*(t1.BID - t2.BID) / t2.BID AS BID_PCT_DIFF
FROM cte t1
INNER JOIN cte t2
ON t1.PERSON = t2.PERSON AND
t1.rn = 1 AND t2.rn = 2;
This output looks correct, because person A went from a bid of 4 to 12, which is an increase of 8, or 200%, and person B went from a bid of 1 to 2, which is a 100% increase.
I created a demo below in SQL Server, because I always have difficulties getting Oracle demos to work. But my query is just ANSI SQL and should run the same on either SQL Server or Oracle.
Demo
Good thing you are using Oracle 12. This way you can use the MATCH_RECOGNIZE clause, which is perfect for your problem.
I calculate the CHANGE column in the MATCH_RECOGNIZE clause, using the LAST() function with the optional second argument, which is a logical offset within the set of rows mapped to a specific pattern variable. I format the CHANGE column in the SELECT clause - I use a favorite hack, using the "currency" symbol to attach the percent sign... you can modify the formatting any way you want, without affecting the calculation (which is hidden in the MATCH_RECOGNIZE clause).
with auction_data ( txn_time, seq, person, product, transaction, bid ) as (
select timestamp '2017-11-22 15:41:10', '20' , 'A', 1, 'BID' , 12 from dual union all
select timestamp '2017-11-22 15:35:10', '10C', 'A', 1, 'CXLBID', NULL from dual union all
select timestamp '2017-11-22 15:34:25', '10' , 'A', 1, 'BID' , 10 from dual union all
select timestamp '2017-11-22 15:35:40', '6' , 'A', 2, 'BID' , 4 from dual union all
select timestamp '2017-11-22 15:34:50', '1C' , 'A', 2, 'CXLBID', NULL from dual union all
select timestamp '2017-11-22 15:34:20', '1' , 'A', 2, 'BID' , 5 from dual union all
select timestamp '2017-11-22 15:35:45', '6' , 'B', 2, 'BID' , 2 from dual union all
select timestamp '2017-11-22 15:34:55', '1C' , 'B', 2, 'CXLBID', NULL from dual union all
select timestamp '2017-11-22 15:34:25', '1' , 'B', 2, 'BID' , 1 from dual
)
-- End of simulated inputs (for testing only, not part of the solution).
select txn_time, seq, person, product, transaction, bid,
to_char( 100 * (change - 1), '999D0L', 'nls_currency=''%''') as change
from auction_data
match_recognize(
partition by person, product
order by txn_time
measures case when classifier() = 'B' then bid / last(B.bid, 1) end as change
all rows per match
pattern ( (B|A)* )
define B as B.transaction = 'BID'
);
TXN_TIME SEQ PERSON PRODUCT TRANSACTION BID CHANGE
------------------- --- ------ ---------- ----------- ---------- ----------------
2017-11-22 15:34:25 10 A 1 BID 10
2017-11-22 15:35:10 10C A 1 CXLBID
2017-11-22 15:41:10 20 A 1 BID 12 20.0%
2017-11-22 15:34:20 1 A 2 BID 5
2017-11-22 15:34:50 1C A 2 CXLBID
2017-11-22 15:35:40 6 A 2 BID 4 -20.0%
2017-11-22 15:34:25 1 B 2 BID 1
2017-11-22 15:34:55 1C B 2 CXLBID
2017-11-22 15:35:45 6 B 2 BID 2 100.0%
I have data in a self-join hierarchical table where Continents have many Countries have many Regions have many States have many Cities.
Self-joining table structure:
|-------------------------------------------------------------|
| ID | Name | Type | ParentID | IsTopLevel |
|-------------------------------------------------------------|
| 1 | North America | Continent | NULL | 1 |
| 12 | United States | Country | 1 | 0 |
| 113 | Midwest | Region | 12 | 0 |
| 155 | Kansas | State | 113 | 0 |
| 225 | Topeka | City | 155 | 0 |
| 2 | South America | Continent | NULL | 1 |
| 22 | Argentina | Country | 2 | 0 |
| 223 | Southern | Region | 22 | 0 |
| 255 | La Pampa | State | 223 | 0 |
| 777 | Santa Rosa | City | 255 | 0 |
|-------------------------------------------------------------|
I have been able to successfully use a recursive CTE to get the tree structure and depth of each node. Where I am failing is using a pivot to create a nice list of all bottom locations and their corresponding parents at each level.
The expected results:
|------------------------------------------------------------------------------------|
| Continent | Country | Region | State | City | Bottom_Level_ID |
|------------------------------------------------------------------------------------|
| North America | United States | Midwest | Kansas | Topeka | 234 |
| South America | Argentina | Southern | La Pampa | Santa Rosa | 777 |
|------------------------------------------------------------------------------------|
There are a few key points I should clarify.
Every single entry has a bottom level and a top level. There are no
cases where all five Types are not present for a given location.
If I filled out this data, I'd have 50 entries for North America at the
State level, so you can imagine how immense this table is at the
City level for every continent on the planet. Billions of rows.
The reason this is a necessity is because I need to be able to join onto a historical table of all addresses a person has lived at, and journey up the tree. I figure if I have the LocationID from that table, I can just LEFT JOIN onto a View of this query and nab the appropriate columns.
This is an old database, 2005, and I don't have sysadmin or control of the schema.
My CTE Code
--CTE
;WITH Tree
AS (
SELECT ID, Name, ParentID, Type, 1 as Depth
FROM LocationTable
WHERE IsTopLevel = 1
UNION ALL
SELECT L.ID, L.Name, L.ParentID, L.Type, T.Depth+1
FROM Tree T
JOIN LocationTable L
ON L.ParentGUID = T.GUID
)
Good solid data, in a mostly useful format. BUT then I got to thinking about it and isn't the table structure already in this format, so why would I bother doing a depth tree search if I wasn't going to join the entries together at the same time?
Anyway, here was the rest.
The Pivot Attempt
;WITH Tree
AS (
SELECT ID, Name, ParentID, Type
FROM LocationTable
WHERE IsTopLevel = 1
UNION ALL
SELECT L.ID, L.Name, L.ParentID, L.Type
FROM Tree T
JOIN LocationTable L
ON L.ParentGUID = T.GUID
)
select *
from Tree
pivot (
max(Name)
for Type in ([Continent],[Country],[Region],[State],[City])
) pvt
And now I have everything by Type in a column, with nulls for everything else. As I have struggled with before, I need to filter/join the CTE data before I attempt my pivot, but I have no idea where to start with that piece. Everything I have tried is soooooooooo sloooooooow.
Everytime I think I understand CTEs and Pivot, something new makes me extremely humbled. Please help me. ; ;
If your structure is as clean as you describe it (no gaps, 5 levels always) you might go the easy way:
This data really demands for a classical 1:n-table-tree, where your Countries, States etc. live in their own tables and link to their parent record
Make sure there's an index on ParentID and ID!
DECLARE #tbl TABLE(ID INT,Name VARCHAR(100),Type VARCHAR(100),ParentID INT,IsTopLevel BIT);
INSERT INTO #tbl VALUES
(1,'North America','Continent',NULL,1)
,(12,'United States','Country',1,0)
,(113,'Midwest','Region',12,0)
,(155,'Kansas','State',113,0)
,(225,'Topeka','City',155,0)
,(2,'South America','Continent',NULL,1)
,(22,'Argentina','Country',2,0)
,(223,'Southern','Region',22,0)
,(255,'La Pampa','State',223,0)
,(777,'Santa Rosa','City',255,0);
SELECT Level1.Name AS Continent
,Level2.Name AS Country
,Level3.Name AS Region
,Level4.Name AS State
,Level5.Name AS City
,Level5.ID AS Bottom_Level_ID
FROM #tbl AS Level1
INNER JOIN #tbl AS Level2 ON Level1.ID=Level2.ParentID
INNER JOIN #tbl AS Level3 ON Level2.ID=Level3.ParentID
INNER JOIN #tbl AS Level4 ON Level3.ID=Level4.ParentID
INNER JOIN #tbl AS Level5 ON Level4.ID=Level5.ParentID
WHERE Level1.ParentID IS NULL
The result
Continent Country Region State City Bottom_Level_ID
North America United States Midwest Kansas Topeka 225
South America Argentina Southern La Pampa Santa Rosa 777
Another solution with CTE could be :
;WITH Tree
AS (
SELECT cast(NULL as varchar(100)) as C1, cast(NULL as varchar(100)) as C2, cast(NULL as varchar(100)) as C3, cast(NULL as varchar(100)) as C4, Name as C5, ID as B_Level
FROM LocationTable
WHERE IsTopLevel = 1
UNION ALL
SELECT T.C2, T.C3, T.C4, T.C5, L.Name, L.ID
FROM Tree T
JOIN LocationTable L
ON L.ParentID = T.B_Level
)
select *
from Tree
where C1 is not null
Is there a unpivot equivalent function in PostgreSQL?
Create an example table:
CREATE TEMP TABLE foo (id int, a text, b text, c text);
INSERT INTO foo VALUES (1, 'ant', 'cat', 'chimp'), (2, 'grape', 'mint', 'basil');
You can 'unpivot' or 'uncrosstab' using UNION ALL:
SELECT id,
'a' AS colname,
a AS thing
FROM foo
UNION ALL
SELECT id,
'b' AS colname,
b AS thing
FROM foo
UNION ALL
SELECT id,
'c' AS colname,
c AS thing
FROM foo
ORDER BY id;
This runs 3 different subqueries on foo, one for each column we want to unpivot, and returns, in one table, every record from each of the subqueries.
But that will scan the table N times, where N is the number of columns you want to unpivot. This is inefficient, and a big problem when, for example, you're working with a very large table that takes a long time to scan.
Instead, use:
SELECT id,
unnest(array['a', 'b', 'c']) AS colname,
unnest(array[a, b, c]) AS thing
FROM foo
ORDER BY id;
This is easier to write, and it will only scan the table once.
array[a, b, c] returns an array object, with the values of a, b, and c as it's elements.
unnest(array[a, b, c]) breaks the results into one row for each of the array's elements.
You could use VALUES() and JOIN LATERAL to unpivot the columns.
Sample data:
CREATE TABLE test(id int, a INT, b INT, c INT);
INSERT INTO test(id,a,b,c) VALUES (1,11,12,13),(2,21,22,23),(3,31,32,33);
Query:
SELECT t.id, s.col_name, s.col_value
FROM test t
JOIN LATERAL(VALUES('a',t.a),('b',t.b),('c',t.c)) s(col_name, col_value) ON TRUE;
DBFiddle Demo
Using this approach it is possible to unpivot multiple groups of columns at once.
EDIT
Using Zack's suggestion:
SELECT t.id, col_name, col_value
FROM test t
CROSS JOIN LATERAL (VALUES('a', t.a),('b', t.b),('c',t.c)) s(col_name, col_value);
<=>
SELECT t.id, col_name, col_value
FROM test t
,LATERAL (VALUES('a', t.a),('b', t.b),('c',t.c)) s(col_name, col_value);
db<>fiddle demo
Great article by Thomas Kellerer found here
Unpivot with Postgres
Sometimes it’s necessary to normalize de-normalized tables - the opposite of a “crosstab” or “pivot” operation. Postgres does not support an UNPIVOT operator like Oracle or SQL Server, but simulating it, is very simple.
Take the following table that stores aggregated values per quarter:
create table customer_turnover
(
customer_id integer,
q1 integer,
q2 integer,
q3 integer,
q4 integer
);
And the following sample data:
customer_id | q1 | q2 | q3 | q4
------------+-----+-----+-----+----
1 | 100 | 210 | 203 | 304
2 | 150 | 118 | 422 | 257
3 | 220 | 311 | 271 | 269
But we want the quarters to be rows (as they should be in a normalized data model).
In Oracle or SQL Server this could be achieved with the UNPIVOT operator, but that is not available in Postgres. However Postgres’ ability to use the VALUES clause like a table makes this actually quite easy:
select c.customer_id, t.*
from customer_turnover c
cross join lateral (
values
(c.q1, 'Q1'),
(c.q2, 'Q2'),
(c.q3, 'Q3'),
(c.q4, 'Q4')
) as t(turnover, quarter)
order by customer_id, quarter;
will return the following result:
customer_id | turnover | quarter
------------+----------+--------
1 | 100 | Q1
1 | 210 | Q2
1 | 203 | Q3
1 | 304 | Q4
2 | 150 | Q1
2 | 118 | Q2
2 | 422 | Q3
2 | 257 | Q4
3 | 220 | Q1
3 | 311 | Q2
3 | 271 | Q3
3 | 269 | Q4
The equivalent query with the standard UNPIVOT operator would be:
select customer_id, turnover, quarter
from customer_turnover c
UNPIVOT (turnover for quarter in (q1 as 'Q1',
q2 as 'Q2',
q3 as 'Q3',
q4 as 'Q4'))
order by customer_id, quarter;
FYI for those of us looking for how to unpivot in RedShift.
The long form solution given by Stew appears to be the only way to accomplish this.
For those who cannot see it there, here is the text pasted below:
We do not have built-in functions that will do pivot or unpivot. However,
you can always write SQL to do that.
create table sales (regionid integer, q1 integer, q2 integer, q3 integer, q4 integer);
insert into sales values (1,10,12,14,16), (2,20,22,24,26);
select * from sales order by regionid;
regionid | q1 | q2 | q3 | q4
----------+----+----+----+----
1 | 10 | 12 | 14 | 16
2 | 20 | 22 | 24 | 26
(2 rows)
pivot query
create table sales_pivoted (regionid, quarter, sales)
as
select regionid, 'Q1', q1 from sales
UNION ALL
select regionid, 'Q2', q2 from sales
UNION ALL
select regionid, 'Q3', q3 from sales
UNION ALL
select regionid, 'Q4', q4 from sales
;
select * from sales_pivoted order by regionid, quarter;
regionid | quarter | sales
----------+---------+-------
1 | Q1 | 10
1 | Q2 | 12
1 | Q3 | 14
1 | Q4 | 16
2 | Q1 | 20
2 | Q2 | 22
2 | Q3 | 24
2 | Q4 | 26
(8 rows)
unpivot query
select regionid, sum(Q1) as Q1, sum(Q2) as Q2, sum(Q3) as Q3, sum(Q4) as Q4
from
(select regionid,
case quarter when 'Q1' then sales else 0 end as Q1,
case quarter when 'Q2' then sales else 0 end as Q2,
case quarter when 'Q3' then sales else 0 end as Q3,
case quarter when 'Q4' then sales else 0 end as Q4
from sales_pivoted)
group by regionid
order by regionid;
regionid | q1 | q2 | q3 | q4
----------+----+----+----+----
1 | 10 | 12 | 14 | 16
2 | 20 | 22 | 24 | 26
(2 rows)
Hope this helps, Neil
Pulling slightly modified content from the link in the comment from #a_horse_with_no_name into an answer because it works:
Installing Hstore
If you don't have hstore installed and are running PostgreSQL 9.1+, you can use the handy
CREATE EXTENSION hstore;
For lower versions, look for the hstore.sql file in share/contrib and run in your database.
Assuming that your source (e.g., wide data) table has one 'id' column, named id_field, and any number of 'value' columns, all of the same type, the following will create an unpivoted view of that table.
CREATE VIEW vw_unpivot AS
SELECT id_field, (h).key AS column_name, (h).value AS column_value
FROM (
SELECT id_field, each(hstore(foo) - 'id_field'::text) AS h
FROM zcta5 as foo
) AS unpiv ;
This works with any number of 'value' columns. All of the resulting values will be text, unless you cast, e.g., (h).value::numeric.
Just use JSON:
with data (id, name) as (
values (1, 'a'), (2, 'b')
)
select t.*
from data, lateral jsonb_each_text(to_jsonb(data)) with ordinality as t
order by data.id, t.ordinality;
This yields
|key |value|ordinality|
|----|-----|----------|
|id |1 |1 |
|name|a |2 |
|id |2 |1 |
|name|b |2 |
dbfiddle
I wrote a horrible unpivot function for PostgreSQL. It's rather slow but it at least returns results like you'd expect an unpivot operation to.
https://cgsrv1.arrc.csiro.au/blog/2010/05/14/unpivotuncrosstab-in-postgresql/
Hopefully you can find it useful..
Depending on what you want to do... something like this can be helpful.
with wide_table as (
select 1 a, 2 b, 3 c
union all
select 4 a, 5 b, 6 c
)
select unnest(array[a,b,c]) from wide_table
You can use FROM UNNEST() array handling to UnPivot a dataset, tandem with a correlated subquery (works w/ PG 9.4).
FROM UNNEST() is more powerful & flexible than the typical method of using FROM (VALUES .... ) to unpivot datasets. This is b/c FROM UNNEST() is variadic (with n-ary arity). By using a correlated subquery the need for the lateral ORDINAL clause is eliminated, & Postgres keeps the resulting parallel columnar sets in the proper ordinal sequence.
This is, BTW, FAST -- in practical use spawning 8 million rows in < 15 seconds on a 24-core system.
WITH _students AS ( /** CTE **/
SELECT * FROM
( SELECT 'jane'::TEXT ,'doe'::TEXT , 1::INT
UNION
SELECT 'john'::TEXT ,'doe'::TEXT , 2::INT
UNION
SELECT 'jerry'::TEXT ,'roe'::TEXT , 3::INT
UNION
SELECT 'jodi'::TEXT ,'roe'::TEXT , 4::INT
) s ( fn, ln, id )
) /** end WITH **/
SELECT s.id
, ax.fanm -- field labels, now expanded to two rows
, ax.anm -- field data, now expanded to two rows
, ax.someval -- manually incl. data
, ax.rankednum -- manually assigned ranks
,ax.genser -- auto-generate ranks
FROM _students s
,UNNEST /** MULTI-UNNEST() BLOCK **/
(
( SELECT ARRAY[ fn, ln ]::text[] AS anm -- expanded into two rows by outer UNNEST()
/** CORRELATED SUBQUERY **/
FROM _students s2 WHERE s2.id = s.id -- outer relation
)
,( /** ordinal relationship preserved in variadic UNNEST() **/
SELECT ARRAY[ 'first name', 'last name' ]::text[] -- exp. into 2 rows
AS fanm
)
,( SELECT ARRAY[ 'z','x','y'] -- only 3 rows gen'd, but ordinal rela. kept
AS someval
)
,( SELECT ARRAY[ 1,2,3,4,5 ] -- 5 rows gen'd, ordinal rela. kept.
AS rankednum
)
,( SELECT ARRAY( /** you may go wild ... **/
SELECT generate_series(1, 15, 3 )
AS genser
)
)
) ax ( anm, fanm, someval, rankednum , genser )
;
RESULT SET:
+--------+----------------+-----------+----------+---------+-------
| id | fanm | anm | someval |rankednum| [ etc. ]
+--------+----------------+-----------+----------+---------+-------
| 2 | first name | john | z | 1 | .
| 2 | last name | doe | y | 2 | .
| 2 | [null] | [null] | x | 3 | .
| 2 | [null] | [null] | [null] | 4 | .
| 2 | [null] | [null] | [null] | 5 | .
| 1 | first name | jane | z | 1 | .
| 1 | last name | doe | y | 2 | .
| 1 | | | x | 3 | .
| 1 | | | | 4 | .
| 1 | | | | 5 | .
| 4 | first name | jodi | z | 1 | .
| 4 | last name | roe | y | 2 | .
| 4 | | | x | 3 | .
| 4 | | | | 4 | .
| 4 | | | | 5 | .
| 3 | first name | jerry | z | 1 | .
| 3 | last name | roe | y | 2 | .
| 3 | | | x | 3 | .
| 3 | | | | 4 | .
| 3 | | | | 5 | .
+--------+----------------+-----------+----------+---------+ ----
Here's a way that combines the hstore and CROSS JOIN approaches from other answers.
It's a modified version of my answer to a similar question, which is itself based on the method at https://blog.sql-workbench.eu/post/dynamic-unpivot/ and another answer to that question.
-- Example wide data with a column for each year...
WITH example_wide_data("id", "2001", "2002", "2003", "2004") AS (
VALUES
(1, 4, 5, 6, 7),
(2, 8, 9, 10, 11)
)
-- that is tided to have "year" and "value" columns
SELECT
id,
r.key AS year,
r.value AS value
FROM
example_wide_data w
CROSS JOIN
each(hstore(w.*)) AS r(key, value)
WHERE
-- This chooses columns that look like years
-- In other cases you might need a different condition
r.key ~ '^[0-9]{4}$';
It has a few benefits over other solutions:
By using hstore and not jsonb, it hopefully minimises issues with type conversions (although hstore does convert everything to text)
The columns don't need to be hard coded or known in advance. Here, columns are chosen by a regex on the name, but you could use any SQL logic based on the name, or even the value.
It doesn't require PL/pgSQL - it's all SQL