You can do THAT without Perl?

You Can Do ThatWithout Perl?

Lists and Recursion and Trees, Oh My!YAPC10, Pittsburgh 2009

Subject: The YAPC::NA 2009 Survey

Lists:Windowing Functions

What are Windowing Functions?

PGCon2009, 21-22 May 2009 Ottawa

What are the Windowing Functions?

• Similar to classical aggregates but does more! • Provides access to set of rows from the current row• Introduced SQL:2003 and more detail in SQL:2008

– SQL:2008 (http://wiscorp.com/sql200n.zip) 02: SQL/Foundation• 4.14.9 Window tables• 4.15.3 Window functions• 6.10 <window function>• 7.8 <window clause>

• Supported by Oracle, SQL Server, Sybase and DB2– No open source RDBMS so far except PostgreSQL (Firebird trying)

• Used in OLAP mainly but also useful in OLTP– Analysis and reporting by rankings, cumulative aggregates

How they work and what you get

How they work and what do you get

• Windowed table– Operates on a windowed table– Returns a value for each row– Returned value is calculated from the rows in the window

• You can use…– New window functions– Existing aggregate functions– User-defined window functions– User-defined aggregate functions

• Completely new concept!– With Windowing Functions, you can reach outside the current row

[Aggregates] SELECT key, SUM(val) FROM tbl GROUP BY key;

[Windowing Functions] SELECT key, SUM(val) OVER (PARTITION BY key) FROM tbl;

depname empno salary

develop 10 5200sales 1 5000personnel 5 3500sales 4 4800sales 6 550personnel 2 3900develop 7 4200develop 9 4500sales 3 4800develop 8 6000develop 11 5200

Who is the highest paid relatively compared with the department average?

depname empno salary avg diff

develop 8 6000 5020 980sales 6 5500 5025 475personnel 2 3900 3700 200develop 11 5200 5020 180develop 10 5200 5020 180sales 1 5000 5025 -25personnel 5 3500 3700 -200sales 4 4800 5025 -225sales 3 4800 5025 -225develop 9 4500 5020 -520develop 7 4200 5020 -820

SELECT depname, empno, salary, avg(salary) OVER (PARTITION BY depname)::int, salary - avg(salary) OVER (PARTITION BY depname)::int AS diffFROM empsalary ORDER BY diff DESC

Anatomy of a Window• Represents set of rows abstractly as:• A partition

– Specified by PARTITION BY clause– Never moves– Can contain:

• A frame– Specified by ORDER BY clause and frame clause– Defined in a partition– Moves within a partition– Never goes across two partitions

• Some functions take values from a partition. Others take them from a frame.

A partition• Specified by PARTITION BY clause in OVER()• Allows to subdivide the table, much like

GROUP BY clause• Without a PARTITION BY clause, the whole

table is in a single partition

func(args)

OVER ( )

partition-clause order-clause frame-

clause

PARTITION BY expr, …

A frame• Specified by ORDER BY clause and frame

clause in OVER()• Allows to tell how far the set is applied• Also defines ordering of the set• Without order and frame clauses, the whole of

partition is a single frame

func(args)

OVER ( )

clause

ORDER BY expr [ASC|DESC] [NULLS FIRST|LAST], …

(ROWS|RANGE) BETWEEN UNBOUNDED (PRECEDING|FOLLOWING) AND CURRENT ROW

The WINDOW clause• You can specify common window definitions

in one place and name it, for convenience• You can use the window at the same query

WINDOW w AS

clause

SELECT target_list, … wfunc() OVER wFROM table_list, …WHERE qual_list, ….GROUP BY groupkey_list, ….HAVING groupqual_list, …

Built-in Windowing Functions

List of built-in Windowing Functions

• row_number()• rank()• dense_rank()• percent_rank()• cume_dist()• ntile()

• lag()• lead()• first_value()• last_value()• nth_value()

row_number()• Returns number of the current row

val row_number()5 15 23 31 4

SELECT val, row_number() OVER (ORDER BY val DESC) FROM tbl;

Note: row_number() always incremented values independent of frame

rank()• Returns rank of the current row with gap

val rank()5 15 13 31 4

SELECT val, rank() OVER (ORDER BY val DESC) FROM tbl;

Note: rank() OVER(*empty*) returns 1 for all rows, since all rowsare peers to each other

dense_rank()• Returns rank of the current row without gap

val dense_rank()5 15 13 21 3

SELECT val, dense_rank() OVER (ORDER BY val DESC) FROM tbl;

no gap

Note: dense_rank() OVER(*empty*) returns 1 for all rows, since all rows are peers to each other

percent_rank()

• Returns relative rank; (rank() – 1) / (total row – 1)

val percent_rank()5 05 03 0.6666666666666671 1

SELECT val, percent_rank() OVER (ORDER BY val DESC) FROM tbl;

Note: percent_rank() OVER(*empty*) returns 0 for all rows, since all rows are peers to each other

values are always between 0 and 1 inclusive.

cume_dist()

• Returns relative rank; (# of preced. or peers) / (total row)

val cume_dist()5 0.55 0.53 0.751 1

SELECT val, cume_dist() OVER (ORDER BY val DESC) FROM tbl;

Note: cume_dist() OVER(*empty*) returns 1 for all rows, since all rowsare peers to each other

= 2 / 4

= 3 / 4

= 4 / 4

The result can be emulated by “count(*) OVER (ORDER BY val DESC) / count(*) OVER ()”

ntile()• Returns dividing bucket number

val ntile(3)5 15 13 21 3

SELECT val, ntile(3) OVER (ORDER BY val DESC) FROM tbl;

Note: ntile() OVER (*empty*) returns same values as above, sincentile() doesn’t care the frame but works against the partition

The results are the divided positions, but if there’s remainder addrow from the head

4 % 3 = 1

lag()• Returns value of row above

val lag(val)5 NULL5 53 51 3

SELECT val, lag(val) OVER (ORDER BY val DESC) FROM tbl;

Note: lag() only acts on a partition.

lead()• Returns value of the row below

val lead(val)5 55 33 11 NULL

SELECT val, lead(val) OVER (ORDER BY val DESC) FROM tbl;

Note: lead() acts against a partition.

first_value()• Returns the first value of the frame

val first_value(val)5 55 53 51 5

SELECT val, first_value(val) OVER (ORDER BY val DESC) FROM tbl;

last_value()• Returns the last value of the frame

val last_value(val)5 15 13 11 1

SELECT val, last_value(val) OVER (ORDER BY val DESC ROWS BETWEEN UNBOUNDED PRECEDINGAND UNBOUNDED FOLLOWING) FROM tbl;

Note: frame clause is necessary since you have a frame betweenthe first row and the current row by only the order clause

nth_value()• Returns the n-th value of the frame

val nth_value(val, val)5 NULL5 NULL3 31 5

SELECT val, nth_value(val, val) OVER (ORDER BY val DESC ROWS BETWEEN UNBOUNDED PRECEDINGAND UNBOUNDED FOLLOWING) FROM tbl;

Note: frame clause is necessary since you have a frame betweenthe first row and the current row by only the order clause

aggregates(all peers)• Returns the same values along the frame

val sum(val)5 145 143 141 14

Note: all rows are the peers to each other

SELECT val, sum(val) OVER () FROM tbl;

cumulative aggregates• Returns different values along the frame

val sum(val)5 105 103 131 14

Note: row#1 and row#2 return the same value since they are the peers.the result of row#3 is sum(val of row#1…#3)

SELECT val, sum(val) OVER (ORDER BY val DESC) FROM tbl;

Recursion and Trees, Oh My:Common Table Expressions

Thanks to:

• The ISO SQL Standards Committee

• Yoshiyuki Asaba

• Ishii Tatsuo

• Jeff Davis

• Gregory Stark

• Tom Lane

• etc., etc., etc.

Recursion

Recursion in General

•Initial Condition

•Recursion step

•Termination condition

E1 List TableCREATE TABLE employee( id INTEGER NOT NULL, boss_id INTEGER, UNIQUE(id, boss_id)/*, etc., etc. */);

INSERT INTO employee(id, boss_id)VALUES(1,NULL), /* El capo di tutti capi */(2,1),(3,1),(4,1),(5,2),(6,2),(7,2),(8,3),(9,3),(10,4),(11,5),(12,5),(13,6),(14,7),(15,8),(1,9);

Tree Query InitiationWITH RECURSIVE t(node, path) AS ( SELECT id, ARRAY[id] FROM employee WHERE boss_id IS NULL /* Initiation Step */UNION ALL SELECT e1.id, t.path || ARRAY[e1.id] FROM employee e1 JOIN t ON (e1.boss_id = t.node) WHERE id NOT IN (t.path))SELECT CASE WHEN array_upper(path,1)>1 THEN '+-' ELSE '' END || REPEAT('--', array_upper(path,1)-2) || node AS "Branch"FROM tORDER BY path;

Tree Query Recursion

WITH RECURSIVE t(node, path) AS ( SELECT id, ARRAY[id] FROM employee WHERE boss_id IS NULLUNION ALL SELECT e1.id, t.path || ARRAY[e1.id] /* Recursion */ FROM employee e1 JOIN t ON (e1.boss_id = t.node) WHERE id NOT IN (t.path))SELECT CASE WHEN array_upper(path,1)>1 THEN '+-' ELSE '' END || REPEAT('--', array_upper(path,1)-2) || node AS "Branch"FROM tORDER BY path;

Tree Query Termination

WITH RECURSIVE t(node, path) AS ( SELECT id, ARRAY[id] FROM employee WHERE boss_id IS NULLUNION ALL SELECT e1.id, t.path || ARRAY[e1.id] FROM employee e1 JOIN t ON (e1.boss_id = t.node) WHERE id NOT IN (t.path) /* Termination Condition */)SELECT CASE WHEN array_upper(path,1)>1 THEN '+-' ELSE '' END || REPEAT('--', array_upper(path,1)-2) || node AS "Branch"FROM tORDER BY path;

Tree Query Display

WITH RECURSIVE t(node, path) AS ( SELECT id, ARRAY[id] FROM employee WHERE boss_id IS NULLUNION ALL SELECT e1.id, t.path || ARRAY[e1.id] FROM employee e1 JOIN t ON (e1.boss_id = t.node) WHERE id NOT IN (t.path))SELECT CASE WHEN array_upper(path,1)>1 THEN '+-' ELSE '' END || REPEAT('--', array_upper(path,1)-2) || node AS "Branch" /* Display */FROM tORDER BY path;

Tree Query Initiation Branch ---------- 1 +-2 +---5 +-----11 +-----12 +---6 +-----13 +---7 +-----14 +-3 +---8 +-----15 +---9 +-4 +---10 +-9(16 rows)

Travelling Salesman ProblemGiven a number of cities and the costs of travelling from any city to any other city, what is the least-cost round-trip route that visits each city exactly once and then returns to the starting city?

TSP Schema

CREATE TABLE pairs ( from_city TEXT NOT NULL, to_city TEXT NOT NULL, distance INTEGER NOT NULL, PRIMARY KEY(from_city, to_city), CHECK (from_city < to_city));

TSP DataINSERT INTO pairsVALUES ('Bari','Bologna',672), ('Bari','Bolzano',939), ('Bari','Firenze',723), ('Bari','Genova',944), ('Bari','Milan',881), ('Bari','Napoli',257), ('Bari','Palermo',708), ('Bari','Reggio Calabria',464), ....

TSP Program:Symmetric Setup

WITH RECURSIVE both_ways( from_city, to_city, distance) /* Working Table */AS ( SELECT from_city, to_city, distance FROM pairsUNION ALL SELECT to_city AS "from_city", from_city AS "to_city", distance FROM pairs),

WITH RECURSIVE both_ways( from_city, to_city, distance)AS (/* Distances One Way */ SELECT from_city, to_city, distance FROM pairsUNION ALL SELECT to_city AS "from_city", from_city AS "to_city", distance FROM pairs),

WITH RECURSIVE both_ways( from_city, to_city, distance)AS ( SELECT from_city, to_city, distance FROM pairsUNION ALL /* Distances Other Way */ SELECT to_city AS "from_city", from_city AS "to_city", distance FROM pairs),

TSP Program:Path Initialization Step

paths ( from_city, to_city, distance, path)AS ( SELECT from_city, to_city, distance, ARRAY[from_city] AS "path" FROM both_ways b1 WHERE b1.from_city = 'Roma'UNION ALL

TSP Program:Path Recursion Step

SELECT b2.from_city, b2.to_city, p.distance + b2.distance, p.path || b2.from_city FROM both_ways b2 JOIN paths p ON ( p.to_city = b2.from_city AND b2.from_city <> ALL (p.path[ 2:array_upper(p.path,1) ]) /* Prevent re-tracing */ AND array_upper(p.path,1) < 6 ))

TSP Program:Timely Termination Step

SELECT b2.from_city, b2.to_city, p.distance + b2.distance, p.path || b2.from_city FROM both_ways b2 JOIN paths p ON ( p.to_city = b2.from_city AND b2.from_city <> ALL (p.path[ 2:array_upper(p.path,1) ]) /* Prevent re-tracing */ AND array_upper(p.path,1) < 6 /* Timely Termination */ ))

TSP Program:Filter and Display

SELECT path || to_city AS "path", distanceFROM pathsWHERE to_city = 'Roma'AND ARRAY['Milan','Firenze','Napoli'] <@ pathORDER BY distance, pathLIMIT 1;

TSP Program:Filter and Display

davidfetter@tsp=# \i travelling_salesman.sql path | distance ----------------------------------+---------- {Roma,Firenze,Milan,Napoli,Roma} | 1553(1 row)

Time: 11679.503 ms

TSP Program:Gritty Details

QUERY PLAN ------------------------------------------------------------------------------------------------------------------------------------------ Limit (cost=44.14..44.15 rows=1 width=68) (actual time=16131.326..16131.327 rows=1 loops=1) CTE both_ways -> Append (cost=0.00..4.64 rows=132 width=19) (actual time=0.011..0.090 rows=132 loops=1) -> Seq Scan on pairs (cost=0.00..1.66 rows=66 width=19) (actual time=0.010..0.026 rows=66 loops=1) -> Seq Scan on pairs (cost=0.00..1.66 rows=66 width=19) (actual time=0.004..0.029 rows=66 loops=1) CTE paths -> Recursive Union (cost=0.00..38.72 rows=31 width=104) (actual time=0.102..10631.385 rows=1092883 loops=1) -> CTE Scan on both_ways b1 (cost=0.00..2.97 rows=1 width=68) (actual time=0.100..0.251 rows=11 loops=1) Filter: (from_city = 'Roma'::text) -> Hash Join (cost=0.29..3.51 rows=3 width=104) (actual time=180.125..958.459 rows=182145 loops=6) Hash Cond: (b2.from_city = p.to_city) Join Filter: (b2.from_city <> ALL (p.path[2:array_upper(p.path, 1)])) -> CTE Scan on both_ways b2 (cost=0.00..2.64 rows=132 width=68) (actual time=0.001..0.056 rows=132 loops=5) -> Hash (cost=0.25..0.25 rows=3 width=68) (actual time=172.292..172.292 rows=22427 loops=6) -> WorkTable Scan on paths p (cost=0.00..0.25 rows=3 width=68) (actual time=123.167..146.659 rows=22427 loops=6) Filter: (array_upper(path, 1) < 6) -> Sort (cost=0.79..0.79 rows=1 width=68) (actual time=16131.324..16131.324 rows=1 loops=1) Sort Key: paths.distance, ((paths.path || paths.to_city)) Sort Method: top-N heapsort Memory: 17kB -> CTE Scan on paths (cost=0.00..0.78 rows=1 width=68) (actual time=36.621..16127.841 rows=4146 loops=1) Filter: (('{Milan,Firenze,Napoli}'::text[] <@ path) AND (to_city = 'Roma'::text)) Total runtime: 16180.335 ms(22 rows)

Hausdorf-Besicovich-WTF Dimension

WITH RECURSIVEZ(Ix, Iy, Cx, Cy, X, Y, I)AS ( SELECT Ix, Iy, X::float, Y::float, X::float, Y::float, 0 FROM (SELECT -2.2 + 0.031 * i, i FROM generate_series(0,101) AS i) AS xgen(x,ix) CROSS JOIN (SELECT -1.5 + 0.031 * i, i FROM generate_series(0,101) AS i) AS ygen(y,iy) UNION ALL SELECT Ix, Iy, Cx, Cy, X * X - Y * Y + Cx AS X, Y * X * 2 + Cy, I + 1 FROM Z WHERE X * X + Y * Y < 16::float AND I < 100),

Filter

Zt (Ix, Iy, I) AS ( SELECT Ix, Iy, MAX(I) AS I FROM Z GROUP BY Iy, Ix ORDER BY Iy, Ix)

Display

SELECT array_to_string( array_agg( SUBSTRING( ' .,,,-----++++%%%%@@@@#### ', GREATEST(I,1) ),'')FROM ZtGROUP BY IyORDER BY Iy;

Questions?Comments?Straitjackets?

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