Welcome to this exploration of
Advanced SQL - Window Functions Part 4.
** To learn frame type - ROWS, RANGE, and GROUPS, refer to the
part 1 of this blog series
** To Discover more about EXCLUDE, PARTITION BY, LAG and LEAD, refer
to the part 2 of this blog series
** To Discover more about FIRST_VALUE, LAST_VALUE, NTH_VALUE,
ROW_NUMBER, DENSE_RANK, RANK, NTILE, refer to the part 3 of this
blog series
RECAP:
What are window functions?
With SQL:2003, the ISO SQL Standard introduced window functions, a new mode of row-based computation:
Input Output
Aggregate functions(sum, avg, count) group of rows → row (one per group)
window function row vicinity → row (one per row)
- Window functions operate on a set of rows ( commonly known as window or frame or row vicinity) and return a single value for each row.
- The term window describes the set of rows on which the function operates.
- A window function uses values from the rows in a window.
In the Part 1 of this blog series, we learnt how to create a frame
using frame_types - ROWS, RANGE and GROUPS.
In the part 2 of this blog series, we learnt about different
functions like EXCLUDE, PARTITION BY, LAG and LEAD
In the part 3 of this blog series, we learnt about different
functions like FIRST_VALUE, LAST_VALUE, NTH_VALUE, ROW_NUMBER,
DENSE_RANK, RANK, NTILE
Now we'll see more about the other functionality that window
functions provide.
USE CASE - What is the likelihood of nice weather on weekends?
Input - Daily weather sensor readings consisting of day, weekday,
temperature, rain
The weather is fine on day d if—on d and the two days prior—the minimum temperature is above
15°C and
the overall rainfall is
less than 600ml/m².
DROP TABLE IF EXISTS sensors;
CREATE TABLE sensors (
day int PRIMARY KEY, -- day of month
weekday text, -- day of week (Mon...Sun)
temp float, -- temperature in °C
rain float); -- rainfall in ml
INSERT INTO sensors(day, weekday, temp, rain) VALUES
( 1, 'Thu', 13, 0),
( 2, 'Fri', 10, 800),
( 3, 'Sat', 12, 300),
( 4, 'Sun', 16, 100),
( 5, 'Mon', 20, 400),
( 6, 'Tue', 20, 80),
( 7, 'Wed', 18, 500),
( 8, 'Thu', 14, 0),
( 9, 'Fri', 10, 0),
(10, 'Sat', 12, 500),
(11, 'Sun', 14, 300),
(12, 'Mon', 14, 800),
(13, 'Tue', 16, 0),
(14, 'Wed', 15, 0),
(15, 'Thu', 18, 100),
(16, 'Fri', 17, 100),
(17, 'Sat', 15, 0),
(18, 'Sun', 16, 300),
(19, 'Mon', 16, 400),
(20, 'Tue', 19, 200),
(21, 'Wed', 19, 100),
(22, 'Thu', 18, 0),
(23, 'Fri', 17, 0),
(24, 'Sat', 16, 200);
As per the problem statement, the weather is fine on day d if—on d and the two days prior—the minimum temperature is above 15°C and the overall rainfall is less than 600ml/m².
Pseudo code
Step 1
- Collect the weather data for each day and 2 days prior as a frame.
- Find the minimum temperature over the 3 days in the window
- Find the overall rainfall over the 3 days in the window
weather data for each day and 2 days prior
- ORDER BY s.dayROWS
BETWEEN 2 PRECEDING AND CURRENT ROW
minimum temperature over the 3 days in the window -
MIN(s.temp) OVER three_days
overall rainfall over the 3 days in the window
SUM(s.rain) OVER three_days
Combining all the above into a query will be
SELECT s.day, s.weekday,
MIN(s.temp) OVER three_days AS temp,
SUM(s.rain) OVER three_days AS rain
FROM sensors AS s
WINDOW three_days AS (ORDER BY s.day ROWS BETWEEN 2 PRECEDING AND CURRENT ROW)
Step 2:
Over the collected data for current day and 2 prior, check if
minimum temperature is above 15°C and total rainfall is less than
600 ml/m2
SELECT s.day, s.weekday,
CASE WHEN s.temp >= 15 and s.rain <= 600
THEN 'sunny'
ELSE 'rainy'
END AS condition
FROM three_day_sensors AS s
Step 3:
- Now that we have determined whether a day is sunny or rainy. From this data collect only the weekend data (saturday and sunday)
- Group all the weekends and weekdays data separately
- For each group, count the number of days it was sunny and calculate its percentage.
SELECT w.weekday IN ('Sat', 'Sun') AS "weekend?",
(COUNT(*) FILTER (WHERE w.condition = '☀') * 100.0 /
COUNT(*)) :: int AS "% fine"
FROM weather AS w
GROUP BY "weekend?";
FInal Query will be
WITH
three_day_sensors(day, weekday, temp, rain) AS (
SELECT s.day, s.weekday,
MIN(s.temp) OVER three_days AS temp,
SUM(s.rain) OVER three_days AS rain
FROM sensors AS s
WINDOW three_days AS (ORDER BY s.day ROWS BETWEEN 2 PRECEDING AND CURRENT ROW)
),
weather(day, weekday, condition) AS (
SELECT s.day, s.weekday,
CASE WHEN s.temp >= 15 and s.rain <= 600
THEN 'sunny'
ELSE 'rainy'
END AS condition
FROM three_day_sensors AS s
)
SELECT w.weekday IN ('Sat', 'Sun') AS "weekend?",
(COUNT(*) FILTER (WHERE w.condition = 'sunny') * 100.0 /
COUNT(*)) :: int AS "% fine"
FROM weather AS w
GROUP BY "weekend?";
OUTPUT -
weekend? | % fine
---------+--------
f | 29
t | 43
Weekdays have pleasant weather 29% of the time and weekends have pleasant weather 43% of the time.
USE CASE - Sessionization
Input:
A table that contains timestamps of 2 user's logs. For a particular
user, if the difference between 2 timestamps is more than 30
seconds, it should have a new session id
DROP TABLE IF EXISTS log;
CREATE TABLE log(uid text NOT NULL,ts timestamp NOT NULL);
INSERT INTO log(uid, ts) VALUES
('User1', '05-25-2020 07:25:12'), -- assumes datestyle = 'ISO, MDY'
('User1', '05-25-2020 07:25:18'),
('User1', '05-25-2020 07:25:21'),
('User2', '05-25-2020 08:01:55'),
('User2', '05-25-2020 08:05:07'),
('User2', '05-25-2020 08:05:30'),
('User2', '05-25-2020 08:05:53'),
('User2', '05-25-2020 08:06:19'), -- ⎱ sessions by User2 and User1 within
('User1', '05-25-2020 08:06:30'), -- ⎰ 30 seconds, still *two* sessions ⇒ partition by uid ⁑
('User1', '05-25-2020 08:06:42'),
('User1', '05-25-2020 18:32:07'),
('User1', '05-25-2020 18:32:27'),
('User1', '05-25-2020 18:32:44'),
('User1', '05-25-2020 18:33:00'),
('User2', '05-25-2020 22:20:06'),
('User2', '05-25-2020 22:20:16');
Pseudo code
1. Create 2 groups, for user 1 and user 2, and order the timestamps
in ascending order.
PARTITION BY uid ORDER BY ts
Partition by is used to separate user1 and user2 logs and the
timestamps are ordered.
2. If the time difference between the current row and its previous
row is greater than 30 secs then mark the session as 1 else mark it
as 0.
Note : 1 marks the beginning of new session
Current.ts > LAG(ts, 1, '-infinity) OVER window_function
SELECT l.*,
CASE WHEN l.ts > LAG (l.ts, 1, '-infinity') OVER (PARTITION BY l.uid ORDER BY l.ts) + :'inactivity'
THEN 1
ELSE 0
END AS sos
FROM log AS l
ORDER BY l.uid, l.ts -- ← for presentation purposes only
OUTPUT -
uid | ts | sos
-------+---------------------+---------------
User1 | 2020-05-25 07:25:12 | 1
User1 | 2020-05-25 07:25:18 | 0
User1 | 2020-05-25 07:25:21 | 0
User1 | 2020-05-25 08:06:30 | 1
User1 | 2020-05-25 08:06:42 | 0
User1 | 2020-05-25 18:32:07 | 1
User1 | 2020-05-25 18:32:27 | 0
User1 | 2020-05-25 18:32:44 | 0
User1 | 2020-05-25 18:33:00 | 0
User2 | 2020-05-25 08:01:55 | 1
User2 | 2020-05-25 08:05:07 | 1
User2 | 2020-05-25 08:05:30 | 0
User2 | 2020-05-25 08:05:53 | 0
User2 | 2020-05-25 08:06:19 | 0
User2 | 2020-05-25 22:20:06 | 1
User2 | 2020-05-25 22:20:16 | 0
3. Perform a running sum over the start of session tag to assign session IDs
SELECT t.*,
SUM (t.sos) OVER (PARTITION BY t.uid ORDER BY t.ts) AS session
FROM tagged AS t
ORDER BY t,uid, t.ts
uid | ts | sos | session
--------+---------------------+-----+------------
User1 | 2020-05-25 07:25:12 | 1 | 1
User1 | 2020-05-25 07:25:18 | 0 | 1
User1 | 2020-05-25 07:25:21 | 0 | 1
User1 | 2020-05-25 08:06:30 | 1 | 2
User1 | 2020-05-25 08:06:42 | 0 | 2
User1 | 2020-05-25 18:32:07 | 1 | 3
User1 | 2020-05-25 18:32:27 | 0 | 3
User1 | 2020-05-25 18:32:44 | 0 | 3
User1 | 2020-05-25 18:33:00 | 0 | 3
User2 | 2020-05-25 08:01:55 | 1 | 1
User2 | 2020-05-25 08:05:07 | 1 | 2
User2 | 2020-05-25 08:05:30 | 0 | 2
User2 | 2020-05-25 08:05:53 | 0 | 2
User2 | 2020-05-25 08:06:19 | 0 | 2
User2 | 2020-05-25 22:20:06 | 1 | 3
User2 | 2020-05-25 22:20:16 | 0 | 3
4. With this information, we can calculate the analytical data such as time per session, total sessions etc.
SELECT s.uid,
s.session,
MAX(s.ts) - MIN(s.ts) AS duration
FROM sessionized AS s
GROUP BY s.uid, s.session
ORDER BY s.uid, s.session
uid | session | duration
------+---------+----------
User1 | 1 | 00:00:09
User1 | 2 | 00:00:12
User1 | 3 | 00:00:53
User2 | 1 | 00:00:00
User2 | 2 | 00:01:12
User2 | 3 | 00:00:10
FINAL QUERY
\set inactivity '30 seconds'
WITH
tagged(uid, ts, sos) AS (
SELECT l.*,
CASE WHEN l.ts >
LAG (l.ts, 1, '-infinity') OVER (PARTITION BY l.uid ORDER BY l.ts) + :'inactivity'
THEN 1
ELSE 0
END AS sos
FROM log AS l
ORDER BY l.uid, l.ts
),
sessionized(uid, ts, sos, session) AS (
SELECT t.*,
SUM (t.sos) OVER (PARTITION BY t.uid ORDER BY t.ts) AS session
FROM tagged AS t
ORDER BY t,uid, t.ts
),
measured(uid, session, duration) AS (
SELECT s.uid,
s.session,
MAX(s.ts) - MIN(s.ts) AS duration
FROM sessionized AS s
GROUP BY s.uid, s.session
ORDER BY s.uid, s.session
)
TABLE measured;
USE CASE - Identify Consecutive Ranges
Input -
Table of random numbers - example - [5,2,14,3,1,42,6,10,7,13] shall
be converted to [1-3,5-7,10,13&14,42].
Input - [5,2,14,3,1,42,6,10,7,13]
On order by/ sorting - [1,2,3,5,6,7,10,13,14,42 ]
Reducing -
[ 1,2,3 ] to 1-3 [5,6,7] to 5-7 [13,14] to 13&14 42 - 42
Approach -
5. Sort the numbers
6. Create row numbers to the table using ROW_NUMBER()
7. Subtract the number with ROW NUMBER.
8. Similar values are grouped together.
┌─────
│ ref │
├─────
│ 1 │ - 1 = 0 ⎫
│ 2 │ - 2 = 0 ⎬ range 0
│ 3 │ - 3 = 0 __ ⎭
│ 5 │ - 4 = 1 ⎫
│ 6 │ - 5 = 1 ⎬ range 1
│ 7 │ - 6 = 1 __ ⎭
│ 10 │ - 7 = 3 __ } range 3
│ 13 │ - 8 = 5 ⎱ range 5
│ 14 │ - 9 = 5 __ ⎰
│ 42 │ - 10 = 32 } range 32
└─────
DROP TABLE IF EXISTS citations;
CREATE TABLE citations(ref int PRIMARY KEY);
INSERT INTO citations VALUES
(5), (2), (14), (3), (1), (42), (6), (10), (7), (13);
1. Sort the numbers
2. Create row numbers to the table using ROW_NUMBER()
3. Subtract the number with ROW NUMBER.
ref - ROW_NUMBER() OVER (ORDER BY c.ref) AS range
SELECT c.ref,
c.ref - ROW_NUMBER() OVER (ORDER BY c.ref) AS range
FROM citations AS c
ref | range
----+-------
1 | 0
2 | 0
3 | 0
5 | 1
6 | 1
7 | 1
10 | 3
13 | 5
14 | 5
42 | 32
From a given range, find the min and max values. These are the first and last values of the range
SELECT r.range, MIN(r.ref) AS first, MAX(r.ref) AS last
FROM ranges AS r
GROUP BY r.range
range | first | last
-------+-------+------
3 | 10 | 10
5 | 13 | 14
0 | 1 | 3
32 | 42 | 42
1 | 5 | 7
Next step is to add - or & between first and last values based on
their difference.
FINAL QUERY
WITH ranges(ref, range) AS (
SELECT c.ref,
c.ref - ROW_NUMBER() OVER (ORDER BY c.ref) AS range
FROM citations AS c
),
outputs(range, first, last) AS (
SELECT r.range, MIN(r.ref) AS first, MAX(r.ref) AS last
FROM ranges AS r
GROUP BY r.range
)
SELECT string_agg(CASE o.last - o.first
WHEN 0 THEN o.first :: text
WHEN 1 THEN o.first || '&' || o.last
ELSE o.first || '-' || o.last
END,
','
ORDER BY o.range) AS citations
FROM outputs AS o;
citations
-----------------------------
1-3,5-7,10,13&14,42
