Database SQL Primer (Part 2) [ Window Functions ]

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Window Functions

Description: A very lucid description of window functions may be found in [2]:

After the database server has completed all of the steps necessary to evaluate a query, including joining, filtering, grouping, and sorting, the result set is complete and ready to be returned to the caller. Imagine if you could pause the query execution at this point and take a walk through the result set while it is still held in memory; what types of analysis might you want to do? If your result set contains sales data, perhaps you might want to generate rankings for salespeople or regions, or calculate percentage differences between one time period and another. If you are generating results for a financial report, perhaps you would like to calculate subtotals for each report section, and a grand total for the final section. Using analytic functions, you can do all of these things and more.

We may find a somewhat more dry description in [3]:

Once you understand the concept of grouping and using aggregates in SQL, understanding window functions is easy. Window functions, like aggregate functions, perform an aggregation on a defined set (a group) of rows, but rather than returning one value per group, window functions can return multiple values for each group. The group of rows to perform the aggregation on is the window.

Window functions certainly sound cool and promising. But given the fuss at the beginning of this study guide about query execution order, it makes sense we might ask the same question of window functions: When are window functions executed or their actions performed? We may find our answer in [4]:

It is important to note that window functions are performed as the last step in SQL processing prior to the ORDER BY clause.

This will now make sense if we look back at the mental model previously discussed for how to think about SELECT queries (relevant portion now in bold):

FROM/JOIN (and all associated ON conditions)
WHERE
GROUP BY
HAVING
SELECT (including window functions)
DISTINCT
ORDER BY
LIMIT/OFFSET

Example: Consider what is currently the top voted answer to problem 178. Rank Scores by StefanPochmann (he actually provides several different answers):

-- Answer 1 (uses correlated subquery)
SELECT
  Score,
  (SELECT COUNT(DISTINCT Score) FROM Scores WHERE Score >= S.Score) 'Rank'
FROM Scores S
ORDER BY Score DESC

-- Answer 2 (nested subquery)
SELECT
  Score,
  (SELECT COUNT(*) FROM (SELECT DISTINCT Score S FROM Scores) tmp WHERE S >= Score) 'Rank'
FROM Scores
ORDER BY Score DESC

-- Answer 3 (non-equi join)
SELECT S.Score, COUNT(DISTINCT t.score) 'Rank'
FROM Scores S JOIN Scores T ON S.Score <= T.score
GROUP BY S.Id
ORDER BY S.Score DESC

These are all great, perfectly legitimate answers. Each one has its own degree of sophistication. But I think nearly everyone would agree on thing: each answer is rather complex. Is there an easier way to craft an answer for this problem? Window functions to the rescue!

The DENSE_RANK() window function makes quick work of this problem:

SELECT
  S.Score,
  DENSE_RANK() OVER (ORDER BY S.Score DESC) AS 'Rank'
FROM
  Scores S;

Of course, not every problem will be so neatly solved by simply using a window function, but this example should give you an idea of just how powerful window functions can be.

Window Function Table Reference: The table below describes nonaggregate window functions that, for each row from a query, perform a calculation using rows related to that row. Most aggregate functions also can be used as window functions, namely AVG(), COUNT(), MAX(), MIN(), and SUM(). The "Docs" link for each function points to that function's reference in the official MySQL docs while the "Tutorial" link for each function points to that function's reference on the MySQL tutorial site. See the official MySQL docs for more detailed information.

Function	Docs	Tutorial	Description
`CUME_DIST()`	Docs	Tutorial	Cumulative distribution value
`DENSE_RANK()`	Docs	Tutorial	Rank of current row within its partition, without gaps
`FIRST_VALUE()`	Docs	Tutorial	Value of argument from first row of window frame
`LAG()`	Docs	Tutorial	Value of argument from row lagging current row within partition
`LAST_VALUE()`	Docs	Tutorial	Value of argument from last row of window frame
`LEAD()`	Docs	Tutorial	Value of argument from row leading current row within partition
`NTH_VALUE()`	Docs	Tutorial	Value of argument from N-th row of window frame
`NTILE()`	Docs	Tutorial	Bucket number of current row within its partition.
`PERCENT_RANK()`	Docs	Tutorial	Percentage rank value
`RANK()`	Docs	Tutorial	Rank of current row within its partition, with gaps
`ROW_NUMBER()`	Docs	Tutorial	Number of current row within its partition

Window Function Problem Reference: A variety of problems appear below for which a window function was used as part of the answer. The window function being used/emphasized is highlighted as a bullet point, the problem may be accessed by clicking on the problem title, and a solution to the problem (to highlight how the window function is being used) may be seen by clicking to expand the widget.

The listing below is not comprehensive. Many of the solutions revealed on click also do not comply with the style guidelines highlighted earlier in this study guide (remember, these were all written over the span of 2 weeks). When time permits, I may revisit these solutions and try to clean them up.

AVG()

615. Average Salary: Departments VS Company

SELECT
  DISTINCT DATE_FORMAT(X.pay_date, "%Y-%m") AS pay_month,
  X.department_id,
  (
    CASE
      WHEN X.sal_comp = 0 THEN "same"
      WHEN X.sal_comp < 0 THEN "lower"
      WHEN X.sal_comp > 0 THEN "higher"
    END
  ) AS comparison
FROM
  (
    SELECT
      S.pay_date,
      E.department_id,
      AVG(S.amount) OVER (PARTITION BY MONTH(S.pay_date), E.department_id)
        - AVG(S.amount) OVER (PARTITION BY MONTH(S.pay_date)) AS sal_comp
    FROM
      salary S
      INNER JOIN employee E ON S.employee_id = E.employee_id
  ) X;

COUNT()

1303. Find the Team Size

SELECT
  E.employee_id,
  COUNT(E.team_id) OVER(PARTITION BY E.team_id) AS team_size
FROM
  Employee E
ORDER BY
  E.employee_id;

DENSE_RANK()

178. Rank Scores

SELECT
  S.Score,
  DENSE_RANK() OVER (ORDER BY S.Score DESC) AS 'Rank'
FROM
  Scores S;

184. Department Highest Salary

SELECT
  X.Department,
  X.Employee,
  X.Salary
FROM
  (
    SELECT
      D.Name AS Department,
      E.Name AS Employee,
      E.Salary,
      DENSE_RANK() OVER (
        PARTITION BY D.Name
        ORDER BY
          E.Salary DESC
      ) AS dep_sal_rank
    FROM
      Employee E
      INNER JOIN Department D ON E.DepartmentId = D.Id
  ) X
WHERE
  X.dep_sal_rank = 1;

185. Department Top Three Salaries

SELECT
  X.Department,
  X.Employee,
  X.Salary
FROM
  (
    SELECT
      D.Name AS Department,
      E.Name AS Employee,
      E.Salary,
      DENSE_RANK() OVER (
        PARTITION BY D.Name
        ORDER BY
          E.Salary DESC
      ) AS dep_sal_rank
    FROM
      Employee E
      INNER JOIN Department D ON E.DepartmentId = D.Id
  ) X
WHERE
  X.dep_sal_rank <= 3;

1875. Group Employees of the Same Salary

WITH non_distinct_salaries AS (
  SELECT salary FROM Employees GROUP BY salary HAVING COUNT(*) > 1
)
SELECT
  E.employee_id,
  E.name,
  E.salary,
  DENSE_RANK() OVER(ORDER BY E.salary) AS team_id
FROM
  Employees E
WHERE
  E.salary IN (SELECT * FROM non_distinct_salaries)
ORDER BY
  team_id, E.employee_id;

LAG()

1709. Biggest Window Between Visits

WITH visit_gaps AS (
  SELECT
    UV.user_id,
    DATEDIFF(
      LAG(UV.visit_date, 1, '2021-01-01') OVER(
        PARTITION BY UV.user_id ORDER BY UV.visit_date DESC
      ),
      UV.visit_date
    )
    AS visit_gap
  FROM
    UserVisits UV
)
SELECT
  V.user_id,
  MAX(V.visit_gap) AS biggest_window
FROM
  visit_gaps V
GROUP BY
  V.user_id
ORDER BY
  V.user_id;

LEAD()

1811. Find Interview Candidates

WITH medal_winners AS (
  SELECT
    U.name,
    U.mail,
    (CASE WHEN
      LEAD(C.contest_id, 2) OVER(PARTITION BY U.user_id ORDER BY C.contest_id)
        - C.contest_id = 2 THEN 1 ELSE 0
    END) AS three_consec,
    (CASE WHEN U.user_id = C.gold_medal THEN 1 ELSE 0 END) AS gold_winner
  FROM
    Users U
    INNER JOIN Contests C ON U.user_id IN (C.gold_medal, C.silver_medal, C.bronze_medal)
)
SELECT
  MW.name,
  MW.mail
FROM
  medal_winners MW
GROUP BY
  MW.name, MW.mail
HAVING
  SUM(MW.gold_winner) >= 3
  OR SUM(MW.three_consec) >= 1;

MAX()

1084. Sales Analysis III

SELECT DISTINCT
  X.product_id,
  X.product_name
FROM
  (
    SELECT
      P.product_id,
      P.product_name,
      MIN(S.sale_date) OVER(W) AS min_sale_date,
      MAX(S.sale_date) OVER(W) AS max_sale_date
    FROM
      Sales S
      INNER JOIN Product P on S.product_id = P.product_id
    WINDOW W AS (PARTITION BY P.product_id)
  ) X
WHERE
  X.min_sale_date >= '2019-01-01'
    AND X.max_sale_date <= '2019-03-31';

1867. Orders With Maximum Quantity Above Average

WITH order_aggregates AS (
  SELECT
    O.order_id,
    AVG(O.quantity) AS avg_quant,
    MAX(O.quantity) AS max_quant,
    MAX(AVG(O.quantity)) OVER() AS max_avg_quant
  FROM
    OrdersDetails O
  GROUP BY
    O.order_id
)
SELECT
  OA.order_id
FROM
  order_aggregates OA
WHERE
  OA.max_quant > OA.max_avg_quant;

MIN()

1084. Sales Analysis III

SELECT DISTINCT
  X.product_id,
  X.product_name
FROM
  (
    SELECT
      P.product_id,
      P.product_name,
      MIN(S.sale_date) OVER(W) AS min_sale_date,
      MAX(S.sale_date) OVER(W) AS max_sale_date
    FROM
      Sales S
      INNER JOIN Product P on S.product_id = P.product_id
    WINDOW W AS (PARTITION BY P.product_id)
  ) X
WHERE
  X.min_sale_date >= '2019-01-01'
    AND X.max_sale_date <= '2019-03-31';

RANK()

512. Game Play Analysis II

SELECT
  X.player_id,
  X.device_id
FROM
  (
    SELECT
      A.*,
      RANK() OVER (
        PARTITION BY A.player_id
        ORDER BY
          A.event_date
      ) device_login_order
    FROM
      Activity A
  ) X
WHERE
  X.device_login_order = 1;

586. Customer Placing the Largest Number of Orders

WITH customer_order_totals AS (
  SELECT
    O.customer_number,
    COUNT(*) AS total_orders
  FROM
    orders O
  GROUP BY
    O.customer_number
), ranked_customer_orders AS (
  SELECT
    C.customer_number,
    RANK() OVER(ORDER BY C.total_orders DESC) AS ranking
  FROM
    customer_order_totals C
)
SELECT
  R.customer_number
FROM
  ranked_customer_orders R
WHERE
  R.ranking = 1;

1070. Product Sales Analysis III

WITH ranked_sales_by_year AS (
  SELECT
    S.*,
    RANK() OVER(
      PARTITION BY S.product_id ORDER BY S.year
    ) year_ranking
  FROM
    Sales S
)
SELECT
  R.product_id,
  R.year AS first_year,
  R.quantity,
  R.price
FROM
  ranked_sales_by_year R
WHERE
  R.year_ranking = 1;

1076. Project Employees II

SELECT
  X.project_id
FROM
  (
    SELECT
      P.project_id,
      RANK() OVER(ORDER BY COUNT(*) DESC) AS project_ranking
    FROM
      Project P
      INNER JOIN Employee E ON P.employee_id = E.employee_id
    GROUP BY
      P.project_id
  ) X
WHERE
  X.project_ranking = 1;

1077. Project Employees III

SELECT
  X.project_id,
  X.employee_id
FROM
  (
    SELECT
      P.project_id,
      E.employee_id,
      RANK() OVER(
        PARTITION BY P.project_id
        ORDER BY
          E.experience_years DESC
      ) AS years_rank
    FROM
      Project P
      INNER JOIN Employee E on P.employee_id = E.employee_id
  ) X
WHERE
  X.years_rank = 1;

1082. Sales Analysis I

SELECT
  X.seller_id
FROM
  (
    SELECT
      S.seller_id,
      RANK() OVER(ORDER BY SUM(S.price) DESC) seller_ranking
    FROM
      Sales S
    GROUP BY
      S.seller_id
  ) X
WHERE
  X.seller_ranking = 1;

1112. Highest Grade For Each Student

WITH student_rankings AS (
  SELECT
    E.student_id,
    E.course_id,
    E.grade,
    RANK() OVER(
      PARTITION BY E.student_id 
      ORDER BY E.grade DESC, E.course_id
    ) AS ranking
  FROM
    Enrollments E
)
SELECT
  S.student_id,
  S.course_id,
  S.grade
FROM
  student_rankings S
WHERE
  S.ranking = 1
ORDER BY
  S.student_id;

ROW_NUMBER()

180. Consecutive Numbers

WITH grps AS (
  SELECT
    Num,
    ROW_NUMBER() OVER (
      ORDER BY
        Id
    ) - ROW_NUMBER() OVER (
      PARTITION BY Num
      ORDER BY
        Id
    ) AS grp
  FROM
    Logs
)
SELECT
  DISTINCT Num AS ConsecutiveNums
FROM
  grps
GROUP BY
  grp,
  Num
HAVING
  COUNT(grp) >= 3
ORDER BY
  Num;

601. Human Traffic of Stadium

WITH grps AS (
  SELECT
    S.*,
    S.id - ROW_NUMBER() OVER (ORDER BY S.id) AS grp
  FROM
    stadium S
  WHERE
    S.people >= 100
), grp_freqs AS (
  SELECT
    G.*,
    COUNT(*) OVER(PARTITION BY G.grp) AS grp_freq
  FROM
    grps G
)
SELECT
  GF.id,
  GF.visit_date,
  GF.people
FROM
  grp_freqs GF
WHERE
  GF.grp_freq >= 3
ORDER BY
  GF.visit_date;

618. Students Report By Geography

SELECT
  MAX((CASE WHEN X.continent = 'America' THEN X.name ELSE NULL END)) AS "America",
  MAX((CASE WHEN X.continent = 'Asia' THEN X.name ELSE NULL END)) AS "Asia",
  MAX((CASE WHEN X.continent = 'Europe' THEN X.name ELSE NULL END)) AS "Europe"
FROM
  (
    SELECT
      S.continent,
      S.name,
      ROW_NUMBER() OVER(PARTITION BY S.continent ORDER BY S.name) AS rn
    FROM
      student S
  ) X
GROUP BY
  X.rn;

SUM()

534. Game Play Analysis III

SELECT
  A.player_id,
  A.event_date,
  SUM(A.games_played) OVER (
    PARTITION BY 
      A.player_id
    ORDER BY
      A.event_date
  ) AS games_played_so_far
FROM
  Activity A;

579. Find Cumulative Salary of an Employee

WITH RECURSIVE months_range AS (
  SELECT 1 AS month_num
  UNION ALL
  SELECT month_num + 1 FROM months_range WHERE month_num < 12
), emps_and_months AS (
  SELECT
    *
  FROM
    (SELECT DISTINCT Id FROM Employee) X, months_range
), cum_sal_info AS (
  SELECT
    EM.Id,
    EM.month_num AS Month,
    E.Salary,
    SUM(E.Salary) OVER (
      PARTITION BY EM.Id
      ORDER BY
        EM.month_num ROWS BETWEEN 2 PRECEDING
        AND 0 FOLLOWING
    ) AS cum_sal
  FROM
    emps_and_months EM
    LEFT JOIN Employee E
      ON EM.Id = E.Id AND EM.month_num = E.Month
)
SELECT
  C.Id,
  C.Month,
  C.cum_sal AS Salary
FROM
  cum_sal_info C
  INNER JOIN Employee E1
  ON E1.Id = C.Id
    AND E1.Month = C.Month
    AND C.Month < (
      SELECT
        MAX(E2.Month)
      FROM
        Employee E2
      WHERE
        E2.Id = E1.Id
    )
ORDER BY
  C.Id,
  C.Month DESC;

1308. Running Total for Different Genders

SELECT
  S.gender,
  S.day,
  SUM(S.score_points) OVER(
    PARTITION BY S.gender ORDER BY S.day
  ) AS total
FROM
  Scores S
ORDER BY
  S.gender, S.day;

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