Summary: The rank() window function assigns integer ranks to rows within a partition, giving tied rows the same rank and skipping subsequent rank values after ties. Use rank() for top-N-per-group queries, leaderboards, or percentile calculations where gaps in the rank sequence after ties are acceptable. Unlike dense_rank() (consecutive ranks, no gaps) or row_number() (unique integers regardless of ties), rank() reflects tied positions by leaving gaps.

Use rank() to assign ranks to rows within a result set

The rank() window function computes a ranking for each row within a partition of the result set. The rank is determined by the order of rows specified in the ORDER BY clause of the OVER clause. Rows with equal values for the ranking criteria receive the same rank, with the next rank(s) skipped.

This function is useful in scenarios such as finding the top N rows per group, calculating percentiles, or generating leaderboards.

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The rank() function has the following form:

rank() OVER ([PARTITION BY partition_expression] ORDER BY order_expression)

The OVER clause defines the window frame for the function.

• The ORDER BY clause specifies the order in which ranks are assigned to rows.
• The PARTITION BY clause is optional - if specified, it divides the result set into partitions and ranks are assigned within each partition. Otherwise, ranks are computed for each row over the entire result set.

Consider an employees table with columns for employee ID, name, department, and salary. We can use rank() to rank employees within each department by their salary.

WITH sample_data AS (
    SELECT *
    FROM (
        VALUES
            ('Alice', 'Sales', 50000),
            ('Bob', 'Marketing', 55000),
            ('Charlie', 'Sales', 52000),
            ('David', 'IT', 60000),
            ('Eve', 'Marketing', 55000),
            ('Frank', 'IT', 62000)
    ) AS t(employee_name, department, salary)
)
SELECT
    employee_name,
    department,
    salary,
    RANK() OVER (PARTITION BY department ORDER BY salary DESC) AS dept_salary_rank
FROM sample_data
ORDER BY department, dept_salary_rank;

This query ranks employees within each department based on their salary in descending order. Employees with the same salary within a department receive the same rank.

 employee_name | department | salary | dept_salary_rank
---------------+------------+--------+------------------
 Frank         | IT         |  62000 |                1
 David         | IT         |  60000 |                2
 Bob           | Marketing  |  55000 |                1
 Eve           | Marketing  |  55000 |                1
 Charlie       | Sales      |  52000 |                1
 Alice         | Sales      |  50000 |                2
(6 rows)

You can use rank() in a subquery to find the top N rows per group.

WITH products AS (
    SELECT *
    FROM (
        VALUES
            (1, 'A', 100),
            (2, 'A', 80),
            (3, 'B', 200),
            (4, 'B', 180),
            (5, 'B', 150),
            (6, 'C', 120)
    ) AS t(product_id, category, price)
)
SELECT *
FROM (
    SELECT
        product_id,
        category,
        price,
        rank() OVER (PARTITION BY category ORDER BY price DESC) AS rank
    FROM products
) ranked
WHERE rank <= 2;

This query finds the top 2 most expensive products in each category. The subquery ranks products within each category by price, and the outer query filters for rows with a rank less than or equal to 2.

 product_id | category | price | rank
------------+----------+-------+------
          1 | A        |   100 |    1
          2 | A        |    80 |    2
          3 | B        |   200 |    1
          4 | B        |   180 |    2
          6 | C        |   120 |    1
(5 rows)

You can calculate percentiles using the rank() function with some arithmetic.

WITH scores AS (
	SELECT *
	FROM (
        VALUES
            ('Student 1', 85),
            ('Student 2', 92),
            ('Student 3', 78),
            ('Student 4', 90),
            ('Student 5', 88)
	) AS t(student, score)
)
SELECT
	student,
	score,
	rank() OVER (ORDER BY score) AS rank,
	round(100.0 * rank() OVER (ORDER BY score) / (SELECT count(*) FROM scores), 2) AS percentile
FROM scores;

This query calculates the percentile rank for each student based on their score. The percentile is calculated by dividing the rank of each row by the total number of rows and multiplying by 100.

  student  | score | rank | percentile
-----------+-------+------+------------
 Student 3 |    78 |    1 |      20.00
 Student 1 |    85 |    2 |      40.00
 Student 5 |    88 |    3 |      60.00
 Student 4 |    90 |    4 |      80.00
 Student 2 |    92 |    5 |     100.00
(5 rows)

The dense_rank() function is similar to rank(), but it does not skip ranks when there are ties. If multiple rows have the same rank, the next rank will be the next consecutive integer.

WITH scores AS (
    SELECT *
    FROM (
        VALUES
            ('Player 1', 100),
            ('Player 2', 95),
            ('Player 3', 95),
            ('Player 4', 90)
    ) AS t(player, score)
)
SELECT
    player,
    score,
    rank() OVER (ORDER BY score DESC) AS rank,
    dense_rank() OVER (ORDER BY score DESC) AS dense_rank
FROM scores;

This query demonstrates the difference between rank() and dense_rank(). While rank() skips rank 3 due to the tie at rank 2, dense_rank() assigns consecutive ranks.

  player  | score | rank | dense_rank
----------+-------+------+------------
 Player 1 |   100 |    1 |          1
 Player 2 |    95 |    2 |          2
 Player 3 |    95 |    2 |          2
 Player 4 |    90 |    4 |          3
(4 rows)

The row_number() function assigns a unique, sequential integer to each row within the partition of a result set. Unlike rank() and dense_rank(), it does not handle ties.

WITH sales AS (
    SELECT date '2023-01-01' AS sale_date, 1000 AS amount
    UNION ALL
    SELECT date '2023-01-01', 1500
    UNION ALL
    SELECT date '2023-01-02', 1200
    UNION ALL
    SELECT date '2023-01-02', 1200
)
SELECT
    sale_date,
    amount,
    row_number() OVER (PARTITION BY sale_date ORDER BY amount DESC) AS row_num
FROM sales;

This query assigns a unique row number to each sale within a date, ordered by the sale amount descending. Even though there are ties for the date 2023-01-02, each row receives a distinct row number.

 sale_date  | amount | row_num
------------+--------+---------
 2023-01-01 |   1500 |       1
 2023-01-01 |   1000 |       2
 2023-01-02 |   1200 |       1
 2023-01-02 |   1200 |       2
(4 rows)

The rank() and dense_rank() functions handle ties differently. rank() assigns the same rank to tied rows and skips the next rank(s), while dense_rank() assigns the same rank to tied rows but does not skip ranks. Choose the appropriate function based on your requirements.

Like other window functions, rank() performs calculations across a set of rows defined by the OVER clause. This can be computationally expensive, especially for large datasets or complex window definitions.

To optimize performance:

• Include an ORDER BY clause in the OVER clause to avoid sorting the entire dataset.
• Use partitioning (PARTITION BY) to divide the data into smaller chunks when possible.
• Create appropriate indexes on the columns used in the OVER clause.

• PostgreSQL documentation: Window functions
• PostgreSQL documentation: Tutorial on window functions

• dense_rank
• lag
• lead

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