DENSE_RANK (Lakehouse v1)

Returns the rank of a value within a group of values, without gaps in the ranks.

The rank value starts at 1 and continues up sequentially.

If two values are the same, they have the same rank.

Analyze Syntax

func.dense_rank().over(partition_by=[<columns>], order_by=[<columns>])

Analyze Examples

table.department, func.sum(salary), func.dense_rank().over(order_by=func.sum(table.salary).desc()).alias('dense_rank')

| department | total_salary | dense_rank |
|------------|--------------|------------|
| IT         | 172000       | 1          |
| HR         | 160000       | 2          |
| Sales      | 77000        | 3          |

SQL Syntax

DENSE_RANK() OVER ( [ PARTITION BY <expr1> ] ORDER BY <expr2> [ ASC | DESC ] [ <window_frame> ] )

SQL Examples

Create the table

CREATE TABLE employees (
  employee_id INT,
  first_name VARCHAR,
  last_name VARCHAR,
  department VARCHAR,
  salary INT
);

Insert data

INSERT INTO employees (employee_id, first_name, last_name, department, salary) VALUES
  (1, 'John', 'Doe', 'IT', 90000),
  (2, 'Jane', 'Smith', 'HR', 85000),
  (3, 'Mike', 'Johnson', 'IT', 82000),
  (4, 'Sara', 'Williams', 'Sales', 77000),
  (5, 'Tom', 'Brown', 'HR', 75000);

Calculating the total salary per department using DENSE_RANK

SELECT
    department,
    SUM(salary) AS total_salary,
    DENSE_RANK() OVER (ORDER BY SUM(salary) DESC) AS dense_rank
FROM
    employees
GROUP BY
    department;

Result:

department	total_salary	dense_rank
IT	172000	1
HR	160000	2
Sales	77000	3