Practical window function patterns in PostgreSQL - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the ROW_NUMBER() window function do?

It assigns a unique sequential number to each row within a partition of the result set, starting at 1 for the first row.

intermediate

How does RANK() differ from ROW_NUMBER() in window functions?

RANK() assigns the same rank to rows with identical values, causing gaps in ranking numbers, while ROW_NUMBER() always assigns unique sequential numbers without gaps.

beginner

Explain the use of the PARTITION BY clause in window functions.

PARTITION BY divides the result set into groups (partitions) and the window function is applied independently within each group.

beginner

What is the purpose of the ORDER BY clause inside a window function?

ORDER BY defines the order of rows within each partition for the window function to process, affecting functions like ROW_NUMBER(), RANK(), and cumulative sums.

intermediate

How can you calculate a running total using window functions?

Use SUM() as a window function with ORDER BY to accumulate values row by row, for example: SUM(amount) OVER (ORDER BY date ROWS UNBOUNDED PRECEDING).

Which window function assigns unique sequential numbers to rows within a partition?

A RANK()

B DENSE_RANK()

C ROW_NUMBER()

D COUNT()

ROW_NUMBER() always assigns unique sequential numbers starting at 1 within each partition.

What does PARTITION BY do in a window function?

A Filters rows before window function runs

B Groups rows so the window function runs separately on each group

C Sorts rows globally

D Limits the number of rows returned

PARTITION BY divides rows into groups so the window function applies independently to each group.

Which window function will assign the same rank to tied rows but skip ranks after ties?

A RANK()

B DENSE_RANK()

C ROW_NUMBER()

D NTILE()

RANK() assigns the same rank to tied rows and leaves gaps in ranking numbers after ties.

How do you calculate a running total of sales ordered by date?

A SUM(sales) OVER (ORDER BY date ROWS UNBOUNDED PRECEDING)

B SUM(sales)

C COUNT(sales) OVER (PARTITION BY date)

D AVG(sales) OVER (ORDER BY date)

SUM() as a window function with ORDER BY date and ROWS UNBOUNDED PRECEDING calculates a running total.

What happens if you omit ORDER BY in a window function like ROW_NUMBER()?

A Rows are numbered randomly

B Rows are numbered in insertion order

C An error occurs

D Rows are numbered but order is not guaranteed

Without ORDER BY, ROW_NUMBER() assigns numbers but the order is not guaranteed.

Describe how to use window functions to rank employees by salary within each department.

Think about grouping by department and ordering salaries to assign ranks.

Explain how to compute a running total of sales over time using window functions.

Running totals accumulate values in order.

Practice

1. What does the ROW_NUMBER() OVER (PARTITION BY department ORDER BY salary DESC) window function do in a query?

easy

A. Counts the number of employees in each department.

B. Assigns a unique rank to each employee within their department based on salary, starting at 1 for the highest salary.

C. Calculates the total salary for each department.

D. Returns the average salary of all employees.

Solution

1. Step 1: Understand ROW_NUMBER() with PARTITION BY

   The function assigns a unique sequential number to rows within each department group.

2. Step 2: ORDER BY salary DESC ranks highest salary first

   Ordering by salary descending means the highest salary gets row number 1 in each department.

3. Final Answer:

   Assigns a unique rank to each employee within their department based on salary, starting at 1 for the highest salary. -> Option B

4. Quick Check:

   ROW_NUMBER() with PARTITION BY and ORDER BY = ranked rows [OK]

Hint: ROW_NUMBER() numbers rows per group ordered by criteria [OK]

Common Mistakes:

• Confusing ROW_NUMBER() with COUNT() or SUM()
• Ignoring PARTITION BY and thinking it ranks globally
• Assuming it calculates totals or averages

2. Which of the following is the correct syntax to calculate a running total of sales ordered by date using window functions in PostgreSQL?

easy

A. SELECT date, sales, SUM(sales) PARTITION BY date FROM sales_data;

B. SELECT date, sales, SUM(sales) GROUP BY date ORDER BY running_total FROM sales_data;

C. SELECT date, sales, SUM(sales) WHERE ORDER BY date FROM sales_data;

D. SELECT date, sales, SUM(sales) OVER (ORDER BY date) AS running_total FROM sales_data;

Solution

1. Step 1: Identify correct window function syntax

   SUM(sales) OVER (ORDER BY date) calculates running total by ordering rows by date.

2. Step 2: Check other options for syntax errors

   The incorrect options misuse PARTITION BY without OVER(), use WHERE incorrectly, or confuse GROUP BY with window functions.

3. Final Answer:

   SELECT date, sales, SUM(sales) OVER (ORDER BY date) AS running_total FROM sales_data; -> Option D

4. Quick Check:

   SUM() OVER (ORDER BY ...) = running total syntax [OK]

Hint: Use OVER() with ORDER BY for running totals [OK]

Common Mistakes:

• Using PARTITION BY without OVER()
• Trying to use WHERE inside window function
• Confusing GROUP BY with window functions

3. Given the table sales(date, amount) with rows:
('2024-01-01', 100), ('2024-01-02', 150), ('2024-01-03', 200)
What is the output of this query?

SELECT date, amount, SUM(amount) OVER (ORDER BY date ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS running_total FROM sales ORDER BY date;

medium

A. [('2024-01-01', 100, 450), ('2024-01-02', 150, 350), ('2024-01-03', 200, 200)]

B. [('2024-01-01', 100, 100), ('2024-01-02', 150, 150), ('2024-01-03', 200, 200)]

C. [('2024-01-01', 100, 100), ('2024-01-02', 150, 250), ('2024-01-03', 200, 450)]

D. [('2024-01-01', 100, NULL), ('2024-01-02', 150, NULL), ('2024-01-03', 200, NULL)]

Solution

1. Step 1: Understand SUM() OVER with frame

   The frame includes all rows from the start up to current row ordered by date, so it sums cumulatively.

2. Step 2: Calculate running totals row by row

   Row 1: 100; Row 2: 100+150=250; Row 3: 250+200=450.

3. Final Answer:

   [('2024-01-01', 100, 100), ('2024-01-02', 150, 250), ('2024-01-03', 200, 450)] -> Option C

4. Quick Check:

   Running total sums up to current row [OK]

Hint: ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW sums cumulatively [OK]

Common Mistakes:

• Ignoring frame clause and expecting total sum on all rows
• Confusing running total with simple SUM() aggregation
• Assuming NULLs appear without reason

4. You wrote this query to rank employees by salary within each department:

SELECT employee_id, department, salary, RANK() OVER (ORDER BY salary DESC) AS dept_rank FROM employees;

But the query returns the same rank for all employees regardless of department. What is the likely mistake?

medium

A. Missing PARTITION BY clause inside OVER()

B. ORDER BY salary ASC instead of DESC

C. Using RANK() instead of ROW_NUMBER()

D. Not ordering the final result by department

Solution

1. Step 1: Check PARTITION BY usage

   If PARTITION BY is missing or incorrect, ranking happens over entire table, ignoring departments.

2. Step 2: Confirm ORDER BY and function choice

   ORDER BY DESC is correct for highest salary first; RANK() is appropriate for ties; ordering final result doesn't affect ranking.

3. Final Answer:

   Missing PARTITION BY clause inside OVER() -> Option A

4. Quick Check:

   PARTITION BY groups rows for ranking [OK]

Hint: Always include PARTITION BY to rank within groups [OK]

Common Mistakes:

• Forgetting PARTITION BY causes global ranking
• Confusing RANK() and ROW_NUMBER() effects
• Thinking ORDER BY outside OVER() affects ranking

5. You want to find each employee's salary difference compared to the previous employee in the same department ordered by salary descending. Which query correctly uses window functions to achieve this?

hard

A. SELECT employee_id, department, salary, salary - LAG(salary) OVER (PARTITION BY department ORDER BY salary DESC) AS salary_diff FROM employees;

B. SELECT employee_id, department, salary, salary - LEAD(salary) OVER (ORDER BY salary) AS salary_diff FROM employees;

C. SELECT employee_id, department, salary, salary - LAG(salary) FROM employees GROUP BY department ORDER BY salary DESC;

D. SELECT employee_id, department, salary, salary - FIRST_VALUE(salary) OVER (PARTITION BY department) AS salary_diff FROM employees;

Solution

1. Step 1: Use LAG() to get previous row's salary per department

   LAG(salary) OVER (PARTITION BY department ORDER BY salary DESC) gets previous salary in descending order within department.

2. Step 2: Subtract previous salary from current salary

   salary - LAG(salary) calculates difference compared to previous employee.

3. Final Answer:

   SELECT employee_id, department, salary, salary - LAG(salary) OVER (PARTITION BY department ORDER BY salary DESC) AS salary_diff FROM employees; -> Option A

4. Quick Check:

   LAG() gets previous row value for difference [OK]

Hint: Use LAG() with PARTITION BY and ORDER BY for previous row values [OK]

Common Mistakes:

• Using LEAD() instead of LAG() for previous row
• Omitting PARTITION BY causing wrong grouping
• Trying to use aggregate functions without OVER()