Practical window function patterns in PostgreSQL - Mini Project: Build & Apply

Practical Window Function Patterns in PostgreSQL

📖 Scenario: You are working as a data analyst for a retail company. You have a table called sales that records each sale transaction with columns for sale_id, store_id, sale_date, and amount. Your manager wants to analyze sales trends by store and over time.

🎯 Goal: Build a PostgreSQL query using window functions to calculate the running total of sales amount per store ordered by sale date, and also rank each sale within its store by amount.

📋 What You'll Learn

Use the sales table with columns sale_id, store_id, sale_date, and amount

Calculate a running total of amount per store_id ordered by sale_date

Rank each sale within its store_id by amount in descending order

Use window functions SUM() OVER() and RANK() OVER()

💡 Why This Matters

🌍 Real World

Retail analysts often need to track cumulative sales and rank transactions to identify top-performing stores or sales days.

💼 Career

Understanding window functions is essential for data analysts and database professionals to perform advanced data analysis and reporting.

1

Create the sales table and insert sample data

Write SQL statements to create a table called sales with columns sale_id (integer), store_id (integer), sale_date (date), and amount (numeric). Then insert these exact rows: (1, 101, '2024-01-01', 100.00), (2, 101, '2024-01-02', 150.00), (3, 102, '2024-01-01', 200.00), (4, 101, '2024-01-03', 50.00), (5, 102, '2024-01-02', 300.00).

PostgreSQL

-- Create the sales table and insert the sample data
-- Your code here

Hint

Use CREATE TABLE to define the table and INSERT INTO to add rows with exact values.

2

Define the window partition and order for running total

Write a SQL query selecting all columns from sales. Add a window function SUM(amount) OVER (PARTITION BY store_id ORDER BY sale_date) and alias it as running_total to calculate the running total of sales amount per store ordered by sale date.

PostgreSQL

SELECT sale_id, store_id, sale_date, amount,
       -- Add running total calculation here
       SUM(amount) OVER (PARTITION BY store_id ORDER BY sale_date) AS running_total
FROM sales;
-- Your code here

Hint

Use SUM(amount) OVER (PARTITION BY store_id ORDER BY sale_date) to get the running total per store.

3

Add ranking of sales by amount within each store

Extend the previous query by adding a new column that uses the window function RANK() OVER (PARTITION BY store_id ORDER BY amount DESC) and alias it as sale_rank. This ranks sales within each store by amount from highest to lowest.

PostgreSQL

SELECT sale_id, store_id, sale_date, amount,
       SUM(amount) OVER (PARTITION BY store_id ORDER BY sale_date) AS running_total,
       -- Add ranking by amount per store here
       RANK() OVER (PARTITION BY store_id ORDER BY amount DESC) AS sale_rank
FROM sales;
-- Your code here

Hint

Use RANK() OVER (PARTITION BY store_id ORDER BY amount DESC) to rank sales by amount within each store.

4

Complete the query with ordering by store and sale date

Add an ORDER BY store_id, sale_date clause at the end of the query to display results sorted by store and then by sale date.

PostgreSQL

SELECT sale_id, store_id, sale_date, amount,
       SUM(amount) OVER (PARTITION BY store_id ORDER BY sale_date) AS running_total,
       RANK() OVER (PARTITION BY store_id ORDER BY amount DESC) AS sale_rank
FROM sales
ORDER BY store_id, sale_date;
-- Your code here

Hint

Use ORDER BY store_id, sale_date to sort the final output by store and date.

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()