SUM, AVG, COUNT as window functions in PostgreSQL - Mini Project: Build & Apply

Using SUM, AVG, and COUNT as Window Functions in PostgreSQL

📖 Scenario: You work for a small online store. You have a table of sales records showing each sale's sale_id, customer_id, and amount. You want to analyze sales data by calculating running totals, average sales, and counts of sales per customer.

🎯 Goal: Build a PostgreSQL query that uses window functions SUM(), AVG(), and COUNT() to calculate running totals, average sale amounts, and total sales count per customer.

📋 What You'll Learn

Create a table called sales with columns sale_id (integer), customer_id (integer), and amount (numeric).

Insert exactly 5 rows into sales with specified values.

Write a query that uses SUM(amount) OVER (PARTITION BY customer_id ORDER BY sale_id) to get running total per customer.

Use AVG(amount) OVER (PARTITION BY customer_id) to get average sale amount per customer.

Use COUNT(*) OVER (PARTITION BY customer_id) to get total number of sales per customer.

💡 Why This Matters

🌍 Real World

Window functions help analyze data trends over groups without collapsing rows, useful in sales, finance, and reporting.

💼 Career

Understanding window functions is essential for data analysts and database developers to write efficient and insightful SQL queries.

1

Create the sales table and insert data

Create a table called sales with columns sale_id (integer), customer_id (integer), and amount (numeric). Then insert these exact rows: (1, 101, 50.00), (2, 102, 75.00), (3, 101, 100.00), (4, 103, 20.00), (5, 102, 125.00).

PostgreSQL

-- Create the sales table and insert the 5 rows below
-- Your code here

Hint

Use CREATE TABLE to define the table and INSERT INTO with multiple rows to add data.

2

Set up the window partition and order

Write the beginning of a SELECT query from sales that selects sale_id, customer_id, and amount. Add a window clause that partitions by customer_id and orders by sale_id. Assign this window clause the name cust_win.

PostgreSQL

CREATE TABLE sales (
    sale_id INTEGER,
    customer_id INTEGER,
    amount NUMERIC
);

INSERT INTO sales (sale_id, customer_id, amount) VALUES
(1, 101, 50.00),
(2, 102, 75.00),
(3, 101, 100.00),
(4, 103, 20.00),
(5, 102, 125.00);

SELECT sale_id, customer_id, amount
FROM sales
WINDOW cust_win AS (PARTITION BY customer_id ORDER BY sale_id)
-- Your code here

Hint

Use the WINDOW clause to define a named window cust_win with partition and order.

3

Add SUM, AVG, and COUNT window functions

Extend the SELECT query to include these columns: SUM(amount) OVER cust_win AS running_total, AVG(amount) OVER (PARTITION BY customer_id) AS avg_amount, and COUNT(*) OVER (PARTITION BY customer_id) AS sales_count.

PostgreSQL

CREATE TABLE sales (
    sale_id INTEGER,
    customer_id INTEGER,
    amount NUMERIC
);

INSERT INTO sales (sale_id, customer_id, amount) VALUES
(1, 101, 50.00),
(2, 102, 75.00),
(3, 101, 100.00),
(4, 103, 20.00),
(5, 102, 125.00);

SELECT sale_id, customer_id, amount,
-- Add SUM, AVG, and COUNT window functions here
SUM(amount) OVER cust_win AS running_total,
AVG(amount) OVER (PARTITION BY customer_id) AS avg_amount,
COUNT(*) OVER (PARTITION BY customer_id) AS sales_count
FROM sales
WINDOW cust_win AS (PARTITION BY customer_id ORDER BY sale_id);

Hint

Use window functions with OVER and the named window cust_win or inline partition clauses.

4

Complete the query with ORDER BY

Add ORDER BY sale_id at the end of the SELECT query to sort the results by sale_id.

PostgreSQL

CREATE TABLE sales (
    sale_id INTEGER,
    customer_id INTEGER,
    amount NUMERIC
);

INSERT INTO sales (sale_id, customer_id, amount) VALUES
(1, 101, 50.00),
(2, 102, 75.00),
(3, 101, 100.00),
(4, 103, 20.00),
(5, 102, 125.00);

SELECT sale_id, customer_id, amount,
       SUM(amount) OVER cust_win AS running_total,
       AVG(amount) OVER (PARTITION BY customer_id) AS avg_amount,
       COUNT(*) OVER (PARTITION BY customer_id) AS sales_count
FROM sales
WINDOW cust_win AS (PARTITION BY customer_id ORDER BY sale_id)
-- Add ORDER BY sale_id here

Hint

Use ORDER BY sale_id at the end of the query to sort results by sale ID.

Practice

1. What does the COUNT(*) OVER() window function do in PostgreSQL?

easy

A. Counts all rows in the result set without grouping them.

B. Counts only distinct rows in the table.

C. Counts rows and groups them by a specified column.

D. Counts rows but returns only one aggregated result.

Solution

1. Step 1: Understand COUNT(*) OVER() usage

   The function counts all rows in the query result but does not group or collapse rows.

2. Step 2: Difference from GROUP BY

   Unlike GROUP BY, it keeps all rows visible and adds the count as a column.

3. Final Answer:

   Counts all rows in the result set without grouping them. -> Option A

4. Quick Check:

   COUNT(*) OVER() counts all rows [OK]

Hint: Window COUNT(*) counts all rows without grouping [OK]

Common Mistakes:

• Confusing window COUNT with GROUP BY count
• Thinking it returns a single aggregated row
• Assuming it counts distinct values only

2. Which of the following is the correct syntax to calculate a running total of sales using SUM as a window function in PostgreSQL?

easy

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

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

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

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

Solution

1. Step 1: Identify correct window function syntax

   SUM(sales) OVER (ORDER BY date) calculates cumulative sum ordered by date.

2. Step 2: Check other options for syntax errors

   The other options misuse window function syntax or omit OVER keyword.

3. Final Answer:

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

4. Quick Check:

   SUM() OVER (ORDER BY ...) is correct syntax [OK]

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

Common Mistakes:

• Omitting OVER() clause
• Using PARTITION BY without OVER()
• Confusing GROUP BY with window functions

3. Given the table employees with columns department and salary, what will the following query return?

SELECT department, salary, AVG(salary) OVER (PARTITION BY department) AS avg_dept_salary FROM employees;

medium

A. Syntax error due to missing GROUP BY clause.

B. Each row shows salary and overall average salary of all employees.

C. Each row shows salary and average salary of all departments combined.

D. Each row shows salary and average salary of its department.

Solution

1. Step 1: Understand PARTITION BY in AVG window function

   AVG(salary) OVER (PARTITION BY department) calculates average salary per department for each row.

2. Step 2: Confirm output per row

   Each row keeps salary and adds average salary of that row's department without grouping rows.

3. Final Answer:

   Each row shows salary and average salary of its department. -> Option D

4. Quick Check:

   AVG() OVER (PARTITION BY ...) averages per group per row [OK]

Hint: PARTITION BY groups rows but keeps all rows visible [OK]

Common Mistakes:

• Thinking it aggregates to one row per group
• Confusing window AVG with GROUP BY AVG
• Expecting overall average without PARTITION BY

4. Identify the error in this query:

SELECT employee_id, salary, COUNT(*) OVER PARTITION BY department FROM employees;

medium

A. COUNT(*) cannot be used as a window function.

B. Missing parentheses around PARTITION BY clause in OVER().

C. employee_id cannot be selected with window functions.

D. No error; query is correct.

Solution

1. Step 1: Check syntax of window function

   OVER clause must have parentheses: OVER (PARTITION BY department).

2. Step 2: Validate other parts

   COUNT(*) is valid as window function; selecting employee_id is allowed.

3. Final Answer:

   Missing parentheses around PARTITION BY clause in OVER(). -> Option B

4. Quick Check:

   OVER() requires parentheses [OK]

Hint: Always use parentheses after OVER keyword [OK]

Common Mistakes:

• Omitting parentheses in OVER()
• Thinking COUNT(*) can't be window function
• Assuming window functions require GROUP BY

5. You want to find the cumulative count of orders per customer ordered by order date. Which query correctly uses COUNT as a window function to achieve this?

hard

A. SELECT customer_id, order_date, COUNT(*) FROM orders GROUP BY customer_id, order_date ORDER BY order_date;

B. SELECT customer_id, order_date, COUNT(*) OVER (ORDER BY order_date) AS cumulative_orders FROM orders;

C. SELECT customer_id, order_date, COUNT(*) OVER (PARTITION BY customer_id ORDER BY order_date) AS cumulative_orders FROM orders;

D. SELECT customer_id, order_date, COUNT(*) OVER (PARTITION BY order_date ORDER BY customer_id) AS cumulative_orders FROM orders;

Solution

1. Step 1: Understand cumulative count per customer

   COUNT(*) OVER (PARTITION BY customer_id ORDER BY order_date) counts rows per customer in order of order_date cumulatively.

2. Step 2: Check other options for correctness

   SELECT customer_id, order_date, COUNT(*) FROM orders GROUP BY customer_id, order_date ORDER BY order_date; uses GROUP BY, not cumulative count; C lacks partition by customer; D partitions by order_date, not customer.

3. Final Answer:

   SELECT customer_id, order_date, COUNT(*) OVER (PARTITION BY customer_id ORDER BY order_date) AS cumulative_orders FROM orders; -> Option C

4. Quick Check:

   Use PARTITION BY customer_id with ORDER BY for cumulative count [OK]

Hint: Partition by customer and order by date for cumulative count [OK]

Common Mistakes:

• Missing PARTITION BY customer_id
• Using GROUP BY instead of window function
• Partitioning by wrong column