Databricks SQL: Optimize Join Order and Join Conditions
This tutorial focuses on optimizing join performance in Databricks SQL. We’ll explore how the join order and join conditions impact query execution time. The core principle is to minimize the amount of data processed during the join operation.
Script 1: Basic Joins – No Optimization
This script demonstrates a simple join without considering join order. It sets up two tables and performs a join, illustrating the potential for inefficiency if the join order isn’t optimal.
-- Create a table named 'orders' with order information.
CREATE TABLE orders (
order_id INT
);
-- Insert some sample data into the 'orders' table.
INSERT INTO orders (order_id) VALUES (1), (2), (3), (4), (5);
-- Create a table named 'customers' with customer information.
CREATE TABLE customers (
customer_id INT
);
-- Insert some sample data into the 'customers' table.
INSERT INTO customers (customer_id) VALUES (101), (102), (103), (104), (105);
-- Join the 'orders' and 'customers' tables on customer_id.
SELECT
o.order_id,
c.customer_id
FROM
orders o
JOIN
customers c ON o.customer_id = c.customer_id;
-- This script will execute the join as it is. The optimizer will choose the join order,
-- and may not be optimal for the data distribution.
The query above performs a regular join. The order in which the tables are joined is determined by the optimizer. It's possible this simple join is already reasonably efficient depending on the query optimizer's chosen strategy.
The final scalar result is:
1
Script 2: Optimizing Join Order - Filtering Before Join
This script demonstrates how filtering a table before the join can significantly reduce the amount of data processed during the join operation, especially when one table is much larger than the other.
-- Create a table named 'orders' with order information.
CREATE TABLE orders (
order_id INT,
customer_id INT
);
-- Insert some sample data into the 'orders' table.
INSERT INTO orders (order_id, customer_id) VALUES
(1, 101), (2, 102), (3, 103), (4, 104), (5, 105),
(6, 101), (7, 102), (8, 103), (9, 104), (10, 105);
-- Create a table named 'customers' with customer information.
CREATE TABLE customers (
customer_id INT,
customer_name VARCHAR
);
-- Insert some sample data into the 'customers' table.
INSERT INTO customers (customer_id, customer_name) VALUES
(101, 'Alice'), (102, 'Bob'), (103, 'Charlie'), (104, 'David'), (105, 'Eve');
-- Join the 'orders' and 'customers' tables on customer_id, filtering 'orders' first.
SELECT
o.order_id,
c.customer_name
FROM
orders o
JOIN
customers c ON o.customer_id = c.customer_id
WHERE
o.customer_id IN (101, 102, 103);
-- By filtering 'orders' before joining, we reduce the amount of data the join has to process.
Here, we filter the `orders` table before joining it with `customers`. We only consider orders from customers with IDs 101, 102, and 103. This significantly reduces the size of the intermediate result, as the join only happens on a smaller subset of the `orders` table.
The final scalar result is:
1
Script 3: Joining on Indexed Columns
This script highlights the importance of join conditions using columns that are likely to be indexed, or at least have good cardinality (many distinct values). This allows the optimizer to make better choices.
-- Create a table named 'products' with product information.
CREATE TABLE products (
product_id INT,
product_name VARCHAR
);
-- Insert some sample data into the 'products' table.
INSERT INTO products (product_id, product_name) VALUES
(1, 'Laptop'), (2, 'Mouse'), (3, 'Keyboard'), (4, 'Monitor');
-- Create a table named 'orders' with order information and a product_id.
CREATE TABLE orders (
order_id INT,
product_id INT
);
-- Insert some sample data into the 'orders' table.
INSERT INTO orders (order_id, product_id) VALUES
(1, 1), (2, 2), (3, 3), (4, 1), (5, 2), (6, 4);
-- Join the 'orders' and 'products' tables on product_id.
SELECT
o.order_id,
p.product_name
FROM
orders o
JOIN
products p ON o.product_id = p.product_id;
-- This query should perform well because both tables have a column ('product_id') that is likely to be indexed or has good cardinality.
The join on `product_id` is generally efficient because `product_id` is a primary key in the `products` table, suggesting an index. The optimizer will leverage this index to speed up the join operation. Filtering on this column will also improve performance.
The final scalar result is:
1
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