ML model training in Snowflake - Mini Project: Build & Apply

ML Model Training in Snowflake

📖 Scenario: You work as a data engineer at a retail company. Your team wants to train a simple machine learning model inside Snowflake to predict customer churn based on historical data. You will create a table with customer data, configure model parameters, train the model using Snowflake's ML capabilities, and finalize the model for use.

🎯 Goal: Build a Snowflake ML model training pipeline step-by-step. You will create a customer data table, set model configuration, train a logistic regression model, and finalize the model for prediction.

📋 What You'll Learn

Create a table named customer_data with specific columns and data

Add a configuration variable model_params with logistic regression settings

Train a model named churn_model using Snowflake's CREATE MODEL syntax

Finalize the model by granting usage rights to a role

💡 Why This Matters

🌍 Real World

Training ML models directly inside Snowflake helps data teams avoid moving data out, speeding up workflows and improving security.

💼 Career

Data engineers and data scientists often build and manage ML models in cloud data platforms like Snowflake to support business decisions.

1

Create the customer data table

Create a table called customer_data with columns customer_id (integer), age (integer), monthly_spend (float), and churned (boolean). Insert these exact rows: (1, 25, 50.5, false), (2, 40, 80.0, true), (3, 30, 65.0, false).

Snowflake

-- Create the customer_data table and insert data
-- Your code here

Hint

Use CREATE OR REPLACE TABLE to define the table and INSERT INTO to add rows.

2

Set model configuration parameters

Create a variable called model_params as a string containing JSON with these exact key-value pairs: "model_type": "logistic_regression", "max_iterations": 100, and "regularization": 0.1.

Snowflake

CREATE OR REPLACE TABLE customer_data (
  customer_id INT,
  age INT,
  monthly_spend FLOAT,
  churned BOOLEAN
);

INSERT INTO customer_data VALUES
  (1, 25, 50.5, FALSE),
  (2, 40, 80.0, TRUE),
  (3, 30, 65.0, FALSE);

-- Create model_params variable with JSON string
-- Your code here

Hint

Use SET to create a session variable with a JSON string.

3

Train the logistic regression model

Use CREATE OR REPLACE MODEL churn_model to train a model on customer_data. Use churned as the label column. Use age and monthly_spend as input features. Include the model parameters from model_params in the WITH clause.

Snowflake

CREATE OR REPLACE TABLE customer_data (
  customer_id INT,
  age INT,
  monthly_spend FLOAT,
  churned BOOLEAN
);

INSERT INTO customer_data VALUES
  (1, 25, 50.5, FALSE),
  (2, 40, 80.0, TRUE),
  (3, 30, 65.0, FALSE);

SET model_params = '{"model_type": "logistic_regression", "max_iterations": 100, "regularization": 0.1}';

-- Create and train churn_model using customer_data
-- Your code here

Hint

Use CREATE OR REPLACE MODEL with OPTIONS and a SELECT query for training.

4

Finalize the model by granting usage rights

Grant USAGE privilege on the model churn_model to the role DATA_SCIENTIST_ROLE so others can use it for predictions.

Snowflake

CREATE OR REPLACE TABLE customer_data (
  customer_id INT,
  age INT,
  monthly_spend FLOAT,
  churned BOOLEAN
);

INSERT INTO customer_data VALUES
  (1, 25, 50.5, FALSE),
  (2, 40, 80.0, TRUE),
  (3, 30, 65.0, FALSE);

SET model_params = '{"model_type": "logistic_regression", "max_iterations": 100, "regularization": 0.1}';

CREATE OR REPLACE MODEL churn_model
  OPTIONS (
    model_type = 'logistic_regression',
    max_iterations = 100,
    regularization = 0.1
  ) AS
SELECT age, monthly_spend, churned AS label
FROM customer_data;

-- Grant USAGE privilege on churn_model to DATA_SCIENTIST_ROLE
-- Your code here

Hint

Use GRANT USAGE ON MODEL to allow the role to use the model.

Practice

1. What is the main purpose of the CREATE MODEL statement in Snowflake's ML feature?

easy

A. To delete an existing machine learning model

B. To export model results to an external file

C. To make predictions using a trained model

D. To define and train a machine learning model using SQL

Solution

1. Step 1: Understand the role of CREATE MODEL

   This statement is used to define the model type, label column, and features, which means it sets up and trains the model.

2. Step 2: Differentiate from other commands

   PREDICT is used for predictions, not CREATE MODEL. Deleting or exporting models uses other commands.

3. Final Answer:

   To define and train a machine learning model using SQL -> Option D

4. Quick Check:

   CREATE MODEL = train model [OK]

Hint: Remember: CREATE MODEL means build and train [OK]

Common Mistakes:

• Confusing CREATE MODEL with PREDICT
• Thinking CREATE MODEL deletes models
• Assuming CREATE MODEL exports data

2. Which of the following is the correct syntax to create a logistic regression model named my_model in Snowflake?

easy

A. CREATE MODEL my_model AS logistic_regression WITH LABEL target AND FEATURES age, income;

B. CREATE MODEL my_model TYPE = 'logistic_regression' LABEL = target FEATURES = (age, income);

C. CREATE MODEL my_model USING logistic_regression LABEL target FEATURES (age, income);

D. CREATE MODEL my_model TYPE logistic_regression LABEL target FEATURES (age, income);

Solution

1. Step 1: Recall Snowflake's CREATE MODEL syntax

   The correct syntax uses TYPE = 'model_type', LABEL = label_column, and FEATURES = (feature_list) with equal signs and parentheses.

2. Step 2: Compare options

   CREATE MODEL my_model TYPE = 'logistic_regression' LABEL = target FEATURES = (age, income); matches the correct syntax exactly. Others have wrong keywords or missing equal signs or parentheses.

3. Final Answer:

   CREATE MODEL my_model TYPE = 'logistic_regression' LABEL = target FEATURES = (age, income); -> Option B

4. Quick Check:

   Correct syntax uses TYPE =, LABEL =, FEATURES = ( ) [OK]

Hint: Look for equal signs and parentheses in syntax [OK]

Common Mistakes:

• Omitting equal signs (=) in parameters
• Missing parentheses around features
• Using wrong keywords like AS or WITH

3. Given the following Snowflake SQL code:

CREATE MODEL churn_model TYPE = 'random_forest' LABEL = churn FEATURES = (age, balance, transactions);
SELECT * FROM PREDICT(churn_model, (SELECT age, balance, transactions FROM customers_test));

What will be the output of the PREDICT statement?

medium

A. The original customers_test table without predictions

B. An error because PREDICT cannot use SELECT statements

C. A table with predicted churn values for each row in customers_test

D. A summary report of model accuracy

Solution

1. Step 1: Understand the PREDICT statement

   PREDICT applies the trained model to new data and returns predicted values for each input row.

2. Step 2: Analyze the code

   The SELECT statement provides features from customers_test. PREDICT will output predictions for each row based on these features.

3. Final Answer:

   A table with predicted churn values for each row in customers_test -> Option C

4. Quick Check:

   PREDICT outputs predictions per input row [OK]

Hint: PREDICT returns predictions for each input row [OK]

Common Mistakes:

• Thinking PREDICT returns original data only
• Assuming PREDICT gives accuracy metrics
• Believing PREDICT cannot use SELECT

4. You run this Snowflake command:

CREATE MODEL sales_model TYPE = 'linear_regression' LABEL = sales_amount FEATURES = (store_size, location);

But get an error saying "Invalid feature column: location". What is the most likely cause?

medium

A. The column 'location' is not numeric or not suitable as a feature

B. The LABEL column 'sales_amount' does not exist

C. The model type 'linear_regression' is not supported

D. The FEATURES list must be empty for linear regression

Solution

1. Step 1: Understand feature requirements

   Features must be numeric or convertible to numeric for linear regression models.

2. Step 2: Analyze the error message

   "Invalid feature column" suggests 'location' is likely categorical or text, which linear regression cannot use directly.

3. Final Answer:

   The column 'location' is not numeric or not suitable as a feature -> Option A

4. Quick Check:

   Features must be numeric for linear regression [OK]

Hint: Check feature columns are numeric for regression models [OK]

Common Mistakes:

• Assuming LABEL column causes this error
• Thinking model type is unsupported
• Believing features list can be empty

5. You want to train a classification model in Snowflake to predict customer churn. You have a dataset with columns: customer_id, age, monthly_spend, churned (1 if churned, 0 otherwise), and signup_date. Which CREATE MODEL statement correctly trains a model ignoring customer_id and signup_date?

hard

A. CREATE MODEL churn_model TYPE = 'logistic_regression' LABEL = churned FEATURES = (age, monthly_spend);

B. CREATE MODEL churn_model TYPE = 'logistic_regression' LABEL = customer_id FEATURES = (age, monthly_spend, churned);

C. CREATE MODEL churn_model TYPE = 'logistic_regression' LABEL = churned FEATURES = (customer_id, age, monthly_spend, signup_date);

D. CREATE MODEL churn_model TYPE = 'linear_regression' LABEL = churned FEATURES = (age, monthly_spend);

Solution

1. Step 1: Identify correct label and features

   The label is churned because it indicates churn. Features should exclude customer_id and signup_date as they are identifiers or dates not useful for prediction.

2. Step 2: Choose appropriate model type

   Since churn is a binary outcome, logistic regression is suitable, not linear regression.

3. Final Answer:

   CREATE MODEL churn_model TYPE = 'logistic_regression' LABEL = churned FEATURES = (age, monthly_spend); -> Option A

4. Quick Check:

   Label = churned, features exclude IDs/dates, logistic regression for classification [OK]

Hint: Use label as target, exclude IDs/dates from features [OK]

Common Mistakes:

• Using ID or date columns as features
• Choosing wrong label column
• Using linear regression for classification