JobMonitor.register_model()

Registers a trained model in the snowflake model registry. The model is registered under the specified name and version in the specified database and schema that the native app has permissions to.

This method is available only for jobs of type train or train_inference and requires a valid model_run_id. Duplicate version names for a model are not allowed.

⚠️ Note: Ensure the native app has the necessary permissions to the specified database and schema.

-- grant access to resources needed for snowflake experiment tracking

GRANT USAGE ON DATABASE <DATABASE> TO APPLICATION RELATIONALAI;
GRANT USAGE ON SCHEMA <DATABASE>.<SCHEMA> TO APPLICATION RELATIONALAI;
GRANT CREATE MODEL ON SCHEMA <DATABASE>.<SCHEMA> TO APPLICATION RELATIONALAI;

Parameters

Name	Type	Description	Optional
model_name	str	Name under which to register the model.	No
version_name	str	Version under which to register the model.	No
database_name	str	Database where the model will be registered.	No
schema_name	str	Schema name in the database where the model will be registered.	No
comment	str	Optional comment for model registration.	Yes

Example

train_job.register_model(
    model_name="new_model",
    version_name="v3",
    database_name="database_name",
    schema_name="schema_name",
    comment="Production-ready model",
)