SPS-C01 Fragenkatalog - SPS-C01 Tests

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Snowflake Certified SnowPro Specialty - Snowpark SPS-C01 Prüfungsfragen mit Lösungen (Q17-Q22):

17. Frage
You are tasked with optimizing a Snowpark application that performs complex data transformations on a large dataset (1 TB) stored in Snowflake. The application currently uses Snowpark DataFrames and is experiencing slow performance. You suspect the issue might be related to data transfer overhead between the Snowflake engine and the Python environment. Which of the following strategies would be MOST effective in minimizing this overhead and improving performance?

A. Convert the Snowpark DataFrame to a Pandas DataFrame and perform the transformations locally within the Python environment.
B. Utilize User-Defined Functions (UDFs) written in Python to encapsulate the transformations and execute them within the Snowflake engine.
C. Reduce the data volume by applying aggressive filtering and aggregation using Snowpark DataFrame operations before any other transformations, minimizing the amount of data transferred.
D. Implement vectorization techniques within the Snowpark DataFrame operations using built-in functions and optimized expressions where applicable.
E. Increase the virtual warehouse size to the largest available option (e.g., X-Large) to improve processing power within Snowflake, regardless of data transfer costs.

Antwort: B,C,D

Begründung:
UDFs written in Python allow you to push the computation to the Snowflake engine, avoiding the need to transfer large amounts of data to the Python environment, improving performance by reducing data transfer overhead. Pre-filtering and aggregating reduces data before more complex transformations, also reducing data transfer. Vectorization through optimized Snowpark DataFrame methods also reduces data transfer and speeds computation in the Snowflake engine. Converting to Pandas DataFrames negates the benefit of Snowpark for large datasets. Increasing the virtual warehouse size only addresses processing power but does not minimize data transfer.

18. Frage
You have a Python function that performs complex data transformations, too intricate to express directly in Snowpark SQL. You want to register this as a User-Defined Table Function (UDTF) so that it can be used to expand rows in a Snowpark DataFrame. The UDTF takes two arguments: an ID (integer) and a string. It returns a table with three columns: (integer), (string), and 'timestamp' (timestamp). Which of the following code snippets correctly registers this UDTF, making it available for use within Snowpark?

Antwort: B

Begründung:
Option C provides the correct way to define and register UDTFs using the class-based approach in Snowpark. It defines the UDTF class with 'process' and methods. returns the schema using 'table' function correctly. Options A and B use the decorator approach, which is valid for simple UDTFs, but it's less flexible than the class-based approach, especially for managing complex state or schema. Option D uses the class-based approach but incorrectly defines the 'output_schema' when registering. Option E has an incorrect definition of return type.

19. Frage
You have a Snowpark DataFrame named 'products' with columns 'product_id' (INT), 'product_name' (STRING), and 'price' (DOUBLE). You want to apply a transformation to calculate a 'discounted_price' column, which is the 'price' reduced by 10% if the price is greater than $100.00. Which of the following code snippets is the most efficient way to achieve this using Snowpark Python?

Antwort: A,E

Begründung:
The most efficient ways are B and C. Option B directly uses the 'when' and 'otherwise' functions in Snowpark, which are optimized for execution within Snowflake. Option C is similar to B but explicitly uses 'lit' to represent the numeric literal, ensuring proper type handling in Snowpark. IJDFs (Option A) are generally less efficient than built-in functions. Option D attempts to use RDDs, which are not part of the Snowpark API. Option E is not valid Snowpark python syntax. Therefore, B and C are the correct answers.

20. Frage
You have a requirement to process a large number of JSON files stored in a Snowflake stage 'json_stage'. These JSON files contain complex nested structures. You need to extract specific fields from these files using Snowpark Python and load them into a Snowflake table. You want to use 'SnowflakeFile' to read the JSON files and minimize the amount of data loaded into memory. Select all that apply from the following options to efficiently accomplish this task:

A. Use to directly load all JSON files into a Snowpark DataFrame and then use 'select' with path expressions (e.g., 'col('fieldl .nested_field')) to extract the required fields.
B. Create a UDTF that accepts a "SnowflakeFile' object, uses 'json.loadS to parse the JSON content incrementally, extracts the desired fields, and yields rows for insertion into the target table. Use 'session.read.option('PATTERN', ' to generate the initial DataFrame.
C. Create a UDF that accepts a 'SnowflakeFile' object, opens the file, reads the JSON data using 'json.load', extracts the required fields, and returns a JSON string. Create a Snowpark DataFrame using 'session.read.option('PATTERN', ' then call the UDF with the file path, and finally parse the returned JSON string to load data.
D. Write a Python script that downloads all JSON files from the stage using 'SnowflakeFile.get' , iterates through the downloaded files, parses each file, extracts the fields, and inserts the data into the Snowflake table using the Snowflake Python connector.
E. Create a UDF that takes a file path as input, constructs a 'SnowflakeFile' object within the UDF, reads the JSON data using 'json.loadS, extracts the required fields, and returns a Row object or a dictionary. Use 'session.sql('SELECT relative_path FROM to get file paths, create Snowpark Dataframe and then call the UDF on each file path.

Antwort: B

Begründung:
Option C is the most efficient approach. - A UDTF allows for parallel processing of the JSON files within the Snowflake environment. - By directly using 'SnowflakeFile' objects, you avoid unnecessary data transfer outside of Snowflake. - The function incrementally parses the JSON data, minimizing memory usage compared to loading the entire file at once. Option A loads all JSON files into a DataFrame which can lead to memory issues with large JSON files. Option B reads all files using snowflake connector in python, so it is not optimal. Option D downloads all files which would be inefficient. Option E returns a JSON string from UDF then parses it again which is redundant and less efficient.

21. Frage
You have a Snowpark application that processes sensitive data'. To enhance security, you want to use key pair authentication and ensure that the private key is never exposed in plain text within the application or logs. Which of the following strategies offers the most robust protection against accidental key exposure, even in the event of a security breach of the application server itself?

A. Store the private key encrypted using AES encryption with a key derived from a passphrase. Prompt the user for the passphrase each time the application starts.
B. Use a dedicated Hardware Security Module (HSM) or a cloud-based key management service (e.g., AWS KMS, Azure Key Vault) to store and manage the private key. Configure the Snowpark application to retrieve the key only when needed and never store it locally.
C. Store the private key in an environment variable on the application server, ensuring that only the application user has read access.
D. Store the private key encrypted with a symmetric key. Store the symmetric key in a separate environment variable.
E. Store the private key in an encrypted file with restricted access, decrypting it only during session creation and overwriting the memory location immediately afterwards.

Antwort: B

Begründung:
Using a dedicated HSM or cloud-based key management service (C) offers the strongest protection. HSMs and KMSs are specifically designed to securely store and manage cryptographic keys, providing hardware-backed security and access controls. The private key never leaves the HSM/KMS, minimizing the risk of exposure. Option A, storing in an environment variable, provides some protection but is still vulnerable if the server is compromised. Option B simply moves the problem of key management from one key to another. The symmetric key is also vulnerable. Option D requires user interaction and the passphrase could be compromised. Option E improves on environment variables, but is still less secure than HSM and KMS. HSMs/KMSs are FIPS 140-2 Level 3 validated, which is most secure.

22. Frage
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SPS-C01 Fragenkatalog - SPS-C01 Tests

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SPS-C01 Tests, SPS-C01 Testantworten

Snowflake Certified SnowPro Specialty - Snowpark SPS-C01 Prüfungsfragen mit Lösungen (Q17-Q22):

SPS-C01 Bestehen Sie Snowflake Certified SnowPro Specialty - Snowpark! - mit höhere Effizienz und weniger Mühen

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