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| Section | Weight | Objectives |
|---|---|---|
| Topic 1: Performance Optimization and Best Practices | 20% | - Query pushdown and optimization - Caching strategies - Debugging and explain plans - Warehouse sizing for Snowpark - Vectorized UDFs - Minimizing data transfer |
| Topic 2: Snowpark API for Python | 30% | - Reading and writing data - DataFrame creation and manipulation - Working with Semi-structured data - Establishing connections and session management - User-Defined Functions (UDFs) and Stored Procedures |
| Topic 3: Data Transformations and DataFrame Operations | 35% | - Persisting transformed data - Using built-in functions - Filtering, Aggregating, and Joining DataFrames - Window functions - Complex data pipelines |
| Topic 4: Snowpark Concepts | 15% | - Snowpark architecture and core concepts - Snowpark DataFrames and query plans - Snowpark Sessions and connection management - Stored procedures and conditional logic - Client-side vs. Server-side execution - Transformations vs. Actions |
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NEW QUESTION # 371
You are working with a Snowpark DataFrame 'products df' containing product information, including 'product_id', 'price', and 'discount'. You need to update the 'price' column in the 'products' table based on the following logic: If 'discount' is greater than 0.2, reduce the 'price' by 15%. If 'discount' is between 0.1 and 0.2 (inclusive), reduce the 'price' by 5%. Otherwise, keep the 'price' as is. Which of the following Snowpark code snippets efficiently implements this update? Assume 'products' table already exists and is correctly populated.
Answer: E
Explanation:
Option E is the most concise and correct solution. It uses 'with_column' to directly update the 'price' column based on the discount conditions, using nested 'when' functions for the logic and persists the change. Option A, while technically correct, is less efficient because it creates a new column ('new_price'), drops the original 'price' column, and then renames the new column. Option B tries to use an 'update' method which doesn't exist directly on Snowpark DataFrames in that way. Option C works correctly. Option D has an issue that it won't keep the original schema of the table being updated as it is selecting each of the columns.
NEW QUESTION # 372
You are building a Snowpark application that requires you to connect to Snowflake from an environment where directly specifying credentials in the code is not permitted for security reasons. Which of the following are valid and recommended ways to securely pass authentication information to the Snowpark Session?
Answer: A,B,C
Explanation:
Options A, C, and D represent valid ways to handle credentials securely. Environment variables (A) are a standard practice for configurations. Using a secret management service (C) provides the best security posture for production environments. Using the Snowflake CLI (D) is acceptable for development. Storing credentials in a Snowflake stage (B) adds unnecessary complexity and doesn't inherently improve security over other options. Base64 encoding (E) is not a secure method; it's easily decoded and provides a false sense of security. Hardcoding and obfuscating credentials is not recommended.
NEW QUESTION # 373
You have a Snowpark DataFrame 'sales df with columns 'product_id' (INTEGER), 'sale date' (DATE), and 'sale_amount' (DOUBLE). You need to filter the DataFrame to include only sales that occurred in the year 2023 and where the 'sale_amount' is greater than the average 'sale_amount' for that specific 'product id'. Which of the following Snowpark Python code snippets will correctly achieve this?
Answer: A
Explanation:
Option C is the most efficient and correct. It calculates the average sale amount per product ID using a window function and then filters the DataFrame based on both the year and the comparison with the average. It avoids joins and 'collect()' which can be inefficient. Option A doesn't import Window or properly use avg over a window, causing an error. Option B requires a join, which is less efficient. Option D uses "mean' which is available but 'avg' is preferrable. Option E 'collects data to the client which is also generally avoided in Snowpark.
NEW QUESTION # 374
You are using Snowpark Python to process a large dataset. You need to persist a DataFrame to a Snowflake table but want to ensure the operation is as efficient as possible and minimizes the data transfer overhead. The table already exists with the appropriate schema. Which of the following strategies would be the MOST efficient way to write the DataFrame to the existing table?
Answer: E
Explanation:
is generally the most efficient method for appending data to an existing Snowflake table using Snowpark. It directly inserts the data into the table without the overhead of creating a new table or overwriting the existing one. with the default 'append' mode (A) might work, but 'insertlnto' is more explicit and potentially optimized for this specific scenario. (B) would replace the entire table, which is not efficient if you only want to add new data. Creating a temporary table and then using 'CREATE OR REPLACE TABLE AS SELECT (D) involves unnecessary steps and data transfer. Writing to a stage and then using 'COPY INTO' (E) is also less efficient than directly inserting the data using Snowpark.
NEW QUESTION # 375
You have a Snowpark DataFrame named 'products_df' with columns 'product_id' (INT), 'product_name' (VARCHAR), and 'price' (FLOAT). You want to create a new DataFrame called 'discounted_products df that includes all columns from 'products_df' plus a new column named 'discounted_price', which is calculated as the original price minus a discount percentage specified by the variable 'discount_rate' (e.g., 0.1 for 10%). The 'discount_rate' is stored in the database table named 'discount_table'. You want to load the rate to variable. Choose the correct ways to achieve this. (Select all that apply)
Answer: A,B,D
Explanation:
Options A, B and D provide valid ways to fetch 'discount_rate' as a single numerical value. And fetch the data and gets the first value from the first row. Similarly, gets the data and return the first row. However, Option C does not have LIMIT 1 and will not work. Option E fetches one row as one array, thus requires rate[0] to compute discounted_price.
NEW QUESTION # 376
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