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Snowflake SnowPro Advanced: Data Analyst Certification Exam Sample Questions (Q152-Q157):

NEW QUESTION # 152
A Snowflake table 'transactions' stores data about financial transactions. The table includes the following columns: 'transaction_id' (INTEGER), 'account_id' (INTEGER), 'transaction_date' (DATE), and 'transaction_amount' (NUMBER). You need to analyze the moving average of transaction amounts for each account over a 7-day window. The moving average should be calculated for each transaction date, considering the 3 preceding days, the current day, and the 3 following days. You want to show the 'account_id' , 'transaction_date', 'transaction amount', and the calculated 'moving_average". What's the most appropriate and efficient Snowflake query to perform this calculation?

A. Option E
B. Option D
C. Option B
D. Option C
E. Option A

Answer: C

Explanation:
Option B is the correct answer. The 'PARTITION BY account_id' clause ensures that the moving average is calculated separately for each account. The 'ORDER BY transaction_date ASCS clause orders the transactions within each account by date. The 'ROWS BETWEEN 3 PRECEDING AND 3 FOLLOWING' clause defines the window frame as the current row, the 3 preceding rows, and the 3 following rows based on row number (not date ranges). Option A doesn't partition by account_id, so the moving average will be across all accounts which is not what we want. Option C uses 'RANGE instead of which relies on finding all entries within that date range, that may have many entries making the numbers incorrect. Option D only considers preceding rows, not following ones. Option E doesn't have any windowing and therefore the data is calculated to the current point, without consideration for following or previous days.

NEW QUESTION # 153
A data analyst needs to process a large JSON payload stored in a VARIANT column named 'payload' in a table called 'raw events' The payload contains an array of user sessions, each with potentially different attributes. Each session object in the array has a 'sessionld' , 'userld' , and an array of 'eventS. The events array contains objects with 'eventType' and 'timestamp'. The analyst wants to use a table function to flatten this nested structure into a relational format for easier analysis. Which approach is most efficient and correct for extracting and transforming this data?

A. Use LATERAL FLATTEN with multiple levels of nesting, specifying 'path' for each level and directly selecting the desired attributes.
B. Employ a combination of LATERAL FLATTEN and Snowpark DataFrames, using LATERAL FLATTEN to partially flatten the JSON and then Snowpark to handle the remaining complex transformations and data type handling.
C. Load the JSON data into a temporary table, then write a series of complex SQL queries with JOINs and UNNEST operations to flatten the data.
D. Utilize a Snowpark DataFrame transformation with multiple 'explode' operations and schema inference to flatten the nested structure and load data into a new table.
E. Create a recursive UDF (User-Defined Function) in Python to traverse the nested JSON and return a structured result, then call this UDF in a SELECT statement.

Answer: A

Explanation:
Option A is the most efficient and Snowflake-native approach. LATERAL FLATTEN is optimized for handling nested data structures within Snowflake. While other options might work, they introduce overhead (UDF execution), are less efficient (temporary tables and complex SQL), or rely on external frameworks (Snowpark), making them less suitable for this scenario. Specifying the path ensures specific fields are targeted, avoiding unnecessary processing of irrelevant data. LATERAL flatten allows you to join the output of a table function with each row of the input table. This is essential to maintain the context (e.g., userId) from the outer table.

NEW QUESTION # 154
A financial institution uses Snowflake to store customer transaction data'. They need to create a dashboard that visualizes daily transaction volume and average transaction amount for fraud detection purposes. This dashboard needs to be automatically updated every hour. The current dashboard query performance is slow, especially during peak hours. Given that the 'TRANSACTIONS table contains billions of rows, which of the following strategies would BEST optimize both the query performance and the automated update process?

A. Implement caching within the dashboard application to store the query results and only refresh the data once a day to avoid performance issues
B. Create a regular view that calculates daily transaction volume and average transaction amount. Use a Snowflake stored procedure to execute the query and update a separate reporting table hourly.
C. Create a materialized view that pre-aggregates the daily transaction volume and average transaction amount. Schedule a Snowflake task to refresh the materialized view hourly.
D. Increase the warehouse size to X-Large and rely on Snowflake's query optimization engine without any changes to the data model or update process.
E. Create a temporary table that stores daily transaction summaries. Truncate and reload the temporary table hourly using a scheduled Snowflake task.

Answer: C

Explanation:
Materialized views are designed for pre-computation of aggregations, providing significant performance improvements. Scheduling a task to refresh the materialized view ensures automatic updates. Regular views are calculated at query time and would not improve performance. Increasing warehouse size (C) might help, but it's not the most efficient solution. Temporary tables are not persistent and truncation/reload is inefficient. Dashboard caching (E) does not solve underlying Snowflake performance issues.

NEW QUESTION # 155
Why is the Parquet format preferred for complex data sets?

A. It supports efficient compression and encoding schemes
B. It has visually appealing data presentation
C. It randomly alters data for testing
D. It changes data colors for differentiation

Answer: A

NEW QUESTION # 156
You have identified inconsistencies in the data type of the 'ORDER DATE' column across several tables within your Snowflake database. Some tables store it as DATE, while others store it as VARCHAR. You need to create a unified view that presents 'ORDER DATE' consistently as DATE, handling potential conversion errors gracefully. You have to use safe aggregate operations. Which of the following approaches provides the most robust and error-tolerant solution?

Answer: A

Explanation:
The most robust and error-tolerant solution is option E, using 'TRY CAST(ORDER DATE AS DATE)'. This function attempts to convert the 'ORDER_DATE' to a DATE data type, and if the conversion fails (e.g., the VARCHAR value cannot be parsed as a date), it returns NULL without raising an error. This ensures that the view creation and queries against it will not fail due to data type conversion issues. Option A, , is an older function, and ' TRY_CAST' is the preferred, more general function. Option B, 'IS_DATE is not a valid snowflake Function. Option C, ' TO DATE will fail the query if the data in column is not a valid date. Option D 'SAFE_CAST is not a valid snowflake function.

NEW QUESTION # 157
......

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