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Snowflake SnowPro Advanced: Data Scientist Certification Exam 認定 DSA-C03 試験問題 (Q231-Q236):

質問 # 231
A marketing analyst is building a propensity model to predict customer response to a new product launch. The dataset contains a 'City' column with a large number of unique city names. Applying one-hot encoding to this feature would result in a very high-dimensional dataset, potentially leading to the curse of dimensionality. To mitigate this, the analyst decides to combine Label Encoding followed by binarization techniques. Which of the following statements are TRUE regarding the benefits and challenges of this combined approach in Snowflake compared to simply label encoding?

A. Label encoding introduces an arbitrary ordinal relationship between the cities, which may not be appropriate. Binarization alone cannot remove this artifact.
B. Binarizing a label encoded column using a simple threshold (e.g., creating a 'high_city_id' flag) addresses the curse of dimensionality by reducing the number of features to one, but it loses significant information about the individual cities.
C. While label encoding itself adds an ordinal relationship, applying binarization techniques like binary encoding (converting the label to binary representation and splitting into multiple columns) after label encoding will remove the arbitrary ordinal relationship.
D. Binarization following label encoding may enhance model performance if a specific split based on a defined threshold is meaningful for the target variable (e.g., distinguishing between cities above/below a certain average income level related to marketing success).
E. Label encoding followed by binarization will reduce the memory required to store the 'City' feature compared to one-hot encoding, and Snowflake's columnar storage optimizes storage for integer data types used in label encoding.

正解：A、B、D、E

解説：
Option A is true because label encoding converts strings into integers, which are more memory-efficient than storing numerous one-hot encoded columns. Snowflake's columnar storage further optimizes integer storage. Option B is also true; label encoding inherently creates an ordinal relationship that might not be valid for nominal features like city names. Option C is incorrect; simple binarization (e.g., > threshold) of label encoded data doesn't remove the arbitrary ordinal relationship; more complex binarization techniques would be needed. Option D is accurate; binarization reduces dimensionality but sacrifices granularity, leading to information loss. Option E is correct because carefully chosen thresholds might correlate with the target variable and improve predictive power.

質問 # 232
You are performing exploratory data analysis on a large sales dataset in Snowflake using Snowpark. The dataset contains columns such as 'order_id', , and 'profit'. You want to identify the top 5 most profitable products for each month. You have already created a Snowpark DataFrame named 'sales_df. Which of the following Snowpark operations, when combined correctly, will efficiently achieve this?

A. Use 'rank()' partitioned by ordered by 'sum(profit) DESC' , after grouping by and 'product_id' , and aggregating 'sum(profity.
B. Use 'ntile(5)' partitioned by ordered by 'sum(profit) DESC' after grouping by and 'product_id', and aggregating 'sum(profit)'.
C. First, create a temporary table with aggregated monthly profit for each product using SQL. Then, use Snowpark to read the temporary table and apply a window function partitioned by ordered by 'sum(profit) DESC'.
D. Group by and 'product_id' , aggregate 'sum(profit)' , then use partitioned by ordered by 'sum(profit) DESC'.
E. Group by 'product_id', aggregate 'sum(profity, then use partitioned by ordered by 'sum(profit) DESC' within a UDF.

正解：D

解説：
Option A correctly describes the process. First group by month and product to calculate total profit, then use with correct partitioning and ordering to assign a rank within each month based on profit. Options B and C use less efficient ranking functions. Option D groups by product globally, missing the monthly granularity. Option E 'ntile' divides products into 5 buckets which is not what we are looking for.

質問 # 233
A data scientist is analyzing website click-through rates (CTR) for two different ad campaigns. Campaign A ran for two weeks and had 10,000 impressions with 500 clicks. Campaign B also ran for two weeks with 12,000 impressions and 660 clicks. The data scientist wants to determine if there's a statistically significant difference in CTR between the two campaigns. Assume the population standard deviation is unknown and unequal for the two campaigns. Which statistical test is most appropriate to use, and what Snowflake SQL code would be used to approximate the p-value for this test (assume 'clicks_b' , and are already defined Snowflake variables)?

A. A one-sample t-test, because we are comparing the sample mean of campaign A to the sample mean of campaign Snowflake code: 'SELECT t_test_lsamp(clicks_a/impressions_a - clicks_b/impressions_b, 0)'
B. Az-test, because we know the population standard deviation. Snowflake code: 'SELECT normcdf(clicks_a/impressions_a - clicks_b/impressions_b, O, 1)'
C. An independent samples t-test, because we are comparing the means of two independent samples. Snowflake code: SELECT
D. A paired t-test, because we are comparing two related samples over time. Snowflake code: 'SELECT t_test_ind(clicks_a/impressions_a, 'VAR EQUAL-TRUE')
E. An independent samples t-test (Welch's t-test), because we are comparing the means of two independent samples with unequal variances. Snowflake code (approximation using UDF - assuming UDF 'p_value_from_t_stat' exists that calculates p-value from t-statistic and degrees of freedom):

正解：C

解説：
The correct answer is E. Since we're comparing the means of two independent samples (Campaign A and Campaign B) and the population standard deviations are unknown, an independent samples t-test is appropriate. Because the problem stated that the variances are unequal, Welch's t-test provides a more accurate p-value and confidence intervals. The Snowflake function handles independent samples and the 'VAR_EQUAL=FALSE' parameter specifies that the variances should not be assumed to be equal. The other options are incorrect because they use inappropriate tests given the problem conditions. The z-test is not appropriate because the population standard deviations are unknown. A paired t-test is for related samples. A one sample test is to test one mean against a constant not another mean.

質問 # 234
You are working with a large dataset of sensor readings stored in a Snowflake table. You need to perform several complex feature engineering steps, including calculating rolling statistics (e.g., moving average) over a time window for each sensor. You want to use Snowpark Pandas for this task. However, the dataset is too large to fit into the memory of a single Snowpark Pandas worker. How can you efficiently perform the rolling statistics calculation without exceeding memory limits? Select all options that apply.

A. Explore using Snowpark's Pandas user-defined functions (UDFs) with vectorization to apply custom rolling statistics logic directly within Snowflake. UDFs allow you to use Pandas within Snowflake without needing to bring the entire dataset client-side.
B. Utilize the 'window' function in Snowpark SQL to define a window specification for each sensor and calculate the rolling statistics using SQL aggregate functions within Snowflake. Leverage Snowpark to consume the results of the SQL transformation.
C. Increase the memory allocation for the Snowpark Pandas worker nodes to accommodate the entire dataset.
D. Use the 'grouped' method in Snowpark DataFrame to group the data by sensor ID, then download each group as a Pandas DataFrame to the client and perform the rolling statistics calculation locally. Then upload back to Snowflake.
E. Break the Snowpark DataFrame into smaller chunks using 'sample' and 'unionAll', process each chunk with Snowpark Pandas, and then combine the results.

正解：A、B

解説：
Explanation:Options B and D are the most appropriate and efficient solutions for handling large datasets when calculating rolling statistics with Snowpark Pandas. Option B uses the 'window' function in Snowpark SQL. Leverage the 'window' function in Snowpark SQL to define a window specification for each sensor and calculate the rolling statistics using SQL aggregate functions within Snowflake. Option D uses Snowpark's Pandas UDFs. Snowpark's Pandas UDFs with vectorization allow you to bring the processing logic to the data within Snowflake, avoiding the need to move the entire dataset to the client-side and bypassing memory limitations. This approach is generally more scalable and performant for large datasets. Option A is inefficient as it retrieves groups of data from Snowflake to client side before creating the calculations before sending back to snowflake. Option C is correct but complex and not optimal. Option E is possible, but it's not a scalable solution and can be costly.

質問 # 235
You've built a complex machine learning model using scikit-learn and deployed it as a Python UDF in Snowflake. The UDF takes a JSON string as input, containing several numerical features, and returns a predicted probability However, you observe significant performance issues, particularly when processing large batches of data'. Which of the following approaches would be MOST effective in optimizing the performance of this UDF in Snowflake?

A. Rewrite the UDF in Java or Scala to leverage the JVM's performance advantages over Python in Snowflake.
B. Serialize the scikit-learn model using 'joblib' instead of 'pickle' for potentially faster deserialization within the UDF.
C. Increase the warehouse size to improve the overall compute resources available for UDF execution.
D. Use Snowflake's vectorized UDF feature to process data in micro-batches, minimizing the overhead of repeated Python interpreter initialization.
E. Pre-process the input data outside of the UDF using SQL transformations, reducing the amount of data passed to the UDF and simplifying the Python code.

正解：D、E

解説：
Vectorized UDFs (A) are designed specifically for performance optimization by processing data in batches, reducing overhead. Pre- processing data using SQL (D) can significantly reduce the complexity and data volume handled by the UDF. While 'joblib' (B) might offer a slight improvement, it's less impactful than vectorization. Increasing warehouse size (C) helps but doesn't address the underlying inefficiencies of repeated interpreter initialization. Rewriting in Java/Scala (E) is a viable option but requires significant effort and might not be necessary if vectorization and pre-processing are sufficient.

質問 # 236
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DSA-C03合格記 & DSA-C03合格率

正確的-一番優秀なDSA-C03合格記試験-試験の準備方法DSA-C03合格率

Snowflake SnowPro Advanced: Data Scientist Certification Exam 認定 DSA-C03 試験問題 (Q231-Q236):

試験の準備方法-一番優秀なDSA-C03合格記試験-検証するDSA-C03合格率

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