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Snowflake SnowPro Advanced: Data Scientist Certification Exam Sample Questions (Q216-Q221):

NEW QUESTION # 216
You are building a fraud detection model for an e-commerce platform. One of the features is 'purchase_amount', which ranges from $1 to $10,000. The data has a skewed distribution with many small purchases and a few very large ones. You need to normalize this feature for your model, which uses gradient descent. Which normalization technique(s) would be most suitable in Snowflake, considering the data characteristics and the need to handle potential future outliers?

A. Robust scaling using interquartile range (IQR) in a stored procedure with Python:
B. Unit Vector normalization (L2 Normalization) using SQL:
C. Min-Max scaling using the following SQL:
D. Z-score standardization using the following SQL:
E. Power Transformer (e.g., Yeo-Johnson) implemented with Snowpark Python:

Answer: A,E

Explanation:
Options C and D are the most suitable. Robust scaling (C) is effective because it uses the IQR, making it less sensitive to outliers compared to Min-Max scaling (A) or Z-score standardization (B). The Snowflake UDF handles potential outliers by not being dramatically influenced by them. Power Transformer (D) addresses the skewness of the data, also mitigating the impact of outliers. Min-Max scaling (A) is highly sensitive to outliers, making it a poor choice. Z-score standardization (B) can be affected by extreme values in skewed distributions. Unit Vector normalization (E) changes the meaning of the purchase amounts by making the total magnitude 1 , which isn't desirable here.

NEW QUESTION # 217
A data scientist is using association rule mining with the Apriori algorithm on customer purchase data in Snowflake to identify product bundles. After generating the rules, they obtain the following metrics for a specific rule: Support = 0.05, Confidence = 0.7, Lift = 1.2. Consider that the overall purchase probability of the consequent (right-hand side) of the rule is 0.4. Which of the following statements are CORRECT interpretations of these metrics in the context of business recommendations for product bundling?

A. The lift value of 1.2 suggests a strong negative correlation between the antecedent and consequent, indicating that purchasing the antecedent items decreases the likelihood of purchasing the consequent items.
B. The lift value of 1.2 indicates that customers are 20% more likely to purchase the consequent items when they have also purchased the antecedent items, compared to the baseline purchase probability of the consequent items.
C. The confidence of 0.7 indicates that 70% of transactions containing the antecedent also contain the consequent.
D. Customers who purchase the items in the antecedent are 70% more likely to also purchase the items in the consequent, compared to the overall purchase probability of the consequent.
E. The rule applies to 5% of all transactions in the dataset, meaning 5% of the transactions contain both the antecedent and the consequent.

Answer: B,C,E

Explanation:
Option A is correct because support represents the proportion of transactions that contain both the antecedent and the consequent. Option D is correct because confidence represents the proportion of transactions containing the antecedent that also contain the consequent. Option E is correct because lift = confidence / (probability of consequent). Therefore, lift of 1.2 means confidence is 1.2 times the probability of the consequent. Hence 20% more likely than the baseline. Option B is incorrect because lift, not confidence, captures the relative likelihood compared to the baseline. Option C is incorrect because a lift > 1 suggests a positive correlation, not a negative one.

NEW QUESTION # 218
You are training a regression model to predict house prices using a Snowflake dataset. The dataset contains various features, including 'number of_bedrooms', , and You want to use time-based partitioning for your training, validation, and holdout sets. However, you also need to ensure that the dataset is properly shuffled within each time partition to mitigate potential bias introduced by the order of data entry. Which of the following strategies is MOST EFFECTIVE and EFFICIENT for partitioning your data into train, validation, and holdout sets in Snowflake, while also ensuring random shuffling within each partition, and addressing potential data leakage issues?

A. Use Snowflake's SAMPLE clause with a 'REPEATABLE seed for each split (train, validation, holdout), filtering by 'sale_date'. Add an 'ORDER BY RANDOM()' clause within each 'SAMPLE query to shuffle the data within each split. This approach does not guarantee non-overlapping sets and can introduce sampling bias.
B. Create a new column 'split_group' using a CASE statement based on 'sale_date' to assign each row to 'train', 'validation', or 'holdout'. Calculate a random number within each 'split_group' by using OVER (PARTITION BY split_group ORDER BY RANDOM())'. Then create temporary tables for each split using 'CREATE TABLE AS SELECT FROM WHERE split_group = QUALIFY ROW NUMBER() OVER (ORDER BY RANDOM()) (SELECT COUNT( ) FROM transactions WHERE split_group -- ...) (respective split percentage);'
C. Create separate views for train, validation, and holdout sets, filtering by 'sale_date' . Shuffle the entire dataset using 'ORDER BY RANDOM()' before creating the views to ensure randomness across all sets. This does not address shuffling within parition.
D. Create a user-defined function (UDF) in Python that takes a 'sale_date' as input and returns either 'train', 'validation', or 'holdout' based on pre-defined date ranges. Apply this UDF to each row, creating a 'split_group' column. Then, create temporary tables for each split using 'CREATE TABLE AS SELECT ... FROM . WHERE split_group = ... ORDER BY RANDOM()'. UDF overhead and global RANDOM sort make it very slow.
E. Create a new column 'split_group' using a CASE statement based on 'sale_date' to assign each row to 'train', 'validation', or 'holdout'. Then, create temporary tables for each split using 'CREATE TABLE AS SELECT FROM WHERE split_group = ORDER BY RANDOM()'. This can be very slow because of global RANDOM sort and leakage issues with using full dataset for randomness.

Answer: B

Explanation:
Option E is the most effective and efficient because it correctly implements the required partitioning and shuffling while minimizing data leakage and maximizing performance. Here's a breakdown: Time-Based Partitioning: The CASE statement accurately divides the data into train, validation, and holdout sets based on 'sale_date' . Random Shuffling Within Partitions: 'ROW NUMBER() OVER (PARTITION BY split_group ORDER BY RANDOM())' calculates a random row number within each split group (train, validation, holdout). This ensures that the data is shuffled within each time-based partition, mitigating bias introduced by the order of data entry, without introducing data leakage. Prevents Data Leakage: Shuffling the data within each partition prevents data leakage that could occur if you shuffle the entire dataset before partitioning. Efficiency: Avoids expensive operations like UDFs or sorting the entire dataset.lJses window functions efficiently to calculate random row numbers within partitions. Option A is not suitable since It does not address shuffling within parition and the shuffle will be affected by other filtering operations later.Option B is not suitable because RANDOM does not work inside create table and if it did it will cause data leakage, because all splits influence the randomness. Option C is not ideal because SAMPLE does not guarantee non-overlapping data, which would undermine the integrity of train/validation/holdout sets, moreover 'order by random()' will only apply the sampling on a sorted result not generate a random sampling. Option D is not suitable because it uses UDFs. UDFs in Snowflake generally have performance overhead compared to native SQL functions. Also using a global 'ORDER BY can be very slow on large datasets and will also introduce data leakage.

NEW QUESTION # 219
You are using Snowflake Cortex to perform sentiment analysis on customer reviews stored in a table called 'CUSTOMER REVIEWS' The table has a column containing the text of each review. You want to create a user-defined function (UDF) to extract sentiment score between the range of -1 to 1 using the 'snowflake_cortex.sentiment' function in Snowflake Cortex. Which of the following UDF definitions would correctly implement this, allowing it to be called directly on the column?

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

Answer: A

Explanation:
The 'snowflake_cortex.sentiment' function returns a VARIANT containing the sentiment score and sentiment label. To extract the sentiment score as a float, you need to access the 'sentiment_score' field and cast it to FLOAT or NUMBER data type using Option D does this correctly, using the Snowflake's preferred casting syntax.Option A return type is incorrect, where it returns the Variant instead of FLOAT. Option B return type is correct but it doesnt cast the result to Float, which is not correct syntax as result is VARIANT. Option C is incorrect because of TO_NUMBER function. Option E result is the SENTIMENT Label instead sentiment score.

NEW QUESTION # 220
You are building a model to predict loan defaults using data stored in Snowflake. As part of your feature engineering process within a Snowflake Notebook, you need to handle missing values in several columns: 'annual _ income', and You want to use a combination of imputation strategies: replace missing values with the median, 'annual_income' with the mean, and with a constant value of 0.5. You are leveraging the Snowpark DataFrame API. Which of the following code snippets correctly implements this imputation strategy?

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

Answer: A,C

Explanation:
Options A and D both correctly implement the specified imputation strategy. Option A uses 'fillna' method with respective median and mean values, calculated using 'approxQuantile' and mean for missing values.Option B uses 'na.fill' which is used in Spark, and Snowflake is not compatible. Option C calculates the median and mean, but incorrectly tries to use the local Python variables inside F.lit() functions, which are executed on the Snowflake server. Option D uses loops for column selection. Option E tries to apply a literal value within a dictionary being used to fill the missing values. This is not correct, and it's important to ensure that a correct implementation is used.

NEW QUESTION # 221
......

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