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Download Databricks Certified Associate Developer for Apache Spark 3.0 Exam Exam Dumps

NEW QUESTION 36
Which of the following statements about Spark's configuration properties is incorrect?

• A. The maximum number of tasks that an executor can process at the same time is controlled by the spark.executor.cores property.
• B. The default value for spark.sql.autoBroadcastJoinThreshold is 10MB.
• C. The default number of partitions to use when shuffling data for joins or aggregations is 300.
• D. The maximum number of tasks that an executor can process at the same time is controlled by the spark.task.cpus property.
• E. The default number of partitions returned from certain transformations can be controlled by the spark.default.parallelism property.

Answer: C

Explanation:
Explanation
The default number of partitions to use when shuffling data for joins or aggregations is 300.
No, the default value of the applicable property spark.sql.shuffle.partitions is 200.
The maximum number of tasks that an executor can process at the same time is controlled by the spark.executor.cores property.
Correct, see below.
The maximum number of tasks that an executor can process at the same time is controlled by the spark.task.cpus property.
Correct, the maximum number of tasks that an executor can process in parallel depends on both properties spark.task.cpus and spark.executor.cores. This is because the available number of slots is calculated by dividing the number of cores per executor by the number of cores per task. For more info specifically to this point, check out Spark Architecture | Distributed Systems Architecture.
More info: Configuration - Spark 3.1.2 Documentation

NEW QUESTION 37
Which of the following code blocks returns all unique values across all values in columns value and productId in DataFrame transactionsDf in a one-column DataFrame?

• A. tranactionsDf.select('value').join(transactionsDf.select('productId'), col('value')==col('productId'),
  'outer')
• B. transactionsDf.select('value').union(transactionsDf.select('productId')).distinct()
• C. transactionsDf.agg({'value': 'collect_set', 'productId': 'collect_set'})
• D. transactionsDf.select(col('value'), col('productId')).agg({'*': 'count'})
• E. transactionsDf.select('value', 'productId').distinct()

Answer: B

Explanation:
Explanation
transactionsDf.select('value').union(transactionsDf.select('productId')).distinct() Correct. This code block uses a common pattern for finding the unique values across multiple columns: union and distinct. In fact, it is so common that it is even mentioned in the Spark documentation for the union command (link below).
transactionsDf.select('value', 'productId').distinct()
Wrong. This code block returns unique rows, but not unique values.
transactionsDf.agg({'value': 'collect_set', 'productId': 'collect_set'}) Incorrect. This code block will output a one-row, two-column DataFrame where each cell has an array of unique values in the respective column (even omitting any nulls).
transactionsDf.select(col('value'), col('productId')).agg({'*': 'count'}) No. This command will count the number of rows, but will not return unique values.
transactionsDf.select('value').join(transactionsDf.select('productId'), col('value')==col('productId'), 'outer') Wrong. This command will perform an outer join of the value and productId columns. As such, it will return a two-column DataFrame. If you picked this answer, it might be a good idea for you to read up on the difference between union and join, a link is posted below.
More info: pyspark.sql.DataFrame.union - PySpark 3.1.2 documentation, sql - What is the difference between JOIN and UNION? - Stack Overflow Static notebook | Dynamic notebook: See test 3

NEW QUESTION 38
Which of the following code blocks shuffles DataFrame transactionsDf, which has 8 partitions, so that it has
10 partitions?

• A. transactionsDf.coalesce(10)
• B. transactionsDf.coalesce(transactionsDf.getNumPartitions()+2)
• C. transactionsDf.repartition(transactionsDf._partitions+2)
• D. transactionsDf.repartition(transactionsDf.getNumPartitions()+2)
• E. transactionsDf.repartition(transactionsDf.rdd.getNumPartitions()+2)

Answer: E

Explanation:
Explanation
transactionsDf.repartition(transactionsDf.rdd.getNumPartitions()+2)
Correct. The repartition operator is the correct one for increasing the number of partitions. calling getNumPartitions() on DataFrame.rdd returns the current number of partitions.
transactionsDf.coalesce(10)
No, after this command transactionsDf will continue to only have 8 partitions. This is because coalesce() can only decreast the amount of partitions, but not increase it.
transactionsDf.repartition(transactionsDf.getNumPartitions()+2)
Incorrect, there is no getNumPartitions() method for the DataFrame class.
transactionsDf.coalesce(transactionsDf.getNumPartitions()+2)
Wrong, coalesce() can only be used for reducing the number of partitions and there is no getNumPartitions() method for the DataFrame class.
transactionsDf.repartition(transactionsDf._partitions+2)
No, DataFrame has no _partitions attribute. You can find out the current number of partitions of a DataFrame with the DataFrame.rdd.getNumPartitions() method.
More info: pyspark.sql.DataFrame.repartition - PySpark 3.1.2 documentation, pyspark.RDD.getNumPartitions - PySpark 3.1.2 documentation Static notebook | Dynamic notebook: See test 3

NEW QUESTION 39
Which of the following describes properties of a shuffle?

• A. Operations involving shuffles are never evaluated lazily.
• B. A shuffle is one of many actions in Spark.
• C. Shuffles involve only single partitions.
• D. In a shuffle, Spark writes data to disk.
• E. Shuffles belong to a class known as "full transformations".

Answer: D

Explanation:
Explanation
In a shuffle, Spark writes data to disk.
Correct! Spark's architecture dictates that intermediate results during a shuffle are written to disk.
A shuffle is one of many actions in Spark.
Incorrect. A shuffle is a transformation, but not an action.
Shuffles involve only single partitions.
No, shuffles involve multiple partitions. During a shuffle, Spark generates output partitions from multiple input partitions.
Operations involving shuffles are never evaluated lazily.
Wrong. A shuffle is a costly operation and Spark will evaluate it as lazily as other transformations. This is, until a subsequent action triggers its evaluation.
Shuffles belong to a class known as "full transformations".
Not quite. Shuffles belong to a class known as "wide transformations". "Full transformation" is not a relevant term in Spark.
More info: Spark - The Definitive Guide, Chapter 2 and Spark: disk I/O on stage boundaries explanation - Stack Overflow

NEW QUESTION 40
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