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NEW QUESTION 22
Which of the following DataFrame operators is never classified as a wide transformation?
Answer: B
Explanation:
Explanation
As a general rule: After having gone through the practice tests you probably have a good feeling for what classifies as a wide and what classifies as a narrow transformation. If you are unsure, feel free to play around in Spark and display the explanation of the Spark execution plan via DataFrame.[operation, for example sort()].explain(). If repartitioning is involved, it would count as a wide transformation.
DataFrame.select()
Correct! A wide transformation includes a shuffle, meaning that an input partition maps to one or more output partitions. This is expensive and causes traffic across the cluster. With the select() operation however, you pass commands to Spark that tell Spark to perform an operation on a specific slice of any partition. For this, Spark does not need to exchange data across partitions, each partition can be worked on independently. Thus, you do not cause a wide transformation.
DataFrame.repartition()
Incorrect. When you repartition a DataFrame, you redefine partition boundaries. Data will flow across your cluster and end up in different partitions after the repartitioning is completed. This is known as a shuffle and, in turn, is classified as a wide transformation.
DataFrame.aggregate()
No. When you aggregate, you may compare and summarize data across partitions. In the process, data are exchanged across the cluster, and newly formed output partitions depend on one or more input partitions. This is a typical characteristic of a shuffle, meaning that the aggregate operation may classify as a wide transformation.
DataFrame.join()
Wrong. Joining multiple DataFrames usually means that large amounts of data are exchanged across the cluster, as new partitions are formed. This is a shuffle and therefore DataFrame.join() counts as a wide transformation.
DataFrame.sort()
False. When sorting, Spark needs to compare many rows across all partitions to each other. This is an expensive operation, since data is exchanged across the cluster and new partitions are formed as data is reordered. This process classifies as a shuffle and, as a result, DataFrame.sort() counts as wide transformation.
More info: Understanding Apache Spark Shuffle | Philipp Brunenberg
NEW QUESTION 23
Which of the following code blocks performs an inner join of DataFrames transactionsDf and itemsDf on columns productId and itemId, respectively, excluding columns value and storeId from DataFrame transactionsDf and column attributes from DataFrame itemsDf?
2.itemsDf.createOrReplaceTempView('itemsDf')
3.
4.spark.sql("SELECT -value, -storeId FROM transactionsDf INNER JOIN itemsDf ON productId==itemId").drop("attributes")C. 1.transactionsDf \
2. .drop(col('value'), col('storeId')) \
3. .join(itemsDf.drop(col('attributes')), col('productId')==col('itemId'))D. transactionsDf.drop("value", "storeId").join(itemsDf.drop("attributes"),
"transactionsDf.productId==itemsDf.itemId")E. 1.transactionsDf.createOrReplaceTempView('transactionsDf')
2.itemsDf.createOrReplaceTempView('itemsDf')
3.
4.statement = """
5.SELECT * FROM transactionsDf
6.INNER JOIN itemsDf
7.ON transactionsDf.productId==itemsDf.itemId
8."""
9.spark.sql(statement).drop("value", "storeId", "attributes")
Answer: E
Explanation:
Explanation
This question offers you a wide variety of answers for a seemingly simple question. However, this variety reflects the variety of ways that one can express a join in PySpark. You need to understand some SQL syntax to get to the correct answer here.
transactionsDf.createOrReplaceTempView('transactionsDf')
itemsDf.createOrReplaceTempView('itemsDf')
statement = """
SELECT * FROM transactionsDf
INNER JOIN itemsDf
ON transactionsDf.productId==itemsDf.itemId
"""
spark.sql(statement).drop("value", "storeId", "attributes")
Correct - this answer uses SQL correctly to perform the inner join and afterwards drops the unwanted columns. This is totally fine. If you are unfamiliar with the triple-quote """ in Python: This allows you to express strings as multiple lines.
transactionsDf \
drop(col('value'), col('storeId')) \
join(itemsDf.drop(col('attributes')), col('productId')==col('itemId'))
No, this answer option is a trap, since DataFrame.drop() does not accept a list of Column objects. You could use transactionsDf.drop('value', 'storeId') instead.
transactionsDf.drop("value", "storeId").join(itemsDf.drop("attributes"),
"transactionsDf.productId==itemsDf.itemId")
Incorrect - Spark does not evaluate "transactionsDf.productId==itemsDf.itemId" as a valid join expression.
This would work if it would not be a string.
transactionsDf.drop('value', 'storeId').join(itemsDf.select('attributes'), transactionsDf.productId==itemsDf.itemId) Wrong, this statement incorrectly uses itemsDf.select instead of itemsDf.drop.
transactionsDf.createOrReplaceTempView('transactionsDf')
itemsDf.createOrReplaceTempView('itemsDf')
spark.sql("SELECT -value, -storeId FROM transactionsDf INNER JOIN itemsDf ON productId==itemId").drop("attributes") No, here the SQL expression syntax is incorrect. Simply specifying -columnName does not drop a column.
More info: pyspark.sql.DataFrame.join - PySpark 3.1.2 documentation
Static notebook | Dynamic notebook: See test 3
NEW QUESTION 24
Which of the following statements about executors is correct?
Answer: E
Explanation:
Explanation
Executors stop upon application completion by default.
Correct. Executors only persist during the lifetime of an application.
A notable exception to that is when Dynamic Resource Allocation is enabled (which it is not by default). With Dynamic Resource Allocation enabled, executors are terminated when they are idle, independent of whether the application has been completed or not.
An executor can serve multiple applications.
Wrong. An executor is always specific to the application. It is terminated when the application completes (exception see above).
Each node hosts a single executor.
No. Each node can host one or more executors.
Executors store data in memory only.
No. Executors can store data in memory or on disk.
Executors are launched by the driver.
Incorrect. Executors are launched by the cluster manager on behalf of the driver.
More info: Job Scheduling - Spark 3.1.2 Documentation, How Applications are Executed on a Spark Cluster | Anatomy of a Spark Application | InformIT, and Spark Jargon for Starters. This blog is to clear some of the... | by Mageswaran D | Medium
NEW QUESTION 25
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