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Databricks Certified Associate Developer for Apache Spark 3.0 Exam Sample Questions (Q88-Q93):

NEW QUESTION # 88
Which of the following code blocks returns a copy of DataFrame transactionsDf in which column productId has been renamed to productNumber?

A. transactionsDf.withColumnRenamed(productId, productNumber)B. transactionsDf.withColumnRenamed("productId", "productNumber")C. transactionsDf.withColumnRenamed(col(productId), col(productNumber))D. transactionsDf.withColumnRenamed("productNumber", "productId")E. transactionsDf.withColumn("productId", "productNumber")

Answer: B

Explanation:
Explanation
More info: pyspark.sql.DataFrame.withColumnRenamed - PySpark 3.1.2 documentation Static notebook | Dynamic notebook: See test 2

NEW QUESTION # 89
The code block displayed below contains an error. The code block should configure Spark to split data in 20 parts when exchanging data between executors for joins or aggregations. Find the error.
Code block:
spark.conf.set(spark.sql.shuffle.partitions, 20)

A. The code block sets the wrong option.B. The code block sets the incorrect number of parts.C. The code block is missing a parameter.D. The code block uses the wrong command for setting an option.E. The code block expresses the option incorrectly.

Answer: E

Explanation:
Explanation
Correct code block:
spark.conf.set("spark.sql.shuffle.partitions", 20)
The code block expresses the option incorrectly.
Correct! The option should be expressed as a string.
The code block sets the wrong option.
No, spark.sql.shuffle.partitions is the correct option for the use case in the question.
The code block sets the incorrect number of parts.
Wrong, the code block correctly states 20 parts.
The code block uses the wrong command for setting an option.
No, in PySpark spark.conf.set() is the correct command for setting an option.
The code block is missing a parameter.
Incorrect, spark.conf.set() takes two parameters.
More info: Configuration - Spark 3.1.2 Documentation

NEW QUESTION # 90
Which of the following DataFrame operators is never classified as a wide transformation?

A. DataFrame.sort()B. DataFrame.repartition()C. DataFrame.join()D. DataFrame.select()E. DataFrame.aggregate()

Answer: D

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 # 91
The code block shown below should return a two-column DataFrame with columns transactionId and supplier, with combined information from DataFrames itemsDf and transactionsDf. The code block should merge rows in which column productId of DataFrame transactionsDf matches the value of column itemId in DataFrame itemsDf, but only where column storeId of DataFrame transactionsDf does not match column itemId of DataFrame itemsDf. Choose the answer that correctly fills the blanks in the code block to accomplish this.
Code block:
transactionsDf.__1__(itemsDf, __2__).__3__(__4__)

A. 1. join
2. transactionsDf.productId==itemsDf.itemId, how="inner"
3. select
4. "transactionId", "supplier"B. 1. select
2. "transactionId", "supplier"
3. join
4. [transactionsDf.storeId!=itemsDf.itemId, transactionsDf.productId==itemsDf.itemId]C. 1. filter
2. "transactionId", "supplier"
3. join
4. "transactionsDf.storeId!=itemsDf.itemId, transactionsDf.productId==itemsDf.itemId"D. 1. join
2. [transactionsDf.productId==itemsDf.itemId, transactionsDf.storeId!=itemsDf.itemId]
3. select
4. "transactionId", "supplier"E. 1. join
2. transactionsDf.productId==itemsDf.itemId, transactionsDf.storeId!=itemsDf.itemId
3. filter
4. "transactionId", "supplier"

Answer: D

Explanation:
Explanation
This question is pretty complex and, in its complexity, is probably above what you would encounter in the exam. However, reading the question carefully, you can use your logic skills to weed out the wrong answers here.
First, you should examine the join statement which is common to all answers. The first argument of the join() operator (documentation linked below) is the DataFrame to be joined with. Where join is in gap 3, the first argument of gap 4 should therefore be another DataFrame. For none of the questions where join is in the third gap, this is the case. So you can immediately discard two answers.
For all other answers, join is in gap 1, followed by .(itemsDf, according to the code block. Given how the join() operator is called, there are now three remaining candidates.
Looking further at the join() statement, the second argument (on=) expects "a string for the join column name, a list of column names, a join expression (Column), or a list of Columns", according to the documentation. As one answer option includes a list of join expressions (transactionsDf.productId==itemsDf.itemId, transactionsDf.storeId!=itemsDf.itemId) which is unsupported according to the documentation, we can discard that answer, leaving us with two remaining candidates.
Both candidates have valid syntax, but only one of them fulfills the condition in the question "only where column storeId of DataFrame transactionsDf does not match column itemId of DataFrame itemsDf". So, this one remaining answer option has to be the correct one!
As you can see, although sometimes overwhelming at first, even more complex questions can be figured out by rigorously applying the knowledge you can gain from the documentation during the exam.
More info: pyspark.sql.DataFrame.join - PySpark 3.1.2 documentation
Static notebook | Dynamic notebook: See test 3

NEW QUESTION # 92
The code block shown below should return all rows of DataFrame itemsDf that have at least 3 items in column itemNameElements. Choose the answer that correctly fills the blanks in the code block to accomplish this.
Example of DataFrame itemsDf:
1.+------+----------------------------------+-------------------+------------------------------------------+
2.|itemId|itemName |supplier |itemNameElements |
3.+------+----------------------------------+-------------------+------------------------------------------+
4.|1 |Thick Coat for Walking in the Snow|Sports Company Inc.|[Thick, Coat, for, Walking, in, the, Snow]|
5.|2 |Elegant Outdoors Summer Dress |YetiX |[Elegant, Outdoors, Summer, Dress] |
6.|3 |Outdoors Backpack |Sports Company Inc.|[Outdoors, Backpack] |
7.+------+----------------------------------+-------------------+------------------------------------------+ Code block:
itemsDf.__1__(__2__(__3__)__4__)

A. 1. filter
2. size
3. "itemNameElements"
4. >=3
(Correct)B. 1. select
2. count
3. "itemNameElements"
4. >3C. 1. select
2. count
3. col("itemNameElements")
4. >3D. 1. select
2. size
3. "itemNameElements"
4. >3E. 1. filter
2. count
3. itemNameElements
4. >=3

Answer: A

Explanation:
Explanation
Correct code block:
itemsDf.filter(size("itemNameElements")>3)
Output of code block:
+------+----------------------------------+-------------------+------------------------------------------+
|itemId|itemName |supplier |itemNameElements |
+------+----------------------------------+-------------------+------------------------------------------+
|1 |Thick Coat for Walking in the Snow|Sports Company Inc.|[Thick, Coat, for, Walking, in, the, Snow]|
|2 |Elegant Outdoors Summer Dress |YetiX |[Elegant, Outdoors, Summer, Dress] |
+------+----------------------------------+-------------------+------------------------------------------+ The big difficulty with this question is in knowing the difference between count and size (refer to documentation below). size is the correct function to choose here since it returns the number of elements in an array on a per-row basis.
The other consideration for solving this question is the difference between select and filter. Since we want to return the rows in the original DataFrame, filter is the right choice. If we would use select, we would simply get a single-column DataFrame showing which rows match the criteria, like so:
+----------------------------+
|(size(itemNameElements) > 3)|
+----------------------------+
|true |
|true |
|false |
+----------------------------+
More info:
Count documentation: pyspark.sql.functions.count - PySpark 3.1.1 documentation Size documentation: pyspark.sql.functions.size - PySpark 3.1.1 documentation Static notebook | Dynamic notebook: See test 1

NEW QUESTION # 93
......

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