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NEW QUESTION 48
The code block shown below should write DataFrame transactionsDf as a parquet file to path storeDir, using brotli compression and replacing any previously existing file. Choose the answer that correctly fills the blanks in the code block to accomplish this.
transactionsDf.__1__.format("parquet").__2__(__3__).option(__4__, "brotli").__5__(storeDir)

A. 1. save
2. mode
3. "replace"
4. "compression"
5. path
B. 1. write
2. mode
3. "overwrite"
4. "compression"
5. save
(Correct)
C. 1. write
2. mode
3. "overwrite"
4. compression
5. parquet
D. 1. store
2. with
3. "replacement"
4. "compression"
5. path
E. 1. save
2. mode
3. "ignore"
4. "compression"
5. path

Answer: A

Explanation:
Explanation
Correct code block:
transactionsDf.write.format("parquet").mode("overwrite").option("compression", "snappy").save(storeDir) Solving this question requires you to know how to access the DataFrameWriter (link below) from the DataFrame API - through DataFrame.write.
Another nuance here is about knowing the different modes available for writing parquet files that determine Spark's behavior when dealing with existing files. These, together with the compression options are explained in the DataFrameWriter.parquet documentation linked below.
Finally, bracket __5__ poses a certain challenge. You need to know which command you can use to pass down the file path to the DataFrameWriter. Both save and parquet are valid options here.
More info:
- DataFrame.write: pyspark.sql.DataFrame.write - PySpark 3.1.1 documentation
- DataFrameWriter.parquet: pyspark.sql.DataFrameWriter.parquet - PySpark 3.1.1 documentation Static notebook | Dynamic notebook: See test 1

NEW QUESTION 49
Which of the following code blocks returns DataFrame transactionsDf sorted in descending order by column predError, showing missing values last?

A. transactionsDf.sort("predError", ascending=False)
B. transactionsDf.sort(asc_nulls_last("predError"))
C. transactionsDf.orderBy("predError").asc_nulls_last()
D. transactionsDf.desc_nulls_last("predError")
E. transactionsDf.orderBy("predError").desc_nulls_last()

Answer: A

Explanation:
Explanation
transactionsDf.sort("predError", ascending=False)
Correct! When using DataFrame.sort() and setting ascending=False, the DataFrame will be sorted by the specified column in descending order, putting all missing values last. An alternative, although not listed as an answer here, would be transactionsDf.sort(desc_nulls_last("predError")).
transactionsDf.sort(asc_nulls_last("predError"))
Incorrect. While this is valid syntax, the DataFrame will be sorted on column predError in ascending order and not in descending order, putting missing values last.
transactionsDf.desc_nulls_last("predError")
Wrong, this is invalid syntax. There is no method DataFrame.desc_nulls_last() in the Spark API. There is a Spark function desc_nulls_last() however (link see below).
transactionsDf.orderBy("predError").desc_nulls_last()
No. While transactionsDf.orderBy("predError") is correct syntax (although it sorts the DataFrame by column predError in ascending order) and returns a DataFrame, there is no method DataFrame.desc_nulls_last() in the Spark API. There is a Spark function desc_nulls_last() however (link see below).
transactionsDf.orderBy("predError").asc_nulls_last()
Incorrect. There is no method DataFrame.asc_nulls_last() in the Spark API (see above).
More info: pyspark.sql.functions.desc_nulls_last - PySpark 3.1.2 documentation and pyspark.sql.DataFrame.sort - PySpark 3.1.2 documentation (https://bit.ly/3g1JtbI , https://bit.ly/2R90NCS) Static notebook | Dynamic notebook: See test 1 (https://flrs.github.io/spark_practice_tests_code/#1/32.html ,
https://bit.ly/sparkpracticeexams_import_instructions)

NEW QUESTION 50
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. join
2. [transactionsDf.productId==itemsDf.itemId, transactionsDf.storeId!=itemsDf.itemId]
3. select
4. "transactionId", "supplier"
C. 1. filter
2. "transactionId", "supplier"
3. join
4. "transactionsDf.storeId!=itemsDf.itemId, transactionsDf.productId==itemsDf.itemId"
D. 1. select
2. "transactionId", "supplier"
3. join
4. [transactionsDf.storeId!=itemsDf.itemId, transactionsDf.productId==itemsDf.itemId]
E. 1. join
2. transactionsDf.productId==itemsDf.itemId, transactionsDf.storeId!=itemsDf.itemId
3. filter
4. "transactionId", "supplier"

Answer: B

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 51
Which of the following code blocks returns a copy of DataFrame itemsDf where the column supplier has been renamed to manufacturer?

A. itemsDf.withColumnRenamed("supplier", "manufacturer")
B. itemsDf.withColumn(["supplier", "manufacturer"])
C. itemsDf.withColumnsRenamed("supplier", "manufacturer")
D. itemsDf.withColumnRenamed(col("manufacturer"), col("supplier"))
E. itemsDf.withColumn("supplier").alias("manufacturer")

Answer: A

Explanation:
Explanation
itemsDf.withColumnRenamed("supplier", "manufacturer")
Correct! This uses the relatively trivial DataFrame method withColumnRenamed for renaming column supplier to column manufacturer.
Note that the question asks for "a copy of DataFrame itemsDf". This may be confusing if you are not familiar with Spark yet. RDDs (Resilient Distributed Datasets) are the foundation of Spark DataFrames and are immutable. As such, DataFrames are immutable, too. Any command that changes anything in the DataFrame therefore necessarily returns a copy, or a new version, of it that has the changes applied.
itemsDf.withColumnsRenamed("supplier", "manufacturer")
Incorrect. Spark's DataFrame API does not have a withColumnsRenamed() method.
itemsDf.withColumnRenamed(col("manufacturer"), col("supplier"))
No. Watch out - although the col() method works for many methods of the DataFrame API, withColumnRenamed is not one of them. As outlined in the documentation linked below, withColumnRenamed expects strings.
itemsDf.withColumn(["supplier", "manufacturer"])
Wrong. While DataFrame.withColumn() exists in Spark, it has a different purpose than renaming columns.
withColumn is typically used to add columns to DataFrames, taking the name of the new column as a first, and a Column as a second argument. Learn more via the documentation that is linked below.
itemsDf.withColumn("supplier").alias("manufacturer")
No. While DataFrame.withColumn() exists, it requires 2 arguments. Furthermore, the alias() method on DataFrames would not help the cause of renaming a column much. DataFrame.alias() can be useful in addressing the input of join statements. However, this is far outside of the scope of this question. If you are curious nevertheless, check out the link below.
More info: pyspark.sql.DataFrame.withColumnRenamed - PySpark 3.1.1 documentation, pyspark.sql.DataFrame.withColumn - PySpark 3.1.1 documentation, and pyspark.sql.DataFrame.alias - PySpark 3.1.2 documentation (https://bit.ly/3aSB5tm , https://bit.ly/2Tv4rbE , https://bit.ly/2RbhBd2) Static notebook | Dynamic notebook: See test 1 (https://flrs.github.io/spark_practice_tests_code/#1/31.html ,
https://bit.ly/sparkpracticeexams_import_instructions)

NEW QUESTION 52
Which of the following describes a difference between Spark's cluster and client execution modes?

A. In cluster mode, a gateway machine hosts the driver, while it is co-located with the executor in client mode.
B. In cluster mode, the cluster manager resides on a worker node, while it resides on an edge node in client mode.
C. In cluster mode, the Spark driver is not co-located with the cluster manager, while it is co-located in client mode.
D. In cluster mode, the driver resides on a worker node, while it resides on an edge node in client mode.
E. In cluster mode, executor processes run on worker nodes, while they run on gateway nodes in client mode.

Answer: D

Explanation:
Explanation
In cluster mode, the driver resides on a worker node, while it resides on an edge node in client mode.
Correct. The idea of Spark's client mode is that workloads can be executed from an edge node, also known as gateway machine, from outside the cluster. The most common way to execute Spark however is in cluster mode, where the driver resides on a worker node.
In practice, in client mode, there are tight constraints about the data transfer speed relative to the data transfer speed between worker nodes in the cluster. Also, any job in that is executed in client mode will fail if the edge node fails. For these reasons, client mode is usually not used in a production environment.
In cluster mode, the cluster manager resides on a worker node, while it resides on an edge node in client execution mode.
No. In both execution modes, the cluster manager may reside on a worker node, but it does not reside on an edge node in client mode.
In cluster mode, executor processes run on worker nodes, while they run on gateway nodes in client mode.
This is incorrect. Only the driver runs on gateway nodes (also known as "edge nodes") in client mode, but not the executor processes.
In cluster mode, the Spark driver is not co-located with the cluster manager, while it is co-located in client mode.
No, in client mode, the Spark driver is not co-located with the driver. The whole point of client mode is that the driver is outside the cluster and not associated with the resource that manages the cluster (the machine that runs the cluster manager).
In cluster mode, a gateway machine hosts the driver, while it is co-located with the executor in client mode.
No, it is exactly the opposite: There are no gateway machines in cluster mode, but in client mode, they host the driver.

NEW QUESTION 53
......