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NEW QUESTION 44 Which of the following code blocks returns a single-row DataFrame that only has a column corr which shows the Pearson correlation coefficient between columns predError and value in DataFrame transactionsDf?

• A. transactionsDf.select(corr(["predError", "value"]).alias("corr")).first()
• B. transactionsDf.select(corr(col("predError"), col("value")).alias("corr")) (Correct)
• C. transactionsDf.select(corr("predError", "value"))
• D. transactionsDf.select(corr(col("predError"), col("value")).alias("corr")).first()
• E. transactionsDf.select(corr(predError, value).alias("corr"))

Answer: B Explanation: Explanation In difficulty, this question is above what you can expect from the exam. What this question NO: wants to teach you, however, is to pay attention to the useful details included in the documentation. pyspark.sql.corr is not a very common method, but it deals with Spark's data structure in an interesting way. The command takes two columns over multiple rows and returns a single row - similar to an aggregation function. When examining the documentation (linked below), you will find this code example: a = range(20) b = [2 * x for x in range(20)] df = spark.createDataFrame(zip(a, b), ["a", "b"]) df.agg(corr("a", "b").alias('c')).collect() [Row(c=1.0)] See how corr just returns a single row? Once you understand this, you should be suspicious about answers that include first(), since there is no need to just select a single row. A reason to eliminate those answers is that DataFrame.first() returns an object of type Row, but not DataFrame, as requested in the question. transactionsDf.select(corr(col("predError"), col("value")).alias("corr")) Correct! After calculating the Pearson correlation coefficient, the resulting column is correctly renamed to corr. transactionsDf.select(corr(predError, value).alias("corr")) No. In this answer, Python will interpret column names predError and value as variable names. transactionsDf.select(corr(col("predError"), col("value")).alias("corr")).first() Incorrect. first() returns a row, not a DataFrame (see above and linked documentation below). transactionsDf.select(corr("predError", "value")) Wrong. Whie this statement returns a DataFrame in the desired shape, the column will have the name corr(predError, value) and not corr. transactionsDf.select(corr(["predError", "value"]).alias("corr")).first() False. In addition to first() returning a row, this code block also uses the wrong call structure for command corr which takes two arguments (the two columns to correlate). More info: - pyspark.sql.functions.corr - PySpark 3.1.2 documentation - pyspark.sql.DataFrame.first - PySpark 3.1.2 documentation Static notebook | Dynamic notebook: See test 3   NEW QUESTION 45 Which of the following code blocks generally causes a great amount of network traffic?

• A. DataFrame.rdd.map()
• B. DataFrame.select()
• C. DataFrame.collect()
• D. DataFrame.coalesce()
• E. DataFrame.count()

Answer: C Explanation: Explanation DataFrame.collect() sends all data in a DataFrame from executors to the driver, so this generally causes a great amount of network traffic in comparison to the other options listed. DataFrame.coalesce() just reduces the number of partitions and generally aims to reduce network traffic in comparison to a full shuffle. DataFrame.select() is evaluated lazily and, unless followed by an action, does not cause significant network traffic. DataFrame.rdd.map() is evaluated lazily, it does therefore not cause great amounts of network traffic. DataFrame.count() is an action. While it does cause some network traffic, for the same DataFrame, collecting all data in the driver would generally be considered to cause a greater amount of network traffic.   NEW QUESTION 46 Which of the following code blocks concatenates rows of DataFrames transactionsDf and transactionsNewDf, omitting any duplicates?

• A. transactionsDf.union(transactionsNewDf).distinct()
• B. transactionsDf.concat(transactionsNewDf).unique()
• C. transactionsDf.union(transactionsNewDf).unique()
• D. spark.union(transactionsDf, transactionsNewDf).distinct()
• E. transactionsDf.join(transactionsNewDf, how="union").distinct()

Answer: A Explanation: Explanation DataFrame.unique() and DataFrame.concat() do not exist and union() is not a method of the SparkSession. In addition, there is no union option for the join method in the DataFrame.join() statement. More info: pyspark.sql.DataFrame.union - PySpark 3.1.2 documentation Static notebook | Dynamic notebook: See test 2   NEW QUESTION 47 Which of the following describes a difference between Spark's cluster and client execution modes?

• A. In cluster mode, executor processes run on worker nodes, while they run on gateway nodes 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, a gateway machine hosts the driver, while it is co-located with the executor 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 48 ......