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NEW QUESTION 50 Which of the following describes a way for resizing a DataFrame from 16 to 8 partitions in the most efficient way?

• A. Use operation DataFrame.repartition(8) to shuffle the DataFrame and reduce the number of partitions.
• B. Use a wide transformation to reduce the number of partitions. Use operation DataFrame.coalesce(0.5) to halve the number of partitions in the DataFrame.
• C. Use operation DataFrame.coalesce(8) to fully shuffle the DataFrame and reduce the number of partitions.
• D. Use a narrow transformation to reduce the number of partitions.

Answer: D Explanation: Explanation Use a narrow transformation to reduce the number of partitions. Correct! DataFrame.coalesce(n) is a narrow transformation, and in fact the most efficient way to resize the DataFrame of all options listed. One would run DataFrame.coalesce(8) to resize the DataFrame. Use operation DataFrame.coalesce(8) to fully shuffle the DataFrame and reduce the number of partitions. Wrong. The coalesce operation avoids a full shuffle, but will shuffle data if needed. This answer is incorrect because it says "fully shuffle" - this is something the coalesce operation will not do. As a general rule, it will reduce the number of partitions with the very least movement of data possible. More info: distributed computing - Spark - repartition() vs coalesce() - Stack Overflow Use operation DataFrame.coalesce(0.5) to halve the number of partitions in the DataFrame. Incorrect, since the numpartitions parameter needs to be an integer number defining the exact number of partitions desired after the operation. More info: pyspark.sql.DataFrame.coalesce - PySpark 3.1.2 documentation Use operation DataFrame.repartition(8) to shuffle the DataFrame and reduce the number of partitions. No. The repartition operation will fully shuffle the DataFrame. This is not the most efficient way of reducing the number of partitions of all listed options. Use a wide transformation to reduce the number of partitions. No. While possible via the DataFrame.repartition(n) command, the resulting full shuffle is not the most efficient way of reducing the number of partitions.   NEW QUESTION 51 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
• transactionsDf.productId==itemsDf.itemId, how="inner"
• select
• "transactionId", "supplier"
• B. 1. filter
• "transactionId", "supplier"
• join
• "transactionsDf.storeId!=itemsDf.itemId, transactionsDf.productId==itemsDf.itemId"
• C. 1. join
• transactionsDf.productId==itemsDf.itemId, transactionsDf.storeId!=itemsDf.itemId
• filter
• "transactionId", "supplier"
• D. 1. select
• "transactionId", "supplier"
• join
• [transactionsDf.storeId!=itemsDf.itemId, transactionsDf.productId==itemsDf.itemId]
• **E. 1. join
• [transactionsDf.productId==itemsDf.itemId, transactionsDf.storeId!=itemsDf.itemId]
• select
• "transactionId", "supplier"**

Answer: E 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 52 The code block displayed below contains an error. The code block should configure Spark so that DataFrames up to a size of 20 MB will be broadcast to all worker nodes when performing a join. Find the error. Code block:

• A. Spark will only broadcast DataFrames that are much smaller than the default value.
• B. The correct option to write configurations is through spark.config and not spark.conf.
• C. spark.conf.set("spark.sql.autoBroadcastJoinThreshold", 20)
• D. Spark will only apply the limit to threshold joins and not to other joins.
• E. The passed limit has the wrong variable type.
• F. The command is evaluated lazily and needs to be followed by an action.

Answer: A Explanation: Explanation This is question is hard. Let's assess the different answers one-by-one. Spark will only broadcast DataFrames that are much smaller than the default value. This is correct. The default value is 10 MB (10485760 bytes). Since the configuration for spark.sql.autoBroadcastJoinThreshold expects a number in bytes (and not megabytes), the code block sets the limits to merely 20 bytes, instead of the requested 20 * 1024 * 1024 (= 20971520) bytes. The command is evaluated lazily and needs to be followed by an action. No, this command is evaluated right away! Spark will only apply the limit to threshold joins and not to other joins. There are no "threshold joins", so this option does not make any sense. The correct option to write configurations is through spark.config and not spark.conf. No, it is indeed spark.conf! The passed limit has the wrong variable type. The configuration expects the number of bytes, a number, as an input. So, the 20 provided in the code block is fine.   NEW QUESTION 53 Which of the following code blocks returns a new DataFrame in which column attributes of DataFrame itemsDf is renamed to feature0 and column supplier to feature1?

• A. 1.itemsDf.withColumnRenamed("attributes", "feature0") 2.itemsDf.withColumnRenamed("supplier", "feature1")
• B. itemsDf.withColumnRenamed("attributes", "feature0").withColumnRenamed("supplier", "feature1")
• C. itemsDf.withColumnRenamed(col("attributes"), col("feature0"), col("supplier"), col("feature1"))
• D. itemsDf.withColumn("attributes", "feature0").withColumn("supplier", "feature1")
• E. itemsDf.withColumnRenamed(attributes, feature0).withColumnRenamed(supplier, feature1)

Answer: B Explanation: Explanation itemsDf.withColumnRenamed("attributes", "feature0").withColumnRenamed("supplier", "feature1") Correct! Spark's DataFrame.withColumnRenamed syntax makes it relatively easy to change the name of a column. itemsDf.withColumnRenamed(attributes, feature0).withColumnRenamed(supplier, feature1) Incorrect. In this code block, the Python interpreter will try to use attributes and the other column names as variables. Needless to say, they are undefined, and as a result the block will not run. itemsDf.withColumnRenamed(col("attributes"), col("feature0"), col("supplier"), col("feature1")) Wrong. The DataFrame.withColumnRenamed() operator takes exactly two string arguments. So, in this answer both using col() and using four arguments is wrong. itemsDf.withColumnRenamed("attributes", "feature0") itemsDf.withColumnRenamed("supplier", "feature1") No. In this answer, the returned DataFrame will only have column supplier be renamed, since the result of the first line is not written back to itemsDf. itemsDf.withColumn("attributes", "feature0").withColumn("supplier", "feature1") Incorrect. While withColumn works for adding and naming new columns, you cannot use it to rename existing columns. More info: pyspark.sql.DataFrame.withColumnRenamed - PySpark 3.1.2 documentation Static notebook | Dynamic notebook: See test 3   NEW QUESTION 54 Which of the following code blocks saves DataFrame transactionsDf in location /FileStore/transactions.csv as a CSV file and throws an error if a file already exists in the location?

• A. transactionsDf.write.save("/FileStore/transactions.csv")
• B. transactionsDf.write.format("csv").mode("error").path("/FileStore/transactions.csv")
• C. transactionsDf.write("csv").mode("error").save("/FileStore/transactions.csv")
• D. transactionsDf.write.format("csv").mode("error").save("/FileStore/transactions.csv")
• E. transactionsDf.write.format("csv").mode("ignore").path("/FileStore/transactions.csv")

Answer: D Explanation: Explanation Static notebook | Dynamic notebook: See test 1 (https://flrs.github.io/sparkpracticetestscode/#1/28.html , https://bit.ly/sparkpracticeexamsimport_instructions)   NEW QUESTION 55 ......