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NEW QUESTION 44
Which of the following code blocks adds a column predErrorSqrt to DataFrame transactionsDf that is the square root of column predError?
- A. transactionsDf.withColumn("predErrorSqrt", sqrt(predError))
- B. transactionsDf.select(sqrt(predError))
- C. transactionsDf.withColumn("predErrorSqrt", sqrt(col("predError")))
- D. transactionsDf.withColumn("predErrorSqrt", col("predError").sqrt())
- E. transactionsDf.select(sqrt("predError"))
Answer: C
Explanation:
Explanation
transactionsDf.withColumn("predErrorSqrt", sqrt(col("predError")))
Correct. The DataFrame.withColumn() operator is used to add a new column to a DataFrame. It takes two arguments: The name of the new column (here: predErrorSqrt) and a Column expression as the new column. In PySpark, a Column expression means referring to a column using the col("predError") command or by other means, for example by transactionsDf.predError, or even just using the column name as a string, "predError".
The question asks for the square root. sqrt() is a function in pyspark.sql.functions and calculates the square root. It takes a value or a Column as an input. Here it is the predError column of DataFrame transactionsDf expressed through col("predError").
transactionsDf.withColumn("predErrorSqrt", sqrt(predError))
Incorrect. In this expression, sqrt(predError) is incorrect syntax. You cannot refer to predError in this way - to Spark it looks as if you are trying to refer to the non-existent Python variable predError.
You could pass transactionsDf.predError, col("predError") (as in the correct solution), or even just "predError" instead.
transactionsDf.select(sqrt(predError))
Wrong. Here, the explanation just above this one about how to refer to predError applies.
transactionsDf.select(sqrt("predError"))
No. While this is correct syntax, it will return a single-column DataFrame only containing a column showing the square root of column predError. However, the question asks for a column to be added to the original DataFrame transactionsDf.
transactionsDf.withColumn("predErrorSqrt", col("predError").sqrt())
No. The issue with this statement is that column col("predError") has no sqrt() method. sqrt() is a member of pyspark.sql.functions, but not of pyspark.sql.Column.
More info: pyspark.sql.DataFrame.withColumn - PySpark 3.1.2 documentation and pyspark.sql.functions.sqrt - PySpark 3.1.2 documentation Static notebook | Dynamic notebook: See test 2
NEW QUESTION 45
Which of the following is a characteristic of the cluster manager?
- A. Each cluster manager works on a single partition of data.
- B. The cluster manager transforms jobs into DAGs.
- C. The cluster manager does not exist in standalone mode.
- D. In client mode, the cluster manager runs on the edge node.
- E. The cluster manager receives input from the driver through the SparkContext.
Answer: E
Explanation:
Explanation
The cluster manager receives input from the driver through the SparkContext.
Correct. In order for the driver to contact the cluster manager, the driver launches a SparkContext. The driver then asks the cluster manager for resources to launch executors.
In client mode, the cluster manager runs on the edge node.
No. In client mode, the cluster manager is independent of the edge node and runs in the cluster.
The cluster manager does not exist in standalone mode.
Wrong, the cluster manager exists even in standalone mode. Remember, standalone mode is an easy means to deploy Spark across a whole cluster, with some limitations. For example, in standalone mode, no other frameworks can run in parallel with Spark. The cluster manager is part of Spark in standalone deployments however and helps launch and maintain resources across the cluster.
The cluster manager transforms jobs into DAGs.
No, transforming jobs into DAGs is the task of the Spark driver.
Each cluster manager works on a single partition of data.
No. Cluster managers do not work on partitions directly. Their job is to coordinate cluster resources so that they can be requested by and allocated to Spark drivers.
More info: Introduction to Core Spark Concepts * BigData
NEW QUESTION 46
Which of the following statements about broadcast variables is correct?
- A. Broadcast variables are immutable.
- B. Broadcast variables are local to the worker node and not shared across the cluster.
- C. Broadcast variables are commonly used for tables that do not fit into memory.
- D. Broadcast variables are serialized with every single task.
- E. Broadcast variables are occasionally dynamically updated on a per-task basis.
Answer: A
Explanation:
Explanation
Broadcast variables are local to the worker node and not shared across the cluster.
This is wrong because broadcast variables are meant to be shared across the cluster. As such, they are never just local to the worker node, but available to all worker nodes.
Broadcast variables are commonly used for tables that do not fit into memory.
This is wrong because broadcast variables can only be broadcast because they are small and do fit into memory.
Broadcast variables are serialized with every single task.
This is wrong because they are cached on every machine in the cluster, precisely avoiding to have to be serialized with every single task.
Broadcast variables are occasionally dynamically updated on a per-task basis.
This is wrong because broadcast variables are immutable - they are never updated.
More info: Spark - The Definitive Guide, Chapter 14
NEW QUESTION 47
Which of the following code blocks reads in the parquet file stored at location filePath, given that all columns in the parquet file contain only whole numbers and are stored in the most appropriate format for this kind of data?
- A. 1.spark.read.schema([
2. StructField("transactionId", NumberType(), True),
3. StructField("predError", IntegerType(), True)
4. ]).load(filePath) - B. 1.spark.read.schema(
2. StructType([
3. StructField("transactionId", StringType(), True),
4. StructField("predError", IntegerType(), True)]
5. )).parquet(filePath) - C. 1.spark.read.schema([
2. StructField("transactionId", IntegerType(), True),
3. StructField("predError", IntegerType(), True)
4. ]).load(filePath, format="parquet") - D. 1.spark.read.schema(
2. StructType(
3. StructField("transactionId", IntegerType(), True),
4. StructField("predError", IntegerType(), True)
5. )).load(filePath) - E. 1.spark.read.schema(
2. StructType([
3. StructField("transactionId", IntegerType(), True),
4. StructField("predError", IntegerType(), True)]
5. )).format("parquet").load(filePath)
Answer: E
Explanation:
Explanation
The schema passed into schema should be of type StructType or a string, so all entries in which a list is passed are incorrect.
In addition, since all numbers are whole numbers, the IntegerType() data type is the correct option here.
NumberType() is not a valid data type and StringType() would fail, since the parquet file is stored in the "most appropriate format for this kind of data", meaning that it is most likely an IntegerType, and Spark does not convert data types if a schema is provided.
Also note that StructType accepts only a single argument (a list of StructFields). So, passing multiple arguments is invalid.
Finally, Spark needs to know which format the file is in. However, all of the options listed are valid here, since Spark assumes parquet as a default when no file format is specifically passed.
More info: pyspark.sql.DataFrameReader.schema - PySpark 3.1.2 documentation and StructType - PySpark 3.1.2 documentation
NEW QUESTION 48
The code block shown below should add column transactionDateForm to DataFrame transactionsDf. The column should express the unix-format timestamps in column transactionDate as string type like Apr 26 (Sunday). Choose the answer that correctly fills the blanks in the code block to accomplish this.
transactionsDf.__1__(__2__, from_unixtime(__3__, __4__))
- A. 1. withColumn
2. "transactionDateForm"
3. "transactionDate"
4. "MM d (EEE)" - B. 1. withColumn
2. "transactionDateForm"
3. "transactionDate"
4. "MMM d (EEEE)" - C. 1. withColumnRenamed
2. "transactionDate"
3. "transactionDateForm"
4. "MM d (EEE)" - D. 1. withColumn
2. "transactionDateForm"
3. "MMM d (EEEE)"
4. "transactionDate" - E. 1. select
2. "transactionDate"
3. "transactionDateForm"
4. "MMM d (EEEE)"
Answer: B
Explanation:
Explanation
Correct code block:
transactionsDf.withColumn("transactionDateForm", from_unixtime("transactionDate", "MMM d (EEEE)")) The question specifically asks about "adding" a column. In the context of all presented answers, DataFrame.withColumn() is the correct command for this. In theory, DataFrame.select() could also be used for this purpose, if all existing columns are selected and a new one is added.
DataFrame.withColumnRenamed() is not the appropriate command, since it can only rename existing columns, but cannot add a new column or change the value of a column.
Once DataFrame.withColumn() is chosen, you can read in the documentation (see below) that the first input argument to the method should be the column name of the new column.
The final difficulty is the date format. The question indicates that the date format Apr 26 (Sunday) is desired. The answers give "MMM d (EEEE)" and "MM d (EEE)" as options. It can be hard to know the details of the date format that is used in Spark. Specifically, knowing the differences between MMM and MM is probably not something you deal with every day. But, there is an easy way to remember the difference: M (one letter) is usually the shortest form: 4 for April. MM includes padding: 04 for April. MMM (three letters) is the three-letter month abbreviation: Apr for April. And MMMM is the longest possible form: April. Knowing this four-letter sequence helps you select the correct option here.
More info: pyspark.sql.DataFrame.withColumn - PySpark 3.1.2 documentation Static notebook | Dynamic notebook: See test 3
NEW QUESTION 49
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