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Databricks Certified Associate Developer for Apache Spark 3.0 Exam Sample Questions (Q18-Q23):

NEW QUESTION # 18
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.conf.set("spark.sql.autoBroadcastJoinThreshold", 20)
• B. The correct option to write configurations is through spark.config and not spark.conf.
• C. Spark will only apply the limit to threshold joins and not to other joins.
• D. The passed limit has the wrong variable type.
• E. Spark will only broadcast DataFrames that are much smaller than the default value.
• F. The command is evaluated lazily and needs to be followed by an action.

Answer: E 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 # 19
Which of the following code blocks returns a DataFrame that matches the multi-column DataFrame itemsDf, except that integer column itemId has been converted into a string column?

• A. itemsDf.withColumn("itemId", col("itemId").cast("string"))
  (Correct)
• B. spark.cast(itemsDf, "itemId", "string")
• C. itemsDf.withColumn("itemId", col("itemId").convert("string"))
• D. itemsDf.withColumn("itemId", convert("itemId", "string"))
• E. itemsDf.select(cast("itemId", "string"))

Answer: A Explanation:
Explanation
itemsDf.withColumn("itemId", col("itemId").cast("string"))
Correct. You can convert the data type of a column using the cast method of the Column class. Also note that you will have to use the withColumn method on itemsDf for replacing the existing itemId column with the new version that contains strings.
itemsDf.withColumn("itemId", col("itemId").convert("string"))
Incorrect. The Column object that col("itemId") returns does not have a convert method.
itemsDf.withColumn("itemId", convert("itemId", "string"))
Wrong. Spark's spark.sql.functions module does not have a convert method. The question is trying to mislead you by using the word "converted". Type conversion is also called "type casting". This may help you remember to look for a cast method instead of a convert method (see correct answer).
itemsDf.select(astype("itemId", "string"))
False. While astype is a method of Column (and an alias of Column.cast), it is not a method of pyspark.sql.functions (what the code block implies). In addition, the question asks to return a full DataFrame that matches the multi-column DataFrame itemsDf. Selecting just one column from itemsDf as in the code block would just return a single-column DataFrame.
spark.cast(itemsDf, "itemId", "string")
No, the Spark session (called by spark) does not have a cast method. You can find a list of all methods available for the Spark session linked in the documentation below.
More info:
- pyspark.sql.Column.cast - PySpark 3.1.2 documentation
- pyspark.sql.Column.astype - PySpark 3.1.2 documentation
- pyspark.sql.SparkSession - PySpark 3.1.2 documentation
Static notebook | Dynamic notebook: See test 3
NEW QUESTION # 20
Which of the following code blocks returns a copy of DataFrame itemsDf where the column supplier has been renamed to manufacturer?

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

Answer: C 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/sparkpracticetestscode/#1/31.html ,
https://bit.ly/sparkpracticeexamsimportinstructions)
NEW QUESTION # 21
The code block shown below should return a copy of DataFrame transactionsDf with an added column cos.
This column should have the values in column value converted to degrees and having the cosine of those converted values taken, rounded to two decimals. Choose the answer that correctly fills the blanks in the code block to accomplish this.
Code block:
transactionsDf.1(2, round(3(4(5_)),2))

• A. 1. withColumnRenamed
  2. "cos"
  3. cos
  4. degrees
  5. "transactionsDf.value"
• B. 1. withColumn
  2. "cos"
  3. cos
  4. degrees
  5. transactionsDf.value
• C. 1. withColumn
  2. col("cos")
  3. cos
  4. degrees
  5. transactionsDf.value
• D. 1. withColumn
  2. col("cos")
  3. cos
  4. degrees
  5. col("value")
  E
  . 1. withColumn
  2. "cos"
  3. degrees
  4. cos
  5. col("value")

Answer: B Explanation:
Explanation
Correct code block:
transactionsDf.withColumn("cos", round(cos(degrees(transactionsDf.value)),2)) This question is especially confusing because col, "cos" are so similar. Similar-looking answer options can also appear in the exam and, just like in this question, you need to pay attention to the details to identify what the correct answer option is.
The first answer option to throw out is the one that starts with withColumnRenamed: The question NO:
speaks specifically of adding a column. The withColumnRenamed operator only renames an existing column, however, so you cannot use it here.
Next, you will have to decide what should be in gap 2, the first argument of transactionsDf.withColumn().
Looking at the documentation (linked below), you can find out that the first argument of withColumn actually needs to be a string with the name of the column to be added. So, any answer that includes col("cos") as the option for gap 2 can be disregarded.
This leaves you with two possible answers. The real difference between these two answers is where the cos and degree methods are, either in gaps 3 and 4, or vice-versa. From the question you can find out that the new column should have "the values in column value converted to degrees and having the cosine of those converted values taken". This prescribes you a clear order of operations: First, you convert values from column value to degrees and then you take the cosine of those values. So, the inner parenthesis (gap 4) should contain the degree method and then, logically, gap 3 holds the cos method. This leaves you with just one possible correct answer.
More info: pyspark.sql.DataFrame.withColumn - PySpark 3.1.2 documentation Static notebook | Dynamic notebook: See test 3
NEW QUESTION # 22
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. count
  3. itemNameElements
  4. >=3
• B. 1. filter
  2. size
  3. "itemNameElements"
  4. >=3
  (Correct)
• C. 1. select
  2. count
  3. "itemNameElements"
  4. >3
• D. 1. select
  2. count
  3. col("itemNameElements")
  4. >3
• E. 1. select
  2. size
  3. "itemNameElements"
  4. >3

Answer: B 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 # 23
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