Databricks-Certified-Associate-Developer-for-Apache-Spark-3.0 Exam Dumps : Databricks Certified Associate Developer for Apache Spark 3.0 Exam

Explanation

As a general rule: After having gone through the practice tests you probably have a good feeling for what classifies as a wide and what classifies as a narrow transformation. If you are unsure, feel

free to play around in Spark and display the explanation of the Spark execution plan via DataFrame.[operation, for example sort()].explain(). If repartitioning is involved, it would count as a wide

transformation.

DataFrame.select()

Correct! A wide transformation includes a shuffle, meaning that an input partition maps to one or more output partitions. This is expensive and causes traffic across the cluster. With the select()

operation however, you pass commands to Spark that tell Spark to perform an operation on a specific slice of any partition. For this, Spark does not need to exchange data across partitions, each

partition can be worked on independently. Thus, you do not cause a wide transformation.

DataFrame.repartition()

Incorrect. When you repartition a DataFrame, you redefine partition boundaries. Data will flow across your cluster and end up in different partitions after the repartitioning is completed. This is

known as a shuffle and, in turn, is classified as a wide transformation.

DataFrame.aggregate()

No. When you aggregate, you may compare and summarize data across partitions. In the process, data are exchanged across the cluster, and newly formed output partitions depend on one or more

input partitions. This is a typical characteristic of a shuffle, meaning that the aggregate operation may classify as a wide transformation.

DataFrame.join()

Wrong. Joining multiple DataFrames usually means that large amounts of data are exchanged across the cluster, as new partitions are formed. This is a shuffle and therefore DataFrame.join()

counts as a wide transformation.

DataFrame.sort()

False. When sorting, Spark needs to compare many rows across all partitions to each other. This is an expensive operation, since data is exchanged across the cluster and new partitions are

formed as data is reordered. This process classifies as a shuffle and, as a result, DataFrame.sort() counts as wide transformation.

More info: Understanding Apache Spark Shuffle | Philipp Brunenberg

Explanation

Correct code block:

transactionsDf.select("storeId").printSchema()

The difficulty of this QUESTION NO: is that it is hard to solve with the stepwise first-to-last-gap approach that has worked well for similar questions, since the answer options are so different from

one

another. Instead, you might want to eliminate answers by looking for patterns of frequently wrong answers.

A first pattern that you may recognize by now is that column names are not expressed in quotes. For this reason, the answer that includes storeId should be eliminated.

By now, you may have understood that the DataFrame.limit() is useful for returning a specified amount of rows. It has nothing to do with specific columns. For this reason, the answer that resolves to

limit("storeId") can be eliminated.

Given that we are interested in information about the data type, you should QUESTION NO: whether the answer that resolves to limit(1).columns provides you with this information. While

DataFrame.columns is a valid call, it will only report back column names, but not column types. So, you can eliminate this option.

The two remaining options either use the printSchema() or print_schema() command. You may remember that DataFrame.printSchema() is the only valid command of the two. The select("storeId")

part just returns the storeId column of transactionsDf - this works here, since we are only interested in that column's type anyways.

More info: pyspark.sql.DataFrame.printSchema — PySpark 3.1.2 documentation

Static notebook | Dynamic notebook: See test 3, QUESTION NO: 57 (Databricks import instructions)

Explanation

Correct code block:

def count_to_target(target):

if target is None:

return

result = list(range(target))

return result

count_to_target_udf = udf(count_to_target, ArrayType(IntegerType()))

transactionsDf.select(count_to_target_udf('predError'))

Output of correct code block:

+--------------------------+

|count_to_target(predError)|

+--------------------------+

| [0, 1, 2]|

| [0, 1, 2, 3, 4, 5]|

| [0, 1, 2]|

| null|

| null|

| [0, 1, 2]|

+--------------------------+

This QUESTION NO: is not exactly easy. You need to be familiar with the syntax around UDFs (user-defined functions). Specifically, in this QUESTION NO: it is important to pass the correct types

to the udf

method - returning an array of a specific type rather than just a single type means you need to think harder about type implications than usual.

Remember that in Spark, you always pass types in an instantiated way like ArrayType(IntegerType()), not like ArrayType(IntegerType). The parentheses () are the key here - make sure you do not

forget those.

You should also pay attention that you actually pass the UDF count_to_target_udf, and not the Python method count_to_target to the select() operator.

Finally, null values are always a tricky case with UDFs. So, take care that the code can handle them correctly.

More info: How to Turn Python Functions into PySpark Functions (UDF) – Chang Hsin Lee – Committing my thoughts to words.

Static notebook | Dynamic notebook: See test 3, QUESTION NO: 24 (Databricks import instructions)