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Databricks Certified Associate Developer for Apache Spark 3.0 Exam Sample Questions (Q57-Q62):

NEW QUESTION # 57
The code block displayed below contains an error. The code block should produce a DataFrame with color as the only column and three rows with color values of red, blue, and green, respectively.
Find the error.
Code block:
1.spark.createDataFrame([("red",), ("blue",), ("green",)], "color")
Instead of calling spark.createDataFrame, just DataFrame should be called.

• A. The colors red, blue, and green should be expressed as a simple Python list, and not a list of tuples.
• B. The commas in the tuples with the colors should be eliminated.
• C. Instead of color, a data type should be specified.
• D. The "color" expression needs to be wrapped in brackets, so it reads ["color"].

Answer: D

Explanation:
Explanation
Correct code block:
spark.createDataFrame([("red",), ("blue",), ("green",)], ["color"])
The createDataFrame syntax is not exactly straightforward, but luckily the documentation (linked below) provides several examples on how to use it. It also shows an example very similar to the code block presented here which should help you answer this question correctly.
More info: pyspark.sql.SparkSession.createDataFrame - PySpark 3.1.2 documentation Static notebook | Dynamic notebook: See test 2

NEW QUESTION # 58
Which of the following code blocks reads in the two-partition parquet file stored at filePath, making sure all columns are included exactly once even though each partition has a different schema?
Schema of first partition:
1.root
2. |-- transactionId: integer (nullable = true)
3. |-- predError: integer (nullable = true)
4. |-- value: integer (nullable = true)
5. |-- storeId: integer (nullable = true)
6. |-- productId: integer (nullable = true)
7. |-- f: integer (nullable = true)
Schema of second partition:
1.root
2. |-- transactionId: integer (nullable = true)
3. |-- predError: integer (nullable = true)
4. |-- value: integer (nullable = true)
5. |-- storeId: integer (nullable = true)
6. |-- rollId: integer (nullable = true)
7. |-- f: integer (nullable = true)
8. |-- tax_id: integer (nullable = false)

• A. 1.nx = 0
  2.for file in dbutils.fs.ls(filePath):
  3. if not file.name.endswith(".parquet"):
  4. continue
  5. df_temp = spark.read.parquet(file.path)
  6. if nx == 0:
  7. df = df_temp
  8. else:
  9. df = df.join(df_temp, how="outer")
  10. nx = nx+1
  11.df
• B. 1.nx = 0
  2.for file in dbutils.fs.ls(filePath):
  3. if not file.name.endswith(".parquet"):
  4. continue
  5. df_temp = spark.read.parquet(file.path)
  6. if nx == 0:
  7. df = df_temp
  8. else:
  9. df = df.union(df_temp)
  10. nx = nx+1
  11.df
• C. spark.read.parquet(filePath)
• D. spark.read.option("mergeSchema", "true").parquet(filePath)
• E. spark.read.parquet(filePath, mergeSchema='y')

Answer: D

Explanation:
Explanation
This is a very tricky question and involves both knowledge about merging as well as schemas when reading parquet files.
spark.read.option("mergeSchema", "true").parquet(filePath)
Correct. Spark's DataFrameReader's mergeSchema option will work well here, since columns that appear in both partitions have matching data types. Note that mergeSchema would fail if one or more columns with the same name that appear in both partitions would have different data types.
spark.read.parquet(filePath)
Incorrect. While this would read in data from both partitions, only the schema in the parquet file that is read in first would be considered, so some columns that appear only in the second partition (e.g. tax_id) would be lost.
nx = 0
for file in dbutils.fs.ls(filePath):
if not file.name.endswith(".parquet"):
continue
df_temp = spark.read.parquet(file.path)
if nx == 0:
df = df_temp
else:
df = df.union(df_temp)
nx = nx+1
df
Wrong. The key idea of this solution is the DataFrame.union() command. While this command merges all data, it requires that both partitions have the exact same number of columns with identical data types.
spark.read.parquet(filePath, mergeSchema="y")
False. While using the mergeSchema option is the correct way to solve this problem and it can even be called with DataFrameReader.parquet() as in the code block, it accepts the value True as a boolean or string variable. But 'y' is not a valid option.
nx = 0
for file in dbutils.fs.ls(filePath):
if not file.name.endswith(".parquet"):
continue
df_temp = spark.read.parquet(file.path)
if nx == 0:
df = df_temp
else:
df = df.join(df_temp, how="outer")
nx = nx+1
df
No. This provokes a full outer join. While the resulting DataFrame will have all columns of both partitions, columns that appear in both partitions will be duplicated - the question says all columns that are included in the partitions should appear exactly once.
More info: Merging different schemas in Apache Spark | by Thiago Cordon | Data Arena | Medium Static notebook | Dynamic notebook: See test 3

NEW QUESTION # 59
The code block shown below should convert up to 5 rows in DataFrame transactionsDf that have the value 25 in column storeId into a Python list. Choose the answer that correctly fills the blanks in the code block to accomplish this.
Code block:
transactionsDf.__1__(__2__).__3__(__4__)

• A. 1. filter
  2. col("storeId")==25
  3. take
  4. 5
• B. 1. filter
  2. col("storeId")==25
  3. toLocalIterator
  4. 5
• C. 1. select
  2. storeId==25
  3. head
  4. 5
• D. 1. filter
  2. "storeId"==25
  3. collect
  4. 5
• E. 1. filter
  2. col("storeId")==25
  3. collect
  4. 5

Answer: A

Explanation:
Explanation
The correct code block is:
transactionsDf.filter(col("storeId")==25).take(5)
Any of the options with collect will not work because collect does not take any arguments, and in both cases the argument 5 is given.
The option with toLocalIterator will not work because the only argument to toLocalIterator is prefetchPartitions which is a boolean, so passing 5 here does not make sense.
The option using head will not work because the expression passed to select is not proper syntax. It would work if the expression would be col("storeId")==25.
Static notebook | Dynamic notebook: See test 1
(https://flrs.github.io/spark_practice_tests_code/#1/24.html ,
https://bit.ly/sparkpracticeexams_import_instructions)

NEW QUESTION # 60
Which of the following describes Spark's standalone deployment mode?

• A. Standalone mode is how Spark runs on YARN and Mesos clusters.
• B. Standalone mode uses only a single executor per worker per application.
• C. Standalone mode uses a single JVM to run Spark driver and executor processes.
• D. Standalone mode is a viable solution for clusters that run multiple frameworks, not only Spark.
• E. Standalone mode means that the cluster does not contain the driver.

Answer: B

Explanation:
Explanation
Standalone mode uses only a single executor per worker per application.
This is correct and a limitation of Spark's standalone mode.
Standalone mode is a viable solution for clusters that run multiple frameworks.
Incorrect. A limitation of standalone mode is that Apache Spark must be the only framework running on the cluster. If you would want to run multiple frameworks on the same cluster in parallel, for example Apache Spark and Apache Flink, you would consider the YARN deployment mode.
Standalone mode uses a single JVM to run Spark driver and executor processes.
No, this is what local mode does.
Standalone mode is how Spark runs on YARN and Mesos clusters.
No. YARN and Mesos modes are two deployment modes that are different from standalone mode. These modes allow Spark to run alongside other frameworks on a cluster. When Spark is run in standalone mode, only the Spark framework can run on the cluster.
Standalone mode means that the cluster does not contain the driver.
Incorrect, the cluster does not contain the driver in client mode, but in standalone mode the driver runs on a node in the cluster.
More info: Learning Spark, 2nd Edition, Chapter 1

NEW QUESTION # 61
Which of the following code blocks prints out in how many rows the expression Inc. appears in the string-type column supplier of DataFrame itemsDf?

• A. 1.accum=sc.accumulator(0)
  2.
  3.def check_if_inc_in_supplier(row):
  4. if 'Inc.' in row['supplier']:
  5. accum.add(1)
  6.
  7.itemsDf.foreach(check_if_inc_in_supplier)
  8.print(accum.value)
• B. print(itemsDf.foreach(lambda x: 'Inc.' in x))
• C. 1.counter = 0
  2.
  3.def count(x):
  4. if 'Inc.' in x['supplier']:
  5. counter = counter + 1
  6.
  7.itemsDf.foreach(count)
  8.print(counter)
• D. 1.counter = 0
  2.
  3.for index, row in itemsDf.iterrows():
  4. if 'Inc.' in row['supplier']:
  5. counter = counter + 1
  6.
  7.print(counter)
• E. print(itemsDf.foreach(lambda x: 'Inc.' in x).sum())

Answer: A

Explanation:
Explanation
Correct code block:
accum=sc.accumulator(0)
def check_if_inc_in_supplier(row):
if 'Inc.' in row['supplier']:
accum.add(1)
itemsDf.foreach(check_if_inc_in_supplier)
print(accum.value)
To answer this question correctly, you need to know both about the DataFrame.foreach() method and accumulators.
When Spark runs the code, it executes it on the executors. The executors do not have any information about variables outside of their scope. This is whhy simply using a Python variable counter, like in the two examples that start with counter = 0, will not work. You need to tell the executors explicitly that counter is a special shared variable, an Accumulator, which is managed by the driver and can be accessed by all executors for the purpose of adding to it.
If you have used Pandas in the past, you might be familiar with the iterrows() command. Notice that there is no such command in PySpark.
The two examples that start with print do not work, since DataFrame.foreach() does not have a return value.
More info: pyspark.sql.DataFrame.foreach - PySpark 3.1.2 documentation
Static notebook | Dynamic notebook: See test 3

NEW QUESTION # 62
......

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