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Databricks Associate-Developer-Apache-Spark-3.5 Exam Syllabus Topics:

SectionObjectives
Topic 1: DataFrame API with PySpark- DataFrame creation and schema management
- Built-in functions and expressions
- Transformations and actions
Topic 2: Spark SQL- Window functions and aggregations
- SQL queries on DataFrames and tables
Topic 3: Data Ingestion and Storage- Reading and writing data (Parquet, JSON, CSV)
- Delta Lake basics
Topic 4: Apache Spark Fundamentals- Spark architecture and execution model
- RDD vs DataFrame vs Dataset concepts
Topic 5: Structured Streaming Basics- Streaming DataFrames
- Windowed aggregations in streaming
Topic 6: Data Processing and Performance- Optimization techniques
- Caching and persistence strategies
- Joins and data partitioning

Databricks Certified Associate Developer for Apache Spark 3.5 - Python Sample Questions:

1. 19 of 55.
A Spark developer wants to improve the performance of an existing PySpark UDF that runs a hash function not available in the standard Spark functions library.
The existing UDF code is:
import hashlib
from pyspark.sql.types import StringType
def shake_256(raw):
return hashlib.shake_256(raw.encode()).hexdigest(20)
shake_256_udf = udf(shake_256, StringType())
The developer replaces this UDF with a Pandas UDF for better performance:
@pandas_udf(StringType())
def shake_256(raw: str) -> str:
return hashlib.shake_256(raw.encode()).hexdigest(20)
However, the developer receives this error:
TypeError: Unsupported signature: (raw: str) -> str
What should the signature of the shake_256() function be changed to in order to fix this error?

A) def shake_256(raw: pd.Series) -> pd.Series:
B) def shake_256(raw: [pd.Series]) -> pd.Series:
C) def shake_256(raw: [str]) -> [str]:
D) def shake_256(raw: str) -> str:


2. 14 of 55.
A developer created a DataFrame with columns color, fruit, and taste, and wrote the data to a Parquet directory using:
df.write.partitionBy("color", "taste").parquet("/path/to/output")
What is the result of this code?

A) It throws an error if there are null values in either partition column.
B) It appends new partitions to an existing Parquet file.
C) It creates separate directories for each unique combination of color and taste.
D) It stores all data in a single Parquet file.


3. A developer is trying to join two tables, sales.purchases_fct and sales.customer_dim, using the following code:

fact_df = purch_df.join(cust_df, F.col('customer_id') == F.col('custid')) The developer has discovered that customers in the purchases_fct table that do not exist in the customer_dim table are being dropped from the joined table.
Which change should be made to the code to stop these customer records from being dropped?

A) fact_df = cust_df.join(purch_df, F.col('customer_id') == F.col('custid'))
B) fact_df = purch_df.join(cust_df, F.col('cust_id') == F.col('customer_id'))
C) fact_df = purch_df.join(cust_df, F.col('customer_id') == F.col('custid'), 'left')
D) fact_df = purch_df.join(cust_df, F.col('customer_id') == F.col('custid'), 'right_outer')


4. A Spark application suffers from too many small tasks due to excessive partitioning. How can this be fixed without a full shuffle?
Options:

A) Use the repartition() transformation with a lower number of partitions
B) Use the sortBy() transformation to reorganize the data
C) Use the coalesce() transformation with a lower number of partitions
D) Use the distinct() transformation to combine similar partitions


5. A data engineer writes the following code to join two DataFrames df1 and df2:
df1 = spark.read.csv("sales_data.csv") # ~10 GB
df2 = spark.read.csv("product_data.csv") # ~8 MB
result = df1.join(df2, df1.product_id == df2.product_id)

Which join strategy will Spark use?

A) Shuffle join because no broadcast hints were provided
B) Broadcast join, as df2 is smaller than the default broadcast threshold
C) Shuffle join, as the size difference between df1 and df2 is too large for a broadcast join to work efficiently
D) Shuffle join, because AQE is not enabled, and Spark uses a static query plan


Solutions:

Question # 1
Answer: A
Question # 2
Answer: C
Question # 3
Answer: C
Question # 4
Answer: C
Question # 5
Answer: B

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