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

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

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

9 of 55.
Given the code fragment:
import pyspark.pandas as ps
pdf = ps.DataFrame(data)
Which method is used to convert a Pandas API on Spark DataFrame (pyspark.pandas.DataFrame) into a standard PySpark DataFrame (pyspark.sql.DataFrame)?

  • A. pdf.to_dataframe()
  • B. pdf.to_pandas()
  • C. pdf.spark()
  • D. pdf.to_spark()
Reveal Solution  Discussion  0

Correct Answer: D  🗳️

Explanation: Only visible for PracticeVCE members. You can sign-up / login (it's free).

A developer is working with a pandas DataFrame containing user behavior data from a web application.
Which approach should be used for executing a groupBy operation in parallel across all workers in Apache Spark 3.5?
A)
Use the applylnPandas API
B)

C)

  • A. Use a Pandas UDF:
    @pandas_udf("double")
    def mean_func(value: pd.Series) -> float:
    return value.mean()
    df.groupby("user_id").agg(mean_func(df["value"])).show()
  • B. Use the mapInPandas API:
    df.mapInPandas(mean_func, schema="user_id long, value double").show()
  • C. Use the applyInPandas API:
    df.groupby("user_id").applyInPandas(mean_func, schema="user_id long, value double").show()
  • D. Use a regular Spark UDF:
    from pyspark.sql.functions import mean
    df.groupBy("user_id").agg(mean("value")).show()
Reveal Solution  Discussion  0

Correct Answer: C  🗳️

Explanation: Only visible for PracticeVCE members. You can sign-up / login (it's free).

A data analyst builds a Spark application to analyze finance data and performs the following operations: filter, select, groupBy, and coalesce.
Which operation results in a shuffle?

  • A. groupBy
  • B. filter
  • C. coalesce
  • D. select
Reveal Solution  Discussion  0

Correct Answer: A  🗳️

Explanation: Only visible for PracticeVCE members. You can sign-up / login (it's free).

41 of 55.
A data engineer is working on the DataFrame df1 and wants the Name with the highest count to appear first (descending order by count), followed by the next highest, and so on.
The DataFrame has columns:
id | Name | count | timestamp
---------------------------------
1 | USA | 10
2 | India | 20
3 | England | 50
4 | India | 50
5 | France | 20
6 | India | 10
7 | USA | 30
8 | USA | 40
Which code fragment should the engineer use to sort the data in the Name and count columns?

  • A. df1.orderBy(col("count").desc(), col("Name").asc())
  • B. df1.orderBy(col("Name").desc(), col("count").asc())
  • C. df1.sort("Name", "count")
  • D. df1.orderBy("Name", "count")
Reveal Solution  Discussion  0

Correct Answer: A  🗳️

Explanation: Only visible for PracticeVCE members. You can sign-up / login (it's free).

A data engineer needs to write a Streaming DataFrame as Parquet files.
Given the code:

Which code fragment should be inserted to meet the requirement?
A)

B)

C)

D)

Which code fragment should be inserted to meet the requirement?

  • A. .format("parquet")
    .option("location", "path/to/destination/dir")
  • B. .format("parquet")
    .option("path", "path/to/destination/dir")
  • C. .option("format", "parquet")
    .option("location", "path/to/destination/dir")
  • D. CopyEdit
    .option("format", "parquet")
    .option("destination", "path/to/destination/dir")
Reveal Solution  Discussion  0

Correct Answer: B  🗳️

Explanation: Only visible for PracticeVCE members. You can sign-up / login (it's free).

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