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Databricks Certified Associate Developer for Apache Spark 3.5 - Python Sample Questions (Q41-Q46):

NEW QUESTION # 41
A developer needs to produce a Python dictionary using data stored in a small Parquet table, which looks like this:

The resulting Python dictionary must contain a mapping of region-> region id containing the smallest 3 region_idvalues.
Which code fragment meets the requirements?
A)

B)

C)

D)

The resulting Python dictionary must contain a mapping ofregion -> region_idfor the smallest
3region_idvalues.
Which code fragment meets the requirements?

A. regions = dict(
regions_df
.select('region', 'region_id')
.sort(desc('region_id'))
.take(3)
)
B. regions = dict(
regions_df
.select('region_id', 'region')
.limit(3)
.collect()
)
C. regions = dict(
regions_df
.select('region', 'region_id')
.sort('region_id')
.take(3)
)
D. regions = dict(
regions_df
.select('region_id', 'region')
.sort('region_id')
.take(3)
)

Answer: C

Explanation:
Comprehensive and Detailed Explanation From Exact Extract:
The question requires creating a dictionary where keys areregionvalues and values are the correspondingregion_idintegers. Furthermore, it asks to retrieve only the smallest 3region_idvalues.
Key observations:
select('region', 'region_id')puts the column order as expected bydict()- where the first column becomes the key and the second the value.
sort('region_id')ensures sorting in ascending order so the smallest IDs are first.
take(3)retrieves exactly 3 rows.
Wrapping the result indict(...)correctly builds the required Python dictionary:{ 'AFRICA': 0, 'AMERICA': 1,
'ASIA': 2 }.
Incorrect options:
Option B flips the order toregion_idfirst, resulting in a dictionary with integer keys - not what's asked.
Option C uses.limit(3)without sorting, which leads to non-deterministic rows based on partition layout.
Option D sorts in descending order, giving the largest rather than smallestregion_ids.
Hence, Option A meets all the requirements precisely.

NEW QUESTION # 42
A data engineer is working with a large JSON dataset containing order information. The dataset is stored in a distributed file system and needs to be loaded into a Spark DataFrame for analysis. The data engineer wants to ensure that the schema is correctly defined and that the data is read efficiently.
Which approach should the data scientist use to efficiently load the JSON data into a Spark DataFrame with a predefined schema?

A. Use spark.read.format("json").load() and then use DataFrame.withColumn() to cast each column to the desired data type.
B. Define a StructType schema and use spark.read.schema(predefinedSchema).json() to load the data.
C. Use spark.read.json() to load the data, then use DataFrame.printSchema() to view the inferred schema, and finally use DataFrame.cast() to modify column types.
D. Use spark.read.json() with the inferSchema option set to true

Answer: B

Explanation:
The most efficient and correct approach is to define a schema using StructType and pass it tospark.read.
schema(...).
This avoids schema inference overhead and ensures proper data types are enforced during read.
Example:
frompyspark.sql.typesimportStructType, StructField, StringType, DoubleType schema = StructType([ StructField("order_id", StringType(),True), StructField("amount", DoubleType(),True),
])
df = spark.read.schema(schema).json("path/to/json")
- Source:Databricks Guide - Read JSON with predefined schema

NEW QUESTION # 43
A Spark developer wants to improve the performance of an existing PySpark UDF that runs a hash function that is not available in the standard Spark functions library. The existing UDF code is:

import hashlib
import pyspark.sql.functions as sf
from pyspark.sql.types import StringType
def shake_256(raw):
return hashlib.shake_256(raw.encode()).hexdigest(20)
shake_256_udf = sf.udf(shake_256, StringType())
The developer wants to replace this existing UDF with a Pandas UDF to improve performance. The developer changes the definition ofshake_256_udfto this:CopyEdit shake_256_udf = sf.pandas_udf(shake_256, StringType()) However, the developer receives the error:
What should the signature of theshake_256()function be changed to in order to fix this error?

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

Answer: D

Explanation:
Comprehensive and Detailed Explanation From Exact Extract:
When converting a standard PySpark UDF to a Pandas UDF for performance optimization, the function must operate on a Pandas Series as input and return a Pandas Series as output.
In this case, the original function signature:
def shake_256(raw: str) -> str
is scalar - not compatible with Pandas UDFs.
According to the official Spark documentation:
"Pandas UDFs operate onpandas.Seriesand returnpandas.Series. The function definition should be:
def my_udf(s: pd.Series) -> pd.Series:
and it must be registered usingpandas_udf(...)."
Therefore, to fix the error:
The function should be updated to:
def shake_256(df: pd.Series) -> pd.Series:
return df.apply(lambda x: hashlib.shake_256(x.encode()).hexdigest(20))
This will allow Spark to efficiently execute the Pandas UDF in vectorized form, improving performance compared to standard UDFs.
Reference: Apache Spark 3.5 Documentation # User-Defined Functions # Pandas UDFs

NEW QUESTION # 44
A developer is running Spark SQL queries and notices underutilization of resources. Executors are idle, and the number of tasks per stage is low.
What should the developer do to improve cluster utilization?

A. Reduce the value of spark.sql.shuffle.partitions
B. Increase the value of spark.sql.shuffle.partitions
C. Enable dynamic resource allocation to scale resources as needed
D. Increase the size of the dataset to create more partitions

Answer: B

Explanation:
The number of tasks is controlled by the number of partitions. By default, spark.sql.shuffle.partitions is 200. If stages are showing very few tasks (less than total cores), you may not be leveraging full parallelism.
From the Spark tuning guide:
"To improve performance, especially for large clusters, increase spark.sql.shuffle.partitions to create more tasks and parallelism." Thus:
A is correct: increasing shuffle partitions increases parallelism
B is wrong: it further reduces parallelism
C is invalid: increasing dataset size doesn't guarantee more partitions D is irrelevant to task count per stage Final answer: A

NEW QUESTION # 45
What is the difference between df.cache() and df.persist() in Spark DataFrame?

A. Both cache() and persist() can be used to set the default storage level (MEMORY_AND_DISK_SER)
B. cache() - Persists the DataFrame with the default storage level (MEMORY_AND_DISK) and persist() - Can be used to set different storage levels to persist the contents of the DataFrame
C. persist() - Persists the DataFrame with the default storage level (MEMORY_AND_DISK_SER) and cache() - Can be used to set different storage levels to persist the contents of the DataFrame.
D. Both functions perform the same operation. The persist() function provides improved performance as its default storage level is DISK_ONLY.

Answer: B

Explanation:
df.cache() is shorthand for df.persist(StorageLevel.MEMORY_AND_DISK)
df.persist() allows specifying any storage level such as MEMORY_ONLY, DISK_ONLY, MEMORY_AND_DISK_SER, etc.
By default, persist() uses MEMORY_AND_DISK, unless specified otherwise.

NEW QUESTION # 46
......

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Free PDF 2026 High Hit-Rate Databricks Associate-Developer-Apache-Spark-3.5: Databricks Certified Associate Developer for Apache Spark 3.5 - Python Reliable Exam Guide

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Databricks Certified Associate Developer for Apache Spark 3.5 - Python Sample Questions (Q41-Q46):

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