UDFs (User Defined Functions) in Apache Spark - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to register a simple UDF that doubles a number.

Apache Spark

from pyspark.sql.functions import udf
from pyspark.sql.types import IntegerType

def double_num(x):
    return x * 2

double_udf = udf(double_num, [1])

Drag options to blanks, or click blank then click option'

AIntegerType()

BStringType()

CFloatType()

DBooleanType()

Attempts:

3 left

💡 Hint

Common Mistakes

Using StringType instead of IntegerType.

Not specifying the return type at all.

The UDF returns an integer, so we use IntegerType() to specify the return type.

2fill in blank

medium

Complete the code to apply the UDF to a DataFrame column named 'value'.

Apache Spark

from pyspark.sql import SparkSession
from pyspark.sql.functions import col

spark = SparkSession.builder.getOrCreate()
data = [(1,), (2,), (3,)]
df = spark.createDataFrame(data, ['value'])

result_df = df.withColumn('double_value', [1](col('value')))

Drag options to blanks, or click blank then click option'

Adouble_num

Bcol

Cudf

Ddouble_udf

Attempts:

3 left

💡 Hint

Common Mistakes

Using the original Python function instead of the UDF.

Using 'col' instead of the UDF.

We use the registered UDF double_udf to apply the function to the column.

3fill in blank

hard

Fix the error in the UDF registration by choosing the correct return type.

Apache Spark

from pyspark.sql.functions import udf
from pyspark.sql.types import [1]

def is_even(x):
    return x % 2 == 0

even_udf = udf(is_even, BooleanType())

Drag options to blanks, or click blank then click option'

AIntegerType

BBooleanType

CStringType

DFloatType

Attempts:

3 left

💡 Hint

Common Mistakes

Using IntegerType for a boolean result.

Using StringType which is incorrect here.

The function returns True or False, so the return type should be BooleanType.

4fill in blank

hard

Fill both blanks to create a UDF that returns the length of a string and apply it to the 'name' column.

Apache Spark

from pyspark.sql.functions import col
from pyspark.sql.types import [1]

def str_length(s):
    return len(s)

length_udf = udf(str_length, [2])

result_df = df.withColumn('name_length', length_udf(col('name')))

Drag options to blanks, or click blank then click option'

AIntegerType

BStringType

CFloatType

DBooleanType

Attempts:

3 left

💡 Hint

Common Mistakes

Using StringType for the return type.

Using different types for the two blanks.

The function returns the length of a string, which is an integer, so IntegerType is used for both blanks.

5fill in blank

hard

Fill all three blanks to create a UDF that checks if a number is positive, register it, and apply it to the 'score' column.

Apache Spark

from pyspark.sql.functions import col
from pyspark.sql.types import [1]

def is_positive(n):
    return n > 0

positive_udf = udf(is_positive, [2])

result_df = df.withColumn('is_positive', positive_udf([3]('score')))

Drag options to blanks, or click blank then click option'

AIntegerType

BBooleanType

Ccol

DStringType

Attempts:

3 left

💡 Hint

Common Mistakes

Using IntegerType instead of BooleanType.

Not using col() to refer to the column.

The function returns True or False, so BooleanType is the return type. The UDF is applied to the column using col('score').

Practice

1. What is the main purpose of a User Defined Function (UDF) in Apache Spark?

easy

A. To create new Spark sessions

B. To manage cluster resources

C. To optimize Spark SQL queries automatically

D. To apply custom logic to each row of a DataFrame

Solution

1. Step 1: Understand UDF role in Spark

   A UDF lets you add your own logic to process each row in a DataFrame, beyond built-in functions.

2. Step 2: Compare options

   The other options relate to Spark setup or optimization, not custom row logic.

3. Final Answer:

   To apply custom logic to each row of a DataFrame -> Option D

4. Quick Check:

   UDF = custom row logic [OK]

Hint: UDFs add your own row-wise logic in Spark [OK]

Common Mistakes:

• Confusing UDFs with Spark session management
• Thinking UDFs optimize queries automatically
• Assuming UDFs manage cluster resources

2. Which of the following is the correct way to register a UDF in PySpark with a return type of IntegerType?

easy

A. spark.udf.register('my_udf', func, 'IntegerType')

B. spark.udf.register('my_udf', func)

C. spark.udf.register('my_udf', func, IntegerType())

D. spark.udf.create('my_udf', func, IntegerType())

Solution

1. Step 1: Recall UDF registration syntax

   The correct method is spark.udf.register with function name, function, and return type as a DataType object.

2. Step 2: Evaluate options

   spark.udf.register('my_udf', func, IntegerType()) uses IntegerType() correctly. spark.udf.register('my_udf', func) misses return type. spark.udf.create('my_udf', func, IntegerType()) uses a non-existent method. spark.udf.register('my_udf', func, 'int') uses a string instead of IntegerType().

3. Final Answer:

   spark.udf.register('my_udf', func, IntegerType()) -> Option C

4. Quick Check:

   Register UDF with return type object [OK]

Hint: Always specify return type as DataType object when registering UDF [OK]

Common Mistakes:

• Omitting the return type parameter
• Using incorrect method names like create
• Passing return type as string instead of DataType

3. Given the following PySpark code, what will be the output of df.select('name', 'age_plus_one').show()?

from pyspark.sql.functions import udf
from pyspark.sql.types import IntegerType

def add_one(age):
    return age + 1

add_one_udf = udf(add_one, IntegerType())
df = spark.createDataFrame([('Alice', 30), ('Bob', 25)], ['name', 'age'])
df = df.withColumn('age_plus_one', add_one_udf('age'))

medium

A. [[Alice, 31], [Bob, 26]]

B. [[Alice, 30], [Bob, 25]]

C. Error: UDF missing return type

D. [[Alice, 1], [Bob, 1]]

Solution

1. Step 1: Understand UDF logic

   The UDF adds 1 to the 'age' column values: 30 + 1 = 31, 25 + 1 = 26.

2. Step 2: Check DataFrame transformation

   The new column 'age_plus_one' will have values 31 and 26 for Alice and Bob respectively.

3. Final Answer:

   [[Alice, 31], [Bob, 26]] -> Option A

4. Quick Check:

   age + 1 = 31, 26 [OK]

Hint: UDF applies function row-wise, transforming each value [OK]

Common Mistakes:

• Forgetting to specify return type causing errors
• Assuming original age remains unchanged
• Confusing UDF output with input data

4. Identify the error in the following PySpark UDF code snippet:

from pyspark.sql.functions import udf
from pyspark.sql.types import StringType

def greet(name):
    return 'Hello ' + name

greet_udf = udf(greet)
df = spark.createDataFrame([('John',), ('Jane',)], ['name'])
df = df.withColumn('greeting', greet_udf('name'))
df.show()

medium

A. UDF missing return type specification

B. Function greet has wrong parameter type

C. DataFrame creation syntax is incorrect

D. withColumn method is used incorrectly

Solution

1. Step 1: Check UDF registration

   The UDF is created without specifying the return type, which is required.

2. Step 2: Verify other code parts

   Function parameter and DataFrame creation are correct. withColumn usage is proper.

3. Final Answer:

   UDF missing return type specification -> Option A

4. Quick Check:

   Always specify return type when creating UDF [OK]

Hint: Always specify return type when defining UDF [OK]

Common Mistakes:

• Omitting return type causes runtime errors
• Assuming return type is optional
• Misunderstanding function parameter types

5. You want to create a UDF that returns the length of a string column but safely handles null values by returning 0 instead of failing. Which of the following UDF definitions correctly implements this behavior?

hard

A. def safe_length(s):
return len(s)
safe_length_udf = udf(safe_length, IntegerType())

B. def safe_length(s):
if s is None:
return 0
else:
return len(s)
safe_length_udf = udf(safe_length, IntegerType())

C. def safe_length(s):
if s == '':
return 0
return len(s)
safe_length_udf = udf(safe_length, StringType())

D. def safe_length(s):
return s.length()
safe_length_udf = udf(safe_length, IntegerType())

Solution

1. Step 1: Handle null values safely

   def safe_length(s): if s is None: return 0 else: return len(s) safe_length_udf = udf(safe_length, IntegerType()) checks if input is None and returns 0, preventing errors.

2. Step 2: Check return type and logic

   def safe_length(s): if s is None: return 0 else: return len(s) safe_length_udf = udf(safe_length, IntegerType()) returns IntegerType and uses len(s) correctly. Others either miss null check or use wrong return type or method.

3. Final Answer:

   def safe_length(s): if s is None: return 0 else: return len(s) safe_length_udf = udf(safe_length, IntegerType()) -> Option B

4. Quick Check:

   Check nulls first, then return length [OK]

Hint: Check for None before calling len() in UDF [OK]

Common Mistakes:

• Not handling None causing runtime errors
• Returning wrong data type from UDF
• Using string methods incorrectly in Python