Self joins in Apache Spark - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to join the DataFrame to itself using the 'id' column.

Apache Spark

joined_df = a.join(b, a['id'] [1] b['id'])

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

A!=

B==

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' instead of '==' causes no matching rows.

Using '<' or '>' causes incorrect join conditions.

We use '==' to join rows where the 'id' values are equal in both aliases.

2fill in blank

medium

Complete the code to select columns 'a.id' and 'b.value' after the self join.

Apache Spark

result = joined_df.select([1])

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

A['a.id', 'b.value']

Ba.id, b.value

C['id', 'value']

D['a.id', 'value']

Attempts:

3 left

💡 Hint

Common Mistakes

Passing a string with comma-separated columns instead of a list.

Using column names without alias prefixes.

The select method expects a list of column names as strings.

3fill in blank

hard

Fix the error in the join condition to avoid ambiguous column references.

Apache Spark

joined_df = x.join(y, x['id'] [1] )

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

A== y['id']

B==

C!=

D== x['id']

Attempts:

3 left

💡 Hint

Common Mistakes

Using df['id'] on both sides causes ambiguity.

Using '!=' instead of '==' causes wrong join results.

We must specify the alias for the right side column to avoid ambiguity.

4fill in blank

hard

Fill both blanks to create a self join where 'a.manager_id' equals 'b.id'.

Apache Spark

joined_df = a.join(b, a['[1]'] [2] b['id'])

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

Amanager_id

B==

C!=

Demployee_id

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' instead of '==' causes no matches.

Using wrong column names like 'employee_id'.

We join on 'manager_id' from alias 'a' equal to 'id' from alias 'b'.

5fill in blank

hard

Fill all three blanks to create a dictionary of employee names and their manager names.

Apache Spark

emp_mgr = {row['[1]']: row['[2]'] for row in joined_df.select('[3]', 'manager_name').collect()}

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

Aemployee_name

Bmanager_name

Did

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'id' instead of 'employee_name' as key.

Swapping key and value columns.

We map employee_name to manager_name using the selected columns.

Practice

1.

What is the main purpose of a self join in Apache Spark?

easy

A. To join a DataFrame with itself to compare rows within the same dataset

B. To join two different DataFrames based on a common key

C. To merge two DataFrames vertically by stacking rows

D. To filter rows based on a condition

Solution

1. Step 1: Understand the concept of self join

   A self join means joining a DataFrame to itself to compare or relate rows within the same dataset.

2. Step 2: Differentiate from other joins

   Other joins combine different DataFrames or filter rows, but self join specifically uses the same DataFrame twice.

3. Final Answer:

   To join a DataFrame with itself to compare rows within the same dataset -> Option A

4. Quick Check:

   Self join = join DataFrame to itself [OK]

Hint: Self join means joining the same table twice [OK]

Common Mistakes:

• Confusing self join with joining different DataFrames
• Thinking self join stacks rows vertically
• Assuming self join is just filtering rows

2.

Which of the following is the correct way to create aliases for a self join in Spark using DataFrame API?

df1 = df.alias('a')
df2 = df.alias('b')
joined = df1.join(df2, ...)

easy

A. df1 = df.alias('a'); df2 = df.alias('b'); joined = df1.join(df2, 'id')

B. df1 = df; df2 = df; joined = df1.join(df2, 'id')

C. df1 = df.as('a'); df2 = df.as('b'); joined = df1.join(df2, 'id')

D. df1 = df.copy(); df2 = df.copy(); joined = df1.join(df2, 'id')

Solution

1. Step 1: Use alias() method for self join

   In Spark, to join a DataFrame to itself, you create aliases using alias() to distinguish the two references.

2. Step 2: Check the syntax correctness

   df1 = df.alias('a'); df2 = df.alias('b'); joined = df1.join(df2, 'id') uses alias() correctly and joins on a column 'id'. df1 = df.as('a'); df2 = df.as('b'); joined = df1.join(df2, 'id') uses as() which is not a valid DataFrame method in Spark.

3. Final Answer:

   df1 = df.alias('a'); df2 = df.alias('b'); joined = df1.join(df2, 'id') -> Option A

4. Quick Check:

   Use alias() for self join [OK]

Hint: Use alias() to rename DataFrame before self join [OK]

Common Mistakes:

• Using as() instead of alias()
• Not creating aliases and causing ambiguous columns
• Copying DataFrame instead of aliasing

3.

Given the following Spark code, what will be the output count?

from pyspark.sql import SparkSession
spark = SparkSession.builder.getOrCreate()
data = [(1, 'A'), (2, 'B'), (3, 'A')]
df = spark.createDataFrame(data, ['id', 'group'])
df1 = df.alias('df1')
df2 = df.alias('df2')
joined = df1.join(df2, (df1.group == df2.group) & (df1.id != df2.id))
print(joined.count())

medium

A. 2

B. 4

C. 3

D. 6

Solution

1. Step 1: Understand the data and join condition

   Data has 3 rows: (1, 'A'), (2, 'B'), (3, 'A'). The join matches rows with the same group but different ids.

2. Step 2: Count matching pairs

   Group 'A' has ids 1 and 3, so pairs are (1,3) and (3,1) = 2 matches. Group 'B' has only one id, so no match. Total matches = 2 rows. However, since the join is a cross join on the condition, each pair appears twice (once as (1,3) and once as (3,1)), so total count is 4.

3. Final Answer:

   4 -> Option B

4. Quick Check:

   Count matching pairs with condition = 4 [OK]

Hint: Count pairs with same group but different ids [OK]

Common Mistakes:

• Counting only one direction of pairs
• Including rows where id equals
• Ignoring symmetric pairs

4.

Identify the error in this Spark self join code snippet:

df1 = df.alias('a')
df2 = df.alias('b')
joined = df1.join(df2, df1.id == df2.id)
joined.show()

What is the problem?

medium

A. Syntax error in join condition

B. Missing alias() method for DataFrame

C. The join condition causes each row to join with itself, no filtering

D. DataFrame df is not defined

Solution

1. Step 1: Analyze the join condition

   The join condition is df1.id == df2.id, which matches rows with the same id, including the same row joined to itself.

2. Step 2: Identify missing filter to exclude self matches

   There is no condition to exclude rows where the two aliases refer to the same row, so the join includes self matches.

3. Final Answer:

   The join condition causes each row to join with itself, no filtering -> Option C

4. Quick Check:

   Join condition includes self matches [OK]

Hint: Add filter to exclude self matches in self join [OK]

Common Mistakes:

• Not excluding rows where both aliases are same
• Assuming alias() is missing
• Confusing syntax error with logic error

5.

You have a DataFrame df with columns employee_id, manager_id, and department. You want to find pairs of employees who work in the same department and share the same manager but are different employees. Which self join condition is correct?

hard

A. df1.employee_id == df2.employee_id AND df1.manager_id == df2.manager_id

B. df1.employee_id == df2.manager_id AND df1.department == df2.department

C. df1.manager_id != df2.manager_id AND df1.department == df2.department

D. df1.department == df2.department AND df1.manager_id == df2.manager_id AND df1.employee_id != df2.employee_id

Solution

1. Step 1: Understand the requirement

   We want pairs of employees in the same department, with the same manager, but different employee IDs.

2. Step 2: Translate to join condition

   The join condition must check equality of department and manager_id, and inequality of employee_id to avoid self pairs.

3. Final Answer:

   df1.department == df2.department AND df1.manager_id == df2.manager_id AND df1.employee_id != df2.employee_id -> Option D

4. Quick Check:

   Same department & manager, different employees [OK]

Hint: Match same manager and department, exclude same employee [OK]

Common Mistakes:

• Matching employee_id to manager_id incorrectly
• Not excluding same employee pairs
• Checking for different managers instead of same