Left and right joins in Data Analysis Python - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to perform a merge of df1 and df2 on the 'id' column.

Data Analysis Python

result = df1.[1](df2, on='id')

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

Amerge

Bjoin

Cconcat

Dappend

Attempts:

3 left

💡 Hint

Common Mistakes

Using concat or append instead of merge for joining.

Use merge to join two DataFrames on a key column.

2fill in blank

medium

Complete the code to perform a left join using pandas merge with the correct 'how' parameter.

Data Analysis Python

result = pd.merge(df1, df2, on='id', how='[1]')

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

Aright

Binner

Cleft

Douter

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'inner' or 'outer' instead of 'left' for left join.

The how='left' parameter keeps all rows from the left DataFrame.

3fill in blank

hard

Fix the error in the code to perform a right join on 'key' column.

Data Analysis Python

result = pd.merge(df1, df2, on='[1]', how='right')

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

Akey

Bvalue

Cid

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Not quoting the column name or using wrong column name.

The join key column must be specified as a string in quotes.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps words to their lengths only if length is greater than 3.

Data Analysis Python

{word: [1] for word in words if [2]

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

Alen(word)

Blen(word) > 3

Cword.startswith('a')

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using word instead of len(word) for values.

Wrong filter condition.

The dictionary maps each word to its length, filtering words longer than 3 characters.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps uppercase words to their lengths if length is greater than 4.

Data Analysis Python

{ [1]: [2] for word in words if [3] }

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

Aword.upper()

Blen(word)

Clen(word) > 4

Dword.lower()

Attempts:

3 left

💡 Hint

Common Mistakes

Using word.lower() instead of upper.

Wrong filter condition.

The dictionary maps uppercase words to their lengths, filtering words longer than 4 characters.

Practice

1. What does a left join do when combining two tables in data analysis?

easy

A. Keeps all rows from both tables, matching or not.

B. Keeps all rows from the left table and adds matching rows from the right table.

C. Keeps only rows that match in both tables.

D. Keeps all rows from the right table and adds matching rows from the left table.

Solution

1. Step 1: Understand the role of the left table in a left join

   A left join keeps every row from the left table regardless of matches in the right table.

2. Step 2: Understand how matching rows from the right table are added

   Only rows from the right table that match the left table's keys are added; unmatched right rows are ignored.

3. Final Answer:

   Keeps all rows from the left table and adds matching rows from the right table. -> Option B

4. Quick Check:

   Left join = all left rows + matching right rows [OK]

Hint: Left join keeps all left rows, adds right matches [OK]

Common Mistakes:

• Confusing left join with right join
• Thinking left join keeps unmatched right rows
• Believing left join only keeps matching rows

2. Which of the following is the correct syntax for a left join using pandas in Python?

easy

A. pd.merge(df1, df2, how='left', on='key')

B. pd.merge(df1, df2, how='right', on='key')

C. pd.join(df1, df2, how='left')

D. df1.left_join(df2, on='key')

Solution

1. Step 1: Recall pandas merge syntax for joins

   In pandas, the function to join tables is pd.merge(), specifying the join type with the 'how' parameter.

2. Step 2: Identify the correct parameters for a left join

   Using how='left' and on='key' correctly performs a left join on the specified key column.

3. Final Answer:

   pd.merge(df1, df2, how='left', on='key') -> Option A

4. Quick Check:

   pd.merge + how='left' = left join [OK]

Hint: Use pd.merge with how='left' for left join [OK]

Common Mistakes:

• Using pd.join instead of pd.merge
• Wrong 'how' parameter like 'right' or missing
• Calling non-existent left_join method

3. Given these two DataFrames:

df1 = pd.DataFrame({'id': [1, 2, 3], 'value': ['A', 'B', 'C']})
df2 = pd.DataFrame({'id': [2, 3, 4], 'score': [90, 80, 70]})

What is the result of pd.merge(df1, df2, how='left', on='id')?

medium

A. [{'id':1,'value':'A','score':NaN},{'id':2,'value':'B','score':90},{'id':3,'value':'C','score':80}]

B. [{'id':2,'value':'B','score':90},{'id':3,'value':'C','score':80},{'id':4,'value':NaN,'score':70}]

C. [{'id':1,'value':'A','score':NaN},{'id':2,'value':'B','score':90},{'id':3,'value':'C','score':80},{'id':4,'value':NaN,'score':70}]

D. [{'id':1,'value':'A','score':NaN},{'id':2,'value':'B','score':NaN},{'id':3,'value':'C','score':NaN}]

Solution

1. Step 1: Identify the left and right tables

   df1 is the left table with ids 1, 2, 3; df2 is the right table with ids 2, 3, 4.

2. Step 2: Perform left join on 'id'

   All rows from df1 appear. For id=1, no match in df2, so score is NaN. For ids 2 and 3, scores 90 and 80 are added. id=4 from df2 is ignored.

3. Final Answer:

   [{'id':1,'value':'A','score':NaN},{'id':2,'value':'B','score':90},{'id':3,'value':'C','score':80}] -> Option A

4. Quick Check:

   Left join keeps all df1 rows + matching df2 rows [OK]

Hint: Left join keeps all left rows, unmatched right rows ignored [OK]

Common Mistakes:

• Including unmatched right rows in output
• Replacing unmatched left values with NaN incorrectly
• Confusing left and right join results

4. You run this code but get an error:

pd.merge(df1, df2, how='right', key='id')

What is the error and how to fix it?

medium

A. Error: DataFrames must have same columns. Fix by renaming columns.

B. Error: 'how' must be 'left' or 'inner'. Fix by changing 'right' to 'left'.

C. Error: 'key' is not a valid argument. Fix by using 'on='id'' instead.

D. No error, code runs fine.

Solution

1. Step 1: Identify the incorrect parameter name

   The parameter to specify join keys in pd.merge is 'on', not 'key'.

2. Step 2: Correct the parameter name

   Replace 'key='id'' with 'on='id'' to fix the error.

3. Final Answer:

   Error: 'key' is not a valid argument. Fix by using 'on='id'' instead. -> Option C

4. Quick Check:

   Use 'on' for join keys in pd.merge [OK]

Hint: Use 'on' not 'key' for join columns in pd.merge [OK]

Common Mistakes:

• Using 'key' instead of 'on' parameter
• Changing 'how' incorrectly
• Assuming no error occurs

5. You have two DataFrames:

df_customers = pd.DataFrame({'customer_id': [1, 2, 3], 'name': ['Ann', 'Bob', 'Cara']})
df_orders = pd.DataFrame({'order_id': [101, 102], 'customer_id': [2, 4], 'amount': [250, 300]})

You want to list all orders with customer names, including orders without matching customers. Which join should you use and why?

hard

A. Full outer join to keep all customers and all orders.

B. Left join on 'customer_id' to keep all customers and add matching orders.

C. Inner join on 'customer_id' to keep only matching customers and orders.

D. Right join on 'customer_id' to keep all orders and add matching customer names.

Solution

1. Step 1: Identify which table to keep all rows from

   You want all orders listed, even if no matching customer exists, so keep all rows from df_orders.

2. Step 2: Choose join type to keep all right table rows

   df_orders is the right table in the join. Using a right join keeps all rows from df_orders and adds matching customer names from df_customers.

3. Final Answer:

   Right join on 'customer_id' to keep all orders and add matching customer names. -> Option D

4. Quick Check:

   Right join keeps all right table rows [OK]

Hint: Use right join to keep all orders, add matching customers [OK]

Common Mistakes:

• Using left join keeps all customers, not all orders
• Using inner join drops unmatched orders
• Using full join adds unmatched customers unnecessarily