Accessing values using keys in Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of accessing a nested dictionary value?

Given the dictionary data = {'user': {'name': 'Alice', 'age': 30}}, what is the output of print(data['user']['name'])?

Python

data = {'user': {'name': 'Alice', 'age': 30}}
print(data['user']['name'])

A KeyError

B 30

C Alice

D None

💡 Hint

Think about how to access values inside nested dictionaries using keys.

The key 'user' gives the nested dictionary {'name': 'Alice', 'age': 30}. Then accessing 'name' inside it returns 'Alice'.

❓ Predict Output

intermediate

What happens when accessing a missing key directly?

What error will this code produce?

info = {'city': 'Paris', 'country': 'France'}
print(info['population'])

Python

info = {'city': 'Paris', 'country': 'France'}
print(info['population'])

A KeyError

B Paris

C None

D TypeError

💡 Hint

What happens if you try to get a key that does not exist in a dictionary?

Accessing a key that is not present in the dictionary directly raises a KeyError.

❓ Predict Output

advanced

What is the output when using the get() method with a default value?

Consider the dictionary settings = {'theme': 'dark', 'font': 'Arial'}. What does print(settings.get('color', 'blue')) output?

Python

settings = {'theme': 'dark', 'font': 'Arial'}
print(settings.get('color', 'blue'))

A KeyError

B None

C color

D blue

💡 Hint

The get() method returns the default value if the key is missing.

Since 'color' is not a key in settings, get() returns the default 'blue'.

❓ Predict Output

advanced

What is the output when accessing a key with a mutable object as key?

What happens when running this code?

my_dict = {[1, 2]: 'value'}

Python

my_dict = {[1, 2]: 'value'}

A SyntaxError

B TypeError

C KeyError

D {[1, 2]: 'value'}

💡 Hint

Dictionary keys must be immutable types.

Lists are mutable and cannot be used as dictionary keys, so this raises a TypeError.

🧠 Conceptual

expert

How many items are in the dictionary after this code runs?

What is the number of items in result after executing this code?

keys = ['a', 'b', 'a', 'c']
values = [1, 2, 3, 4]
result = {k: v for k, v in zip(keys, values)}

A 3

B 1

C 2

D 4

💡 Hint

Dictionary keys must be unique; later values overwrite earlier ones.

The key 'a' appears twice; the last value (3) overwrites the first (1). So keys are 'a', 'b', 'c' → 3 items.

Practice

1. What is the correct way to access the value associated with the key 'name' in the dictionary person = {'name': 'Alice', 'age': 30}?

easy

A. person.get('age')

B. person.name

C. person['name']

D. person['Alice']

Solution

1. Step 1: Identify the dictionary and key

   The dictionary is person and the key to access is 'name'.

2. Step 2: Use correct syntax to access value by key

   In Python, dictionary values are accessed using square brackets and the key as a string: person['name'].

3. Final Answer:

   person['name'] -> Option C

4. Quick Check:

   Access value by key = person['name'] [OK]

Hint: Use square brackets with the key string to get value [OK]

Common Mistakes:

• Using dot notation like person.name (not valid for dict)
• Using a value instead of key inside brackets
• Accessing a different key than asked

2. Which of the following is the correct syntax to safely get the value for key 'city' from dictionary data without causing an error if the key does not exist?

easy

A. data['city']

B. data.get('city')

C. data.city

D. data['City']

Solution

1. Step 1: Understand safe access in dictionaries

   Using data['city'] causes an error if the key is missing. Using data.get('city') returns None instead.

2. Step 2: Identify correct method for safe access

   The get() method is designed to safely access keys without errors.

3. Final Answer:

   data.get('city') -> Option B

4. Quick Check:

   Safe key access = data.get('city') [OK]

Hint: Use get() to avoid errors if key missing [OK]

Common Mistakes:

• Using square brackets which raise KeyError if key missing
• Using dot notation which is invalid for dict
• Using wrong key case causing KeyError

3. What will be the output of this code?

info = {'a': 1, 'b': 2, 'c': 3}
print(info['b'])

medium

A. 2

B. 1

C. 3

D. KeyError

Solution

1. Step 1: Identify the dictionary and key accessed

   The dictionary info has key 'b' with value 2.

2. Step 2: Access the value using the key

   The code prints info['b'], which is 2.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   info['b'] = 2 [OK]

Hint: Look up the key's value directly in the dictionary [OK]

Common Mistakes:

• Confusing key with value
• Expecting KeyError when key exists
• Mixing up keys and values

4. The following code causes an error. What is the problem?

data = {'x': 10, 'y': 20}
print(data['z'])

medium

A. TypeError because keys must be integers

B. SyntaxError due to wrong brackets

C. No error, prints 0

D. KeyError because 'z' is not in the dictionary

Solution

1. Step 1: Check if key exists in dictionary

   The key 'z' is not present in data.

2. Step 2: Understand error caused by missing key

   Accessing a missing key with square brackets raises a KeyError.

3. Final Answer:

   KeyError because 'z' is not in the dictionary -> Option D

4. Quick Check:

   Missing key access = KeyError [OK]

Hint: Check if key exists before accessing or use get() [OK]

Common Mistakes:

• Assuming missing keys return 0 or None
• Confusing syntax error with runtime error
• Using wrong bracket types

5. Given the dictionary grades = {'Alice': 85, 'Bob': 92, 'Charlie': 78}, which code snippet correctly prints Bob's grade or 'Not found' if Bob is not in the dictionary?

hard

A. print(grades.get('Bob', 'Not found'))

B. print(grades['Bob'] or 'Not found')

C. print(grades['bob'] or 'Not found')

D. print(grades.get('Bob'))

Solution

1. Step 1: Understand the goal

   We want to print Bob's grade if present, otherwise print 'Not found'.

2. Step 2: Analyze each option

   print(grades['Bob'] or 'Not found') will raise KeyError if key missing because access happens before 'or'. print(grades.get('Bob', 'Not found')) uses get() with default value, which is concise and safe. print(grades['bob'] or 'Not found') uses wrong key case and will fail. print(grades.get('Bob')) prints None if key missing, not 'Not found'.

3. Final Answer:

   print(grades.get('Bob', 'Not found')) -> Option A

4. Quick Check:

   Use get() with default for safe access [OK]

Hint: Use get(key, default) to handle missing keys easily [OK]

Common Mistakes:

• Using wrong key case causing missing key
• Not providing default value in get()
• Assuming or operator works for missing keys