Python Practice Set: Data Structures (Related Questions)

Test your ability to choose and use the right data structure (Lists, Tuples, Sets, and Dictionaries) for different scenarios.

1. Comparison & Selection

Q1: Which data structure would you use to store a collection of unique user IDs where order doesn't matter?

Answer: A Set, because sets automatically handle uniqueness and are optimized for membership testing.

Q2: Which data structure is best for storing a fixed geometric coordinate (x, y) that should not be modified?

Answer: A Tuple, because it is immutable and clearly signals that the data should remain constant.

Q3: Match the data structure to its syntax and properties: | Structure | Syntax | Properties | | :--- | :--- | :--- | | List | [] | Ordered, Changeable, Allows Duplicates | | Tuple | () | Ordered, Unchangeable, Allows Duplicates | | Set | {} | Unordered, Unindexed, No Duplicates | | Dictionary | {k:v} | Ordered (3.7+), Changeable, Key-Value Pairs |

2. Conversions & Operations

Q4: How do you quickly remove all duplicates from a list?

Answer: Convert the list to a set and then back to a list. unique_list = list(set(original_list))

Q5: Predict the output:

x = [1, 2, 3]
y = x
y.append(4)
print(x)

Output: [1, 2, 3, 4] (Note: Lists are objects; y = x creates a reference, not a copy. Changing y changes x)

Q6: What is the fastest way to check if an item exists in a large collection?

Answer: Using a Set or Dictionary keys. Membership testing (item in collection) is $O(1)$ for sets/dicts but $O(n)$ for lists.

3. Advanced Challenges

Q7: Explain the difference between list.sort() and sorted(list).

Answer: list.sort() sorts the list in-place and returns None. sorted(list) returns a new sorted list, leaving the original unchanged.

Q8: Give an example of a dictionary where the values are lists.

grades = {
    "Alice": [85, 90, 88],
    "Bob": [70, 75, 80]
}

4. Coding Challenges

Task 1: Frequency Counter

Write a function that counts the frequency of each word in a string using a dictionary.

def count_frequency(text):
    words = text.split()
    freq = {}
    for word in words:
        freq[word] = freq.get(word, 0) + 1
    return freq

print(count_frequency("apple banana apple cherry banana apple"))
# Output: {'apple': 3, 'banana': 2, 'cherry': 1}

Task 2: List of Tuples to Dictionary

Convert a list of tuples [("id", 1), ("name", "Tanis"), ("role", "Admin")] into a dictionary.

data = [("id", 1), ("name", "Tanis"), ("role", "Admin")]
result = dict(data)
print(result) # {'id': 1, 'name': 'Tanis', 'role': 'Admin'}

Task 3: Find Common Elements

Find common elements between two lists using set operations.

list1 = [1, 2, 3, 4, 5]
list2 = [4, 5, 6, 7, 8]

common = list(set(list1) & set(list2))
print(common) # [4, 5]

Task 4: Nested Access

Given the following structure, print the "city" of the second employee.

company = {
    "employees": [
        {"name": "Alice", "location": {"city": "New York"}},
        {"name": "Bob", "location": {"city": "San Francisco"}}
    ]
}

print(company["employees"][1]["location"]["city"])