List vs Tuple vs Set vs Dictionary

Python's four built-in collections overlap, and choosing the right one affects correctness, readability and speed. Should a list of emails be a list or a set? Should a record be a tuple or a dict? This lesson compares lists, tuples, sets and dictionaries side by side — mutability, ordering, duplicates, access patterns and performance — with guidelines and examples for choosing.

Side-by-Side Comparison

The key properties of each collection:

• List [1, 2] — ordered, mutable, duplicates allowed, access by index. Use for sequences that change.
• Tuple (1, 2) — ordered, immutable, duplicates allowed, hashable. Use for fixed records and dict keys.
• Set {1, 2} — unordered, mutable, unique items, fast membership. Use for uniqueness and set algebra.
• Dict {"a": 1} — ordered by insertion, mutable, unique keys, fast lookup by key. Use for mappings and records with named fields.

Performance

Membership tests (x in c) are O(n) for lists and tuples but O(1) average for sets and dict keys. Appending to a list is O(1), inserting at the front is O(n) (use collections.deque). Tuples are slightly smaller and faster to create than lists.

Rules of Thumb

Ordered items that change → list. Fixed group of values → tuple (or NamedTuple/dataclass for named fields). "Is this item present?" or "remove duplicates" → set. "Look up X by Y" → dict. List vs dict: if you often search a list for an item by an id, convert it to a dict keyed by id.

Examples

marks_list = [88, 72, 88, 95]
marks_tuple = (88, 72, 88, 95)
marks_set = {88, 72, 88, 95}
marks_dict = {"Asha": 88, "Ravi": 72, "Meera": 88, "Kiran": 95}

print(marks_list[0], marks_tuple[-1], sorted(marks_set), marks_dict["Kiran"])
print(len(marks_list), len(marks_tuple), len(marks_set), len(marks_dict))

marks_list.append(60)
marks_set.add(60)
marks_dict["Zoya"] = 60
print(marks_list, sorted(marks_set), marks_dict)

88 95 [72, 88, 95] 95
4 4 3 4
[88, 72, 88, 95, 60] [60, 72, 88, 95] {'Asha': 88, 'Ravi': 72, 'Meera': 88, 'Kiran': 95, 'Zoya': 60}

users = [{"id": i, "name": f"user{i}"} for i in range(1, 50_001)]

def find_in_list(user_id):
    for u in users:
        if u["id"] == user_id:
            return u

users_by_id = {u["id"]: u for u in users}

print(find_in_list(49_999)["name"])
print(users_by_id[49_999]["name"])      # O(1) instead of scanning 50,000 items

user49999
user49999