🐍 Python📅 Day 14

Sets in Python

Store unique items with sets, remove duplicates in one step, and combine groups with union, intersection and difference.

🎯

Learning objectives

→Create sets and remove duplicates from a list
→Add and remove items with add(), update(), remove() and discard()
→Combine sets with union, intersection, difference and symmetric difference
→Check subset, superset and disjoint relationships

💡 Key points

A set is an unordered collection of unique items: {3, 1, 2}. Duplicates are dropped automatically.
{} creates an empty dictionary, not a set. Use set() for an empty set.
Sets have no positions, so s[0] and slicing raise TypeError. Loop over a set or use sorted(s) for a fixed order.
Items must be immutable (numbers, strings, tuples), but the set itself can grow and shrink.
remove(x) raises KeyError if x is missing; discard(x) quietly does nothing.
| union (in either), & intersection (in both), - difference (in the first only), ^ symmetric difference (in exactly one).
x in s is fast, because sets use the same hashing idea as dictionary keys.

💻 Code examples(5)

#1Creating sets and removing duplicates

python

nums = {4, 2, 7, 2, 4}
print(nums)
print(len(nums))

roll_nos = [5, 3, 5, 1, 3, 5]
unique = set(roll_nos)
print(unique)
print(sorted(unique))       # a sorted list

empty = set()
print(type(empty), type({}))

Output

{2, 4, 7}
3
{1, 3, 5}
[1, 3, 5]
<class 'set'> <class 'dict'>

set(list) is the quickest way to remove duplicates. sorted() turns the set back into an ordered list.

#2Adding and removing items

python

s = {10, 20, 30}
s.add(40)
s.add(20)               # already there: no change
print(len(s))

s.discard(99)           # missing: no error
s.remove(10)
# s.remove(99)          # KeyError: 99
print(sorted(s))

s.update([50, 60])      # add several at once
print(sorted(s))
s.clear()
print(s)

Output

4
[20, 30, 40]
[20, 30, 40, 50, 60]
set()

add() takes one item and update() takes a list (or any collection). An empty set prints as set(), never as {}.

#3Set operations with operators

python

maths = {"Amit", "Priya", "Ravi", "Neha"}
cs = {"Priya", "Neha", "Karan"}

print(sorted(maths | cs))   # in either club
print(sorted(maths & cs))   # in both
print(sorted(maths - cs))   # maths only
print(sorted(maths ^ cs))   # in exactly one

Output

['Amit', 'Karan', 'Neha', 'Priya', 'Ravi']
['Neha', 'Priya']
['Amit', 'Ravi']
['Amit', 'Karan', 'Ravi']

sorted() is used because a set of strings can print in a different order every time you run the program.

#4Method forms and subset checks

python

a = {1, 2, 3, 4, 5}
b = {4, 5, 6}
print(a.union(b))
print(a.intersection(b))
print(a.difference(b))
print(a.symmetric_difference(b))

small = {1, 2}
print(small.issubset(a), a.issuperset(small))
print(a.isdisjoint({7, 8}))   # nothing shared

Output

{1, 2, 3, 4, 5, 6}
{4, 5}
{1, 2, 3}
{1, 2, 3, 6}
True True
True

Each operator has a method with a readable name. The methods also accept a list, for example a.union([7, 8]).

#5Common letters and repeated entries

python

word1, word2 = "school", "college"
common = set(word1) & set(word2)
print("Common letters:", sorted(common))

entries = [101, 104, 101, 107, 104, 110]
seen = set()
repeats = set()
for roll in entries:
    if roll in seen:
        repeats.add(roll)
    seen.add(roll)
print("Entered twice:", sorted(repeats))
print("Unique entries:", len(seen))

Output

Common letters: ['c', 'l', 'o']
Entered twice: [101, 104]
Unique entries: 4

set(word) breaks a string into its unique letters. Keeping a 'seen' set while looping is a standard way to spot duplicates.

🎯 Practice

Q1. What is the output of print(len({1, 2, 2, 3, 3, 3}))?+

3. The set keeps only 1, 2 and 3.

Q2. How do you create an empty set?+

s = set(). Writing s = {} creates an empty dictionary.

Q3. cricket = {'Amit', 'Ravi', 'Neha'} and football = {'Ravi', 'Karan'}. Who plays cricket but not football?+

print(cricket - football) # Amit and Neha (order may vary)

Q4. Why does s[0] fail for a set?+

Sets are unordered, so their items have no positions. Loop over the set, or convert it first: sorted(s)[0].

Q5. What is the difference between remove() and discard()?+

Both delete an item. If the item is not in the set, remove() raises KeyError but discard() does nothing.

📝 Notes

The four collections side by side

You have now met all four built-in collections:

list [1, 2, 3] — ordered, changeable, duplicates allowed. Reach items by position.
tuple (1, 2, 3) — ordered, cannot change, duplicates allowed.
dict {"a": 1} — key–value pairs, changeable, keys are unique.
set {1, 2, 3} — unordered, changeable, every item unique.

Pick a set when you only care whether something is present, or when you need to remove duplicates or compare two groups.

Why the order looks strange

Python stores set items according to their hash value, not the order you typed them. For small whole numbers this often looks sorted, but for strings the order can change from one run of the program to the next. Never rely on the order of a set; call sorted() when you need one.

frozenset

A frozenset is a set that cannot change, just as a tuple is a list that cannot change:

vowels = frozenset("aeiou")
print("e" in vowels)        # True
# vowels.add("y")           # AttributeError

Because it is immutable, a frozenset can be a dictionary key or an item inside another set.

Common mistakes

Using {} for an empty set — that is a dict.
Putting a list inside a set: {[1, 2]} raises TypeError: unhashable type: 'list'. Use a tuple: {(1, 2)}.
Expecting a fixed order when printing.
Using add() with a list: s.add([4, 5]) fails. Use s.update([4, 5]).

Next: Day 15 — Modules and Packages, for reusing code that you (or Python) have already written.