Dictionaries and Sets

Topic — Two collections that aren't accessed by position. A dict looks things up by key; a set only answers "is this in here?".

	Access by	Ordered	Duplicates
dict	key	insertion order	keys ❌, values ✅
set	membership only	❌	❌

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Dictionaries

A dict stores key: value pairs.

student = {"Name": "Vettri", "Age": 28, "Department": "CSE", "CGPA": 8.9}
print(list(student.values()))

Output

['Vettri', 28, 'CSE', 8.9]

Reading values

d = {"a": 1, "b": 2, "c": 3}
print(d["b"])              # → 2
print(d.get("z"))          # → None
print(d.get("z", 0))       # → 0
print(list(d.keys()))      # → ['a', 'b', 'c']
print(list(d.values()))    # → [1, 2, 3]
print(list(d.items()))     # → [('a', 1), ('b', 2), ('c', 3)]
print(len(d))              # → 3

Output

2
None
0
['a', 'b', 'c']
[1, 2, 3]
[('a', 1), ('b', 2), ('c', 3)]
3

Note .items() gives you a list of tuples — which is why for k, v in d.items(): unpacks cleanly.

[] vs .get()

A missing key with [] is an error:

d = {"a": 1}
print(d["z"])

Error

KeyError: 'z'

print(d.get("z"))      # → None    no error
print(d.get("z", 0))   # → 0       your chosen default
print("z" in d)        # → False   just checking

Form	Missing key
d["z"]	raises KeyError
d.get("z")	returns None
d.get("z", 0)	returns 0
"z" in d	returns False

tip

Use [] when the key must exist — you want the error if it doesn't. Use .get() when absence is normal and you have a sensible default.

Adding, changing, removing

Dicts are mutable. There's no separate "add" and "update" — assigning to a key does both, depending on whether it already exists.

student = {"Name": "Vettri", "Age": 28, "Department": "CSE", "CGPA": 8.9}

student["College"] = "SRM"      # new key → added
print(student)

student["CGPA"] = 9.1           # existing key → updated
print(student)

del student["College"]
print(student)

popped = student.pop("Age")
print("popped", popped, student)

Output

{'Name': 'Vettri', 'Age': 28, 'Department': 'CSE', 'CGPA': 8.9, 'College': 'SRM'}
{'Name': 'Vettri', 'Age': 28, 'Department': 'CSE', 'CGPA': 9.1, 'College': 'SRM'}
{'Name': 'Vettri', 'Age': 28, 'Department': 'CSE', 'CGPA': 9.1}
popped 28 {'Name': 'Vettri', 'Department': 'CSE', 'CGPA': 9.1}

.update() merges another dict in, overwriting on collision. .setdefault() reads a key, inserting a default only if it's absent:

e = {"a": 1}
e.update({"b": 2, "a": 99})
print(e)                        # → {'a': 99, 'b': 2}   'a' overwritten
print(e.setdefault("c", 3), e)  # → 3  — 'c' was missing, so inserted
print(e.setdefault("a", 100), e) # → 99 — 'a' existed, default ignored

Output

{'a': 99, 'b': 2}
3 {'a': 99, 'b': 2, 'c': 3}
99 {'a': 99, 'b': 2, 'c': 3}

Iterating

Looping a dict directly gives keys:

d = {"name": "Vettri", "age": 28}

for k in d:
    print(k)

for v in d.values():
    print(v)

for k, v in d.items():
    print(f"{k}: {v}")

Output

name
age
Vettri
28
name: Vettri
age: 28

Nesting and comprehensions

nd = {"person": {"name": "Alice", "age": 30}}
print(nd["person"]["name"])   # → Alice

Output

Alice

print({x: x * x for x in range(1, 5)})
print({k: v for k, v in {"a": 1, "b": 2, "c": 3}.items() if v > 1})

Output

{1: 1, 2: 4, 3: 9, 4: 16}
{'b': 2, 'c': 3}

Keys must be immutable

for key in ["name", 10, (1, 2), 3.14]:
    print(f"  {key!r:<10} legal")

{[1, 2]: "x"}

Output

  'name'     legal
  10         legal
  (1, 2)     legal
  3.14       legal

Error

TypeError: cannot use 'list' as a dict key (unhashable type: 'list')

Candidate	Legal key?	Why
"name"	✅	str is immutable
10	✅	int is immutable
(1, 2)	✅	tuple is immutable
3.14	✅	float is immutable
[1, 2]	❌	list is mutable → unhashable

Why the rule exists

A dict finds a value by computing a hash of the key and using it to pick a storage slot. If the key could change after insertion, its hash would change, and the value would be stranded in a slot nobody looks in again. Python forbids the situation instead of letting it happen.

(1, 2) is fine — but a tuple containing a list is not, for the same reason tuple immutability is shallow.

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Sets

A set holds unique, unordered values.

Duplicates vanish on creation

s = {1, 2, 2, 3, 3, 3, 4}
print(s)        # → {1, 2, 3, 4}
print(len(s))   # → 4

Output

{1, 2, 3, 4}
4

Seven values in, four out. A set stores each value by its hash, so a repeat lands in a slot that's already occupied and is simply not stored again.

print(sorted(set([3, 1, 2, 1])))   # → [1, 2, 3]

This is the standard way to deduplicate a list — though it loses the original order.

Set order is not something you can rely on

Sets of small integers often print in ascending order, which makes it look ordered. It isn't — that's a side effect of how ints hash. Sets of strings print in a different order on every run, because Python randomises string hashing per process. Use sorted() when order matters.

No indexing

{1, 2, 3}[0]

Error

TypeError: 'set' object is not subscriptable

There is no "first" element, so there's nothing for [0] to mean. Convert to a list first if you need positions.

The empty-set trap

print(type({}).__name__)       # → dict
print(type(set()).__name__)    # → set

Output

dict
set

{} is an empty dict. The only way to write an empty set is set().

Set algebra

This is what sets are really for.

a = {1, 2, 3, 4}
b = {3, 4, 5, 6}

print(a | b)   # → {1, 2, 3, 4, 5, 6}   union — everything
print(a & b)   # → {3, 4}               intersection — in both
print(a - b)   # → {1, 2}               difference — in a only
print(a ^ b)   # → {1, 2, 5, 6}         symmetric difference — not in both

Output

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

Each operator has a method form — a.union(b), a.intersection(b), a.difference(b), a.symmetric_difference(b) — which read better when the other side isn't already a set.

Comparison operators test containment:

print({1, 2} <= {1, 2, 3})           # → True    subset
print({1, 2, 3} >= {1, 2})           # → True    superset
print({1, 5}.isdisjoint({2, 3}))     # → True    nothing in common

Output

True
True
True

Changing a set

c = {"red", "green"}
c.add("blue")
print(sorted(c))        # → ['blue', 'green', 'red']

c.discard("nope")       # not there — no error
print(sorted(c))

c.remove("nope")        # KeyError: 'nope'

Output

['blue', 'green', 'red']
['blue', 'green', 'red']

.discard() is silent on a missing value; .remove() raises KeyError. Pick whichever matches your intent.

Membership is the point

Checking x in some_set is dramatically faster than x in some_list for large collections — a hash lookup instead of scanning every element. If your code does a lot of "have I seen this already?", a set is the right container.

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Booleans in conditions

Comparisons produce bool, which is what if consumes:

x, y = 10, 10
print(x == y)   # → True
print(x != y)   # → False

Output

True
False

Any value can act as a condition — 0, "", [], {}, None and False are falsy, everything else is truthy. Some of those results surprise people (bool(" ") and bool("False") are both True); the full table is in Values and Types.

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Common Mistakes

#	Mistake	Wrong	Right
1	{} for an empty set	it's a dict	set()
2	Indexing a set	s[0] → TypeError	convert: list(s)[0]
3	Relying on set order	varies per run for strings	sorted(s)
4	d["missing"] when absence is normal	KeyError	d.get("missing", default)
5	List as a dict key	TypeError: unhashable	use a tuple
6	.remove() on a maybe-absent value	KeyError	.discard()
7	Expecting a set to keep duplicates	{1, 1, 2} has 2 items	use a list
8	True and 1 as separate set items	{1, True} has 1 item	True == 1, so they collide

That last one follows from bool being a subclass of int — see the boolean surprise.

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Summary

dict

Operation	Syntax
Create	{"k": v} or dict(k=v)
Read	d["k"] / d.get("k", default)
Add or update	d["k"] = v
Remove	del d["k"] / d.pop("k")
Merge	d.update(other)
Keys / values / pairs	.keys() / .values() / .items()
Test a key	"k" in d
Build	{k: v for ... in ...}

set

Operation	Syntax
Create	{1, 2, 3} — empty is set()
Add / remove	.add(x) / .discard(x) / .remove(x)
Union / intersection	a | b / a & b
Difference / symmetric	a - b / a ^ b
Subset / superset	a <= b / a >= b
Deduplicate a list	set(lst)

Key takeaways

• d["k"] raises on a missing key; .get("k", default) doesn't
• Assigning to a key adds it or updates it — there's no separate operation
• .items() yields tuples, which is what makes for k, v in d.items(): work
• Keys must be immutable (hashable) — str, int, float, tuple; never list
• {} is an empty dict; set() is the only way to write an empty set
• Sets drop duplicates and have no order or indexing — sorted() when you need order
• |, &, -, ^ are union, intersection, difference, symmetric difference
• x in some_set is far faster than x in some_list at scale

See also: Tuples for why tuples make valid keys · Lists for the ordered mutable alternative · Data Types for dict and set in context

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Run It Yourself

dicts_and_sets.py

# --- dict ---
student = {"Name": "Vettri", "Age": 28, "Department": "CSE", "CGPA": 8.9}
print(f"{'dict':<14} {student}")
print(f"{'values':<14} {list(student.values())}")
print(f"{'get Name':<14} {student.get('Name')}")
print(f"{'get missing':<14} {student.get('College', 'not set')}")

student["College"] = "SRM"
student["CGPA"] = 9.1
print(f"{'after edits':<14} {student}")

for key, value in student.items():
    print(f"  {key:<12} {value}")

# --- set ---
marks = [85, 90, 75, 90, 85, 95]
unique = set(marks)
print(f"{'raw list':<14} {marks}")
print(f"{'deduplicated':<14} {sorted(unique)}")
print(f"{'count':<14} {len(marks)} -> {len(unique)}")

a = {1, 2, 3, 4}
b = {3, 4, 5, 6}
print(f"{'union':<14} {sorted(a | b)}")
print(f"{'intersection':<14} {sorted(a & b)}")
print(f"{'difference':<14} {sorted(a - b)}")
print(f"{'symmetric':<14} {sorted(a ^ b)}")

Output

dict           {'Name': 'Vettri', 'Age': 28, 'Department': 'CSE', 'CGPA': 8.9}
values         ['Vettri', 28, 'CSE', 8.9]
get Name       Vettri
get missing    not set
after edits    {'Name': 'Vettri', 'Age': 28, 'Department': 'CSE', 'CGPA': 9.1, 'College': 'SRM'}
  Name         Vettri
  Age          28
  Department   CSE
  CGPA         9.1
  College      SRM
raw list       [85, 90, 75, 90, 85, 95]
deduplicated   [75, 85, 90, 95]
count          6 -> 4
union          [1, 2, 3, 4, 5, 6]
intersection   [3, 4]
difference     [1, 2]
symmetric      [1, 2, 5, 6]

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Practice Questions

From the Unit 1 question bank. Tags and marks are explained on the Python index.

Unit 1 § G — Dictionaries, Sets & Booleans

G1. [PROG] Create a dictionary holding Name, Age, Department, CGPA. Print all the values. [3]

G2. [PROG] Using that dictionary: add a new key "College", then update the CGPA, then display the final dictionary. [3]

G3. [OUT] [5]

d = {"a": 1, "b": 2, "c": 3}
print(d["b"])
print(d.get("z"))
print(d.get("z", 0))
print(list(d.keys()))
print(list(d.values()))
print(list(d.items()))
print(len(d))

G4. [OUT] Name the error, and say which line of G3 would have avoided it. [2]

d = {"a": 1}
print(d["z"])

G5. [OUT] [3]

s = {1, 2, 2, 3, 3, 3, 4}
print(s)
print(len(s))

Explain in one sentence why the duplicates disappeared.

G6. [OUT] [4]

a = {1, 2, 3, 4}
b = {3, 4, 5, 6}
print(a | b)
print(a & b)
print(a - b)
print(a ^ b)

G7. [PROG] Accept two numbers from the user and print whether they are equal, using a Boolean expression. [3]

G8. [OUT] Truthiness. Two of these are True and will surprise you. [5]

print(bool(0))
print(bool(1))
print(bool(-1))
print(bool(""))
print(bool(" "))
print(bool([]))
print(bool([0]))
print(bool("False"))
print(bool(None))

G9. [THEORY] A dictionary key must be of an immutable type. Explain why, and say which of these are legal as keys: "name", 10, (1, 2), [1, 2], 3.14. [3]

Viva

1. Why do duplicates vanish from a set?