Dictionaries
Dictionaries store key-value pairs. Keys must be unique and hashable.
# Creating dicts
empty = {}
person = {"name": "Alice", "age": 30, "city": "NYC"}
d = dict(name="Bob", age=25)
from_pairs = dict([("a", 1), ("b", 2)])
# Type check
print(type(person)) # <class 'dict'>
Accessing and Modifying
person = {"name": "Alice", "age": 30}
# Access
print(person["name"]) # Alice
print(person.get("age")) # 30
print(person.get("job", "N/A"))# N/A (default if key missing)
# Modify
person["age"] = 31 # update existing
person["email"] = "a@b.com" # add new key
# Delete
del person["email"]
removed = person.pop("age") # returns removed value
person.clear() # empty the dict
Dictionary Methods
d = {"a": 1, "b": 2, "c": 3}
print(d.keys()) # dict_keys(['a', 'b', 'c'])
print(d.values()) # dict_values([1, 2, 3])
print(d.items()) # dict_items([('a', 1), ('b', 2), ('c', 3)])
# Iteration
for key in d:
print(key, d[key])
for key, value in d.items():
print(f"{key} → {value}")
# Merge dicts
d1 = {"a": 1, "b": 2}
d2 = {"c": 3, "b": 99}
merged = {**d1, **d2} # {'a':1, 'b':99, 'c':3} — d2 wins
d1.update(d2) # in-place merge
Dictionary Comprehensions
# {key_expr: val_expr for item in iterable if condition}
squares = {x: x**2 for x in range(1, 6)}
# {1: 1, 2: 4, 3: 9, 4: 16, 5: 25}
filtered = {k: v for k, v in squares.items() if v > 9}
# {4: 16, 5: 25}
inverted = {v: k for k, v in {"a": 1, "b": 2}.items()}
# {1: 'a', 2: 'b'}
defaultdict and Counter
from collections import defaultdict, Counter
# defaultdict — no KeyError on missing keys
word_count = defaultdict(int)
words = "the cat sat on the mat the cat".split()
for word in words:
word_count[word] += 1
print(dict(word_count))
# Counter — counts hashable items
counter = Counter(words)
print(counter.most_common(2)) # [('the', 3), ('cat', 2)]
Sets
Sets are unordered collections of unique elements.
# Creating sets
empty = set() # NOT {} — that creates an empty dict!
numbers = {1, 2, 3, 4, 5}
from_list = set([1, 2, 2, 3, 3, 3]) # {1, 2, 3} — duplicates removed
print(type(numbers)) # <class 'set'>
Set Operations
a = {1, 2, 3, 4, 5}
b = {3, 4, 5, 6, 7}
print(a | b) # {1,2,3,4,5,6,7} — union
print(a & b) # {3,4,5} — intersection
print(a - b) # {1,2} — difference (in a, not in b)
print(a ^ b) # {1,2,6,7} — symmetric difference
# Methods
a.add(6)
a.remove(1) # raises KeyError if missing
a.discard(99) # safe remove (no error if missing)
a.update({10, 11}) # add multiple
# Subset and superset
{1, 2}.issubset({1, 2, 3}) # True
{1, 2, 3}.issuperset({1, 2}) # True
{1, 2}.isdisjoint({3, 4}) # True (no common elements)
Frozensets
Immutable sets (can be dict keys or set elements):
fs = frozenset([1, 2, 3])
d = {fs: "immutable set as key!"}
Exercises
Exercise 1: Word Frequency
Count the frequency of each character in "abracadabra".
Solution:
from collections import Counter
freq = Counter("abracadabra")
print(freq.most_common())
# [('a', 5), ('b', 2), ('r', 2), ('c', 1), ('d', 1)]
Exercise 2: Anagram Check
Two words are anagrams if they contain the same characters. Check if "listen" and "silent" are anagrams.
Solution:
def is_anagram(s1, s2):
return sorted(s1.lower()) == sorted(s2.lower())
# OR: Counter(s1) == Counter(s2)
print(is_anagram("listen", "silent")) # True
print(is_anagram("hello", "world")) # False
Exercise 3: Group By
Given a list of words, group them by their first letter using a defaultdict.
Solution:
from collections import defaultdict
words = ["apple", "banana", "avocado", "blueberry", "cherry", "apricot"]
grouped = defaultdict(list)
for word in words:
grouped[word[0]].append(word)
for letter, group in sorted(grouped.items()):
print(f"{letter}: {group}")
# a: ['apple', 'avocado', 'apricot']
# b: ['banana', 'blueberry']
# c: ['cherry']
Exercise 4: Set Operations
Find names that appear in list A but NOT in list B.
Solution:
a = {"Alice", "Bob", "Carol", "Dave"}
b = {"Bob", "Dave", "Eve"}
only_in_a = a - b
print(only_in_a) # {'Alice', 'Carol'}