Creating and Accessing Dictionaries

A dictionary stores data as key-value pairs inside curly braces {}. Each key maps to a value, like a real dictionary maps words to definitions. Keys must be unique and immutable (strings, numbers).

# Creating a dictionary
student = {
    "name": "Alice",
    "age": 16,
    "year": 11,
    "subjects": ["Maths", "English", "Science"]
}

# Accessing values
print(student["name"])         # Alice
print(student["age"])          # 16
print(student.get("email", "Not provided"))  # Safe access with default

# Adding and updating
student["email"] = "[email protected]"   # Add new key
student["age"] = 17                     # Update existing key
print(student)

Tip: Use .get(key, default) to avoid errors when a key might not exist.

Dictionary Methods and Iteration

Dictionaries have powerful built-in methods for working with keys, values, and items. You can loop through dictionaries in several ways.

scores = {"Maths": 85, "English": 92, "Science": 78}

# Useful methods
print(scores.keys())    # dict_keys(['Maths', 'English', 'Science'])
print(scores.values())  # dict_values([85, 92, 78])
print(scores.items())   # dict_items([('Maths', 85), ...])

# Iteration
for subject in scores:
    print(f"{subject}: {scores[subject]}")

# Better: iterate over key-value pairs
for subject, mark in scores.items():
    print(f"{subject}: {mark}%")

# Removing items
removed = scores.pop("Science")  # Removes and returns value
print(f"Removed Science: {removed}")
print(f"Remaining: {scores}")

Nested Dictionaries

Dictionaries can contain other dictionaries as values. This creates nested structures useful for organising complex data like student records or contact databases.

# Nested dictionary: class of students
classroom = {
    "Alice": {"age": 16, "grade": "A", "score": 92},
    "Bob": {"age": 17, "grade": "B", "score": 81},
    "Charlie": {"age": 16, "grade": "A", "score": 95}
}

# Accessing nested values
print(classroom["Alice"]["score"])    # 92
print(classroom["Bob"]["grade"])      # B

# Loop through all students
for name, info in classroom.items():
    print(f"{name} (age {info['age']}): {info['grade']} - {info['score']}%")

Key Vocabulary

Dictionary

A collection of key-value pairs enclosed in curly braces {}.

Key

A unique identifier used to look up a value, like "name" in a student record.

Value

The data associated with a key. Can be any data type including lists or other dictionaries.

Nested Dictionary

A dictionary that contains another dictionary as one of its values.

Worked Examples

1

Count letter frequency in a word

word = "mississippi"
frequency = {}

for letter in word:
    if letter in frequency:
        frequency[letter] += 1
    else:
        frequency[letter] = 1

for letter, count in frequency.items():
    print(f"'{letter}': {count}")

Step 1: We start with an empty dictionary and loop through each letter.

Step 2: If the letter exists as a key, increment its count. Otherwise, set it to 1.

2

Create a phone book with add and search

phone_book = {}

# Adding contacts
phone_book["Alice"] = "0412 345 678"
phone_book["Bob"] = "0423 456 789"
phone_book["Charlie"] = "0434 567 890"

# Searching
search_name = "Bob"
if search_name in phone_book:
    print(f"{search_name}: {phone_book[search_name]}")
else:
    print(f"{search_name} not found")

# List all contacts
print(f"\nAll contacts ({len(phone_book)}):")
for name, number in sorted(phone_book.items()):
    print(f"  {name}: {number}")

Step 1: We use names as keys and phone numbers as values.

Step 2: Use in to check if a key exists before accessing it.

3

Calculate class averages from a nested dictionary

students = {
    "Alice": {"maths": 85, "english": 92, "science": 88},
    "Bob": {"maths": 76, "english": 81, "science": 79},
    "Charlie": {"maths": 95, "english": 89, "science": 91}
}

for name, subjects in students.items():
    scores = subjects.values()
    avg = sum(scores) / len(scores)
    print(f"{name}: average = {avg:.1f}%")

Step 1: Each student has a nested dictionary of subject scores.

Step 2: .values() gives us all scores, which we can sum and average.

Knowledge Check

Select the correct answer for each question. Click "Check Answer" to see if you are right.

Question 1

What does {"a": 1, "b": 2}["b"] return?

2 1 "b" {"b": 2}

Question 2

Which method returns all keys from a dictionary?

.values() .keys() .items() .get()

Question 3

What happens if you access a key that does not exist using dict["missing_key"]?

Returns None Returns 0 Raises a KeyError Returns an empty string

Question 4

How do you add a new key-value pair "colour": "blue" to an existing dictionary d?

d.add("colour", "blue") d.append("colour": "blue") d.insert("colour", "blue") d["colour"] = "blue"

Question 5

Which method iterates over both keys AND values together?

.items() .values() .keys() .pairs()

Python Dictionaries