PythonDictionary.org - Master Python Dictionaries

Challenges

Word Frequency Counter

Easy

Dictionary Merge

Easy

Nested Dictionary Access

Medium

Dictionary Comprehension

Medium

Challenge: Word Frequency Counter

Create a function that counts the frequency of each word in a text string and returns the result as a dictionary where keys are the unique words found in the text and values are the number of times each word appears.

Requirements:

Convert all words to lowercase
Remove punctuation from words
Ignore empty strings

Example:

text = "Hello world! Hello Python."
result = word_frequency(text)
print(result)
# Output: {'hello': 2, 'world': 1, 'python': 1}

Visualization Explained:

The visualization on the right shows your dictionary as it's being built. Each key-value pair is represented, making it easy to see how your dictionary evolves as your code executes.

Dictionary Resources

Dictionary Methods

Complete reference of all Python dictionary methods with examples

Dictionary Performance

Understanding time complexity and performance optimization

Dictionary Best Practices

Coding standards and best practices for using dictionaries

Advanced Dictionary Techniques

Learn about defaultdict, Counter, and OrderedDict

Dictionary Methods Reference

dict.get(key[, default])

Returns the value for key if key is in the dictionary, else default.

user = {'name': 'John', 'age': 30}
print(user.get('email', 'Not found'))

dict.items()

Returns a view object that displays a list of dictionary's (key, value) tuple pairs.

user = {'name': 'John', 'age': 30}
for key, value in user.items():
    print(key, value)

dict.keys()

Returns a view object that displays a list of all the keys.

user = {'name': 'John', 'age': 30}
print(list(user.keys()))

dict.values()

Returns a view object that displays a list of all the values.

user = {'name': 'John', 'age': 30}
print(list(user.values()))

dict.update([other])

Updates the dictionary with the key/value pairs from other.

user = {'name': 'John'}
user.update({'age': 30, 'city': 'NYC'})
print(user)

dict.pop(key[, default])

Removes and returns the value for key. If key is not found, default is returned.

user = {'name': 'John', 'age': 30}
age = user.pop('age')
email = user.pop('email', None)

Common Dictionary Patterns

Dictionary Comprehension

A concise way to create dictionaries using an expression.

# Create a dictionary of squares
numbers = [1, 2, 3, 4, 5]
squares = {num: num**2 for num in numbers}
print(squares)
# {1: 1, 2: 4, 3: 9, 4: 16, 5: 25}

# Dictionary comprehension with condition
even_squares = {num: num**2 for num in numbers if num % 2 == 0}
print(even_squares)
# {2: 4, 4: 16}

Counting with Dictionaries

Using dictionaries to count occurrences (the basis of the current challenge).

# Method 1: Manual counting
text = "apple banana apple orange banana apple"
counts = {}
for word in text.split():
    counts[word] = counts.get(word, 0) + 1
print(counts)
# {'apple': 3, 'banana': 2, 'orange': 1}

# Method 2: Using collections.Counter
from collections import Counter
counts = Counter(text.split())
print(dict(counts))
# {'apple': 3, 'banana': 2, 'orange': 1}

Nested Dictionaries

Using dictionaries to represent hierarchical data structures.

# Creating a nested dictionary
person = {
    'name': 'John',
    'age': 30,
    'address': {
        'city': 'New York',
        'zip': '10001',
        'coordinates': {
            'lat': 40.7128,
            'lng': -74.0060
        }
    },
    'skills': ['Python', 'JavaScript']
}

# Accessing nested values
print(person['address']['city'])  # New York
print(person['address']['coordinates']['lat'])  # 40.7128

# Safe access with get() for nested dictionaries
city = person.get('address', {}).get('city', 'Unknown')
print(city)  # New York

Dictionary Merging

Different ways to combine dictionaries.

# Method 1: Using update()
dict1 = {'a': 1, 'b': 2}
dict2 = {'b': 3, 'c': 4}
merged = dict1.copy()  # Create a copy to avoid modifying original
merged.update(dict2)
print(merged)  # {'a': 1, 'b': 3, 'c': 4}

# Method 2: Using unpacking (Python 3.5+)
dict1 = {'a': 1, 'b': 2}
dict2 = {'b': 3, 'c': 4}
merged = {**dict1, **dict2}  # Later values override earlier ones
print(merged)  # {'a': 1, 'b': 3, 'c': 4}

# Method 3: Using | operator (Python 3.9+)
dict1 = {'a': 1, 'b': 2}
dict2 = {'b': 3, 'c': 4}
merged = dict1 | dict2  # Similar to unpacking
print(merged)  # {'a': 1, 'b': 3, 'c': 4}