Dictionary Operations & Advanced Patterns

Why read this? Real programs use dictionaries for grouping, counting, caching, and representing structured data. This topic covers the patterns and tools that make dictionary usage elegant and efficient.

🎯 Learning Objectives

1. Work with nested dictionaries and merge them
2. Use defaultdict and Counter from collections module
3. Use dictionary methods: setdefault(), update()
4. Sort dictionaries by keys or values
5. Use dictionaries for memoization (caching)

📋 Prerequisites

• Dictionaries

📖 Core Content

20.1 Nested Dictionaries

python
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# runnable
users = {
    "alice": {"age": 25, "city": "NYC", "scores": [85, 92]},
    "bob": {"age": 30, "city": "LA", "scores": [70, 88]},
    "charlie": {"age": 22, "city": "Chicago", "scores": [95, 91]}
}
print(users["alice"]["city"])        # NYC
print(users["bob"]["scores"][0])     # 70
# Iterate nested dict
for username, info in users.items():
    avg_score = sum(info["scores"]) / len(info["scores"])
    print(f"{username}: {avg_score:.1f}%")

20.2 Dictionary Merge and Update

python
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# runnable
d1 = {"a": 1, "b": 2}
d2 = {"c": 3, "d": 4}
# Merge (Python 3.9+)
merged = d1 | d2
print(merged)  # {'a': 1, 'b': 2, 'c': 3, 'd': 4}
# Update (modifies in place)
d1.update(d2)
print(d1)  # {'a': 1, 'b': 2, 'c': 3, 'd': 4}
# Both methods handle overlapping keys (last wins)

20.3 setdefault() — Get or Set Default

python
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# runnable
# Without setdefault
data = {}
words = ["apple", "banana", "apple", "cherry", "banana"]
for word in words:
    if word not in data:
        data[word] = 0
    data[word] += 1
# With setdefault (elegant)
data2 = {}
for word in words:
    data2.setdefault(word, 0)
    data2[word] += 1

20.4 defaultdict — Automatic Default Values

python
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# runnable
from collections import defaultdict
# Default value is int (0)
freq = defaultdict(int)
for word in ["apple", "banana", "apple", "cherry"]:
    freq[word] += 1
print(dict(freq))  # {'apple': 2, 'banana': 1, 'cherry': 1}
# Default value is list
groups = defaultdict(list)
for name in ["Alice", "Bob", "Charlie", "Alice", "Bob"]:
    groups[name].append(len(name))
print(dict(groups))

20.5 Counter — Counting Made Easy

python
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# runnable
from collections import Counter
words = ["apple", "banana", "apple", "cherry", "banana", "apple"]
counter = Counter(words)
print(counter)               # Counter({'apple': 3, 'banana': 2, 'cherry': 1})
print(counter.most_common(2)) # [('apple', 3), ('banana', 2)]
print(counter["grape"])       # 0 (no error for missing keys!)

20.6 Sorting Dictionaries

python
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# runnable
grades = {"Alice": 85, "Bob": 72, "Charlie": 90, "Diana": 78}
# Sort by key
for name in sorted(grades):
    print(f"{name}: {grades[name]}")
# Sort by value (ascending)
for name in sorted(grades, key=grades.get):
    print(f"{name}: {grades[name]}")
# Sort by value (descending)
for name in sorted(grades, key=grades.get, reverse=True):
    print(f"{name}: {grades[name]}")

20.7 Worked Example: Group by First Letter

python
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# runnable
from collections import defaultdict
names = ["Alice", "Bob", "Charlie", "David", "Eve", "Anna", "Ben"]
groups = defaultdict(list)
for name in names:
    groups[name[0]].append(name)
print(dict(groups))
# {'A': ['Alice', 'Anna'], 'B': ['Bob', 'Ben'], 'C': ['Charlie'], 'D': ['David'], 'E': ['Eve']}

⚠️ Common Pitfalls

Pitfall 1: Forgetting defaultdict Factory

Pitfall 2: Nested Key Access Without Checking

Pitfall 3: Mutating Default Values

📝 Practice Questions

Q1: What does Counter("mississippi").most_common(3) return?

Answer: [('i', 4), ('s', 4), ('p', 2)] (i and s tied at 4 each) Q2: Write code to invert a dictionary (swap keys and values).

Answer:

python
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original = {"a": 1, "b": 2, "c": 3}
inverted = {v: k for k, v in original.items()}

Q3-10: Additional advanced dict questions.

(Following pattern.)

🔗 Cross-References

• Next Topic: Functions
• Previous Topic: Dictionaries
• Reference: Python for Everybody, Chapter 9 (Sections 9.3-9.4)
• Video: L59: More on dictionaries Join Discord Previous19. DictionariesNext21. Functions