Session 9 · Dictionaries: Annachi Kadai#

📖 Based on the blog post Task: Annachi Kadai, Python Dictionary.

The big idea#

A dictionary stores key → value pairs. Instead of asking “what’s in position 2?”, you ask “what’s the price of dal?”. Lookups by key are instant, even with millions of entries.

Everyday example: the annachi kadai (the corner grocery shop). Annachi doesn’t search shelf by shelf; he knows rice → ₹60, dal → ₹120. The item name is the key, the price is the value.

Annachi Kadai's price list A dictionary maps keys to values. You look up a value by its key, not by position. Keys are unique; assigning to an existing key replaces its value. prices = { "rice" 60 "dal" 120 "oil" 180 } prices["dal"] → 120 prices["sugar"] = 45 → add prices["dal"] = 125 → update prices.get("tea", 0) → 0 prices["tea"] → KeyError
A dictionary maps keys to values. You look up a value by its key, not by position. Keys are unique; assigning to an existing key replaces its value.

Creating and reading#

prices = {"rice": 60, "dal": 120, "oil": 180}
print(prices["dal"])
print(len(prices))
print("oil" in prices, "tea" in prices)     # `in` checks the KEYS

120
3
True False

Asking for a missing key with [] raises KeyError. get() returns a default instead:

prices = {"rice": 60, "dal": 120}
print(prices.get("tea"))          # None
print(prices.get("tea", 0))       # your default

None
0

Adding, updating, removing#

prices = {"rice": 60, "dal": 120, "oil": 180}
prices["sugar"] = 45          # new key → added
prices["dal"] = 125           # existing key → updated
del prices["oil"]             # removed
removed = prices.pop("rice")  # removed, and you get the value back
print(prices, removed)

{'dal': 125, 'sugar': 45} 60

Keys are unique: assigning to a key that exists replaces its value. Keys must be immutable (strings, numbers, tuples); values can be anything, even lists or other dictionaries.

Looping#

prices = {"rice": 60, "dal": 120}
for item, price in prices.items():
    print(f"{item:>5}: ₹{price}")
print(list(prices.keys()), list(prices.values()))

 rice: ₹60
  dal: ₹120
['rice', 'dal'] [60, 120]

Since Python 3.7, a dictionary remembers the order you inserted keys.

Counting things: a classic pattern#

fruits = ["apple", "banana", "apple", "orange", "banana", "banana"]
count = {}
for fruit in fruits:
    count[fruit] = count.get(fruit, 0) + 1
print(count)

{'apple': 2, 'banana': 3, 'orange': 1}

Common mistakes#

  • d["missing"] raises KeyError. Use get() or check in first.

  • Lists as keys: {[1, 2]: "x"} raises TypeError (unhashable). Use a tuple.

  • Changing size while looping (for k in d: del d[k]) raises RuntimeError. Loop over list(d) instead.

  • copy() is shallow: nested dictionaries inside are still shared. Use copy.deepcopy() for those.

Hands-on exercises#

The tasks from Task: Annachi Kadai. Each solution sets up what it needs so you can run it alone.

Task 1. Create student with name “Alice”, age 21, major “Computer Science” and print it.

Solution 
student = {"name": "Alice", "age": 21, "major": "Computer Science"}
print(student)

Task 2. Print the values for "name" and "major".

Solution 
student = {"name": "Alice", "age": 21, "major": "Computer Science"}
print(student["name"], student["major"])

Task 3. Add "gpa": 3.8, then update "age" to 22.

Solution 
student = {"name": "Alice", "age": 21, "major": "Computer Science"}
student["gpa"] = 3.8
student["age"] = 22
print(student)

Task 4. Remove "major" with del and print the dictionary.

Solution 
student = {"name": "Alice", "age": 22, "major": "Computer Science"}
del student["major"]
print(student)

Task 5. Check whether the key "age" exists.

Solution 
student = {"name": "Alice", "age": 22}
print("age" in student)      # True

Task 6. Create prices = {"apple": 0.5, "banana": 0.3, "orange": 0.7} and print each pair.

Solution 
prices = {"apple": 0.5, "banana": 0.3, "orange": 0.7}
for fruit, price in prices.items():
    print(fruit, price)

Task 7. Print how many pairs prices has.

Solution 
prices = {"apple": 0.5, "banana": 0.3, "orange": 0.7}
print(len(prices))     # 3

Task 8. Use get() for "gpa", and for "graduation_year" with a default of 2025.

Solution 
student = {"name": "Alice", "gpa": 3.8}
print(student.get("gpa"))                     # 3.8
print(student.get("graduation_year", 2025))   # 2025 (key doesn't exist)

Task 9. Merge extra_info = {"graduation_year": 2025, "hometown": "Springfield"} into student with update().

Solution 
student = {"name": "Alice", "gpa": 3.8}
extra_info = {"graduation_year": 2025, "hometown": "Springfield"}
student.update(extra_info)
print(student)

Task 10. Use a dictionary comprehension to map 1–5 to their squares.

Solution 
squares = {n: n ** 2 for n in range(1, 6)}
print(squares)     # {1: 1, 2: 4, 3: 9, 4: 16, 5: 25}

Task 11. Print the keys and values of prices as two separate lists.

Solution 
prices = {"apple": 0.5, "banana": 0.3, "orange": 0.7}
print(list(prices.keys()))
print(list(prices.values()))

Task 12. Create school with two nested students and print the age of "student2".

Solution 
school = {
    "student1": {"name": "Alice", "age": 21},
    "student2": {"name": "Bob", "age": 22},
}
print(school["student2"]["age"])     # 22

Task 13. Use setdefault() to add "advisor": "Dr. Smith" only if it’s missing.

Solution 
student = {"name": "Alice"}
student.setdefault("advisor", "Dr. Smith")
student.setdefault("advisor", "Dr. Jones")     # ignored: the key exists now
print(student)     # {'name': 'Alice', 'advisor': 'Dr. Smith'}

Task 14. pop() the "hometown" key into a variable and print it.

Solution 
student = {"name": "Alice", "hometown": "Springfield"}
hometown = student.pop("hometown")
print(hometown, student)

Task 15. Empty prices with clear().

Solution 
prices = {"apple": 0.5, "banana": 0.3}
prices.clear()
print(prices)      # {}

Task 16. copy() the student, change the copy’s name to “Charlie”, print both.

Solution 
student = {"name": "Alice", "age": 22}
clone = student.copy()
clone["name"] = "Charlie"
print(student)     # {'name': 'Alice', 'age': 22}
print(clone)       # {'name': 'Charlie', 'age': 22}

Task 17. Build a dictionary from keys = ["name", "age", "major"] and values = ["Eve", 20, "Mathematics"] with zip().

Solution 
keys = ["name", "age", "major"]
values = ["Eve", 20, "Mathematics"]
print(dict(zip(keys, values)))

Task 18. Loop over student with items().

Solution 
student = {"name": "Eve", "age": 20}
for key, value in student.items():
    print(f"{key}: {value}")

Task 19. Count the fruits in ["apple", "banana", "apple", "orange", "banana", "banana"].

Solution 
fruits = ["apple", "banana", "apple", "orange", "banana", "banana"]
fruit_count = {}
for fruit in fruits:
    fruit_count[fruit] = fruit_count.get(fruit, 0) + 1
print(fruit_count)     # {'apple': 2, 'banana': 3, 'orange': 1}

Task 20. Count words with collections.defaultdict.

Solution 
from collections import defaultdict

word_count = defaultdict(int)          # missing keys start at int() = 0
for word in ["hello", "world", "hello", "python"]:
    word_count[word] += 1
print(dict(word_count))     # {'hello': 2, 'world': 1, 'python': 1}

collections.Counter(words) does the same in one line.