Python Dictionary Notes: Keys, Values, Methods and Looping

A dictionary maps keys to values

A dictionary (dict) stores pairs: a key you look up, and the value it leads to. Think of a phone book: the name is the key, the number is the value.

student = {
    "name": "Asha",
    "class": 10,
    "marks": 91,
}
print(student["name"])     # Asha

Dictionaries use curly brackets {} and key: value pairs.
Keys must be unique and unchangeable (usually strings or numbers).
Values can be anything, including lists and other dictionaries.
Since Python 3.7 a dictionary remembers the order items were added.

Reading values safely

print(student["city"])           # KeyError: 'city'
print(student.get("city"))       # None, no error
print(student.get("city", "NA")) # NA -- a default value
print("marks" in student)        # True -- 'in' checks the keys

Add, update and delete

student["city"] = "Ranchi"     # new key: adds it
student["marks"] = 95          # existing key: updates it
student.update({"class": 11, "house": "Blue"})

del student["house"]           # delete a key
marks = student.pop("marks")   # delete and return the value

Looping through a dictionary

prices = {"tea": 10, "samosa": 15, "coffee": 20}

for item in prices:                  # keys
    print(item)

for price in prices.values():        # values
    print(price)

for item, price in prices.items():   # both together
    print(f"{item:<8} Rs {price}")

Method	Gives
`keys()`	all the keys
`values()`	all the values
`items()`	`(key, value)` pairs
`get(k, default)`	the value, or `default` if `k` is missing
`pop(k)`	removes `k` and returns its value
`update(other)`	adds or overwrites pairs from `other`

Counting with a dictionary

Counting how often each thing appears is one of the most common uses of a dictionary.

votes = ["Asha", "Ravi", "Asha", "Meena", "Asha", "Ravi"]
count = {}
for name in votes:
    count[name] = count.get(name, 0) + 1
print(count)       # {'Asha': 3, 'Ravi': 2, 'Meena': 1}

winner = max(count, key=count.get)
print("Winner:", winner)

count.get(name, 0) gives 0 the first time a name appears, so the line works for new and old names alike.

Nested data: a list of dictionaries

A list of dictionaries works like a table: each dictionary is one row.

students = [
    {"name": "Asha", "marks": 91},
    {"name": "Ravi", "marks": 78},
    {"name": "Meena", "marks": 85},
]
for s in students:
    print(s["name"], s["marks"])

topper = max(students, key=lambda s: s["marks"])
print("Topper:", topper["name"])

Nested dictionaries

report = {
    "Asha": {"maths": 95, "science": 88},
    "Ravi": {"maths": 72, "science": 80},
}
print(report["Asha"]["science"])     # 88
for name, subjects in report.items():
    print(name, sum(subjects.values()))

Example: phone book search

phone_book = {"asha": "98765 43210", "ravi": "91234 56780"}
name = input("Search name: ").strip().lower()
print(phone_book.get(name, "Not found"))

Common mistakes

Reading a missing key with d[key] and getting a KeyError; use get() when the key may be missing.
Expecting to find a value with in: 91 in student checks the keys, not the values. Use 91 in student.values().
Using a list as a key: TypeError: unhashable type: 'list'.
Changing the size of a dictionary while looping over it.

Key points

A dictionary stores key: value pairs; keys are unique.
d[k] = v adds or updates; del and pop remove.
Loop with items() to get keys and values together.
d.get(k, 0) + 1 is the standard counting pattern.

Practice questions

1. What does {"a": 1, "a": 2} become?

{"a": 2}. Keys are unique, so the second value replaces the first.

2. Count the letters in the word "banana".

count = {}
for ch in "banana":
    count[ch] = count.get(ch, 0) + 1
print(count)     # {'b': 1, 'a': 3, 'n': 2}

3. Print every item from prices that costs more than Rs 12.

for item, price in prices.items():
    if price > 12:
        print(item)

4. What is the difference between d["x"] and d.get("x") when "x" is missing?

d["x"] raises a KeyError; d.get("x") returns None (or the default you give).

5. From a list of student dictionaries, make a list of only the names.

names = [s["name"] for s in students]