Lesson 14 of 21 · about 6 minutes with the exercises

Dictionaries

A dictionary stores values under names. In a real dictionary you look up a word and find its meaning. In a Python dictionary you look up a key and find its value. Use one whenever your data is a set of labelled facts: a student's details, the price of each item, the number of votes for each choice.

Making a dictionary

Use curly brackets. Each entry is a key, a colon and a value.

student = {"name": "Asha", "class": 8, "city": "Ludhiana"}
print(student)
print(len(student))

Output

{'name': 'Asha', 'class': 8, 'city': 'Ludhiana'}
3

Keys are usually strings. They must be unique: a second entry with the same key replaces the first. Values can be anything, including lists and other dictionaries.

Reading a value

Put the key in square brackets.

student = {"name": "Asha", "class": 8, "city": "Ludhiana"}
print(student["name"])
print(student["class"] + 1)

Output

Asha
9

Asking for a key that is not there stops the program with a KeyError.

student = {"name": "Asha", "class": 8}
print(student["phone"])

What Python says

Traceback (most recent call last):
  File "example.py", line 2, in <module>
    print(student["phone"])
          ~~~~~~~^^^^^^^^^
KeyError: 'phone'

The get method is the safe way to read. It returns None, or a default you choose, when the key is missing.

student = {"name": "Asha", "class": 8}
print(student.get("phone"))
print(student.get("phone", "not given"))
print("phone" in student)

Output

None
not given
False

Adding, changing and deleting

Assigning to a key adds it if it is new and changes it if it exists. There is no separate append.

prices = {"tea": 10, "samosa": 15}
prices["lassi"] = 40
prices["tea"] = 12
print(prices)
del prices["samosa"]
print(prices)
removed = prices.pop("lassi")
print(removed, prices)

Output

{'tea': 12, 'samosa': 15, 'lassi': 40}
{'tea': 12, 'lassi': 40}
40 {'tea': 12}

Looping

Looping over a dictionary gives its keys. Use .items() to get the key and value together, which is what you want most of the time.

prices = {"tea": 12, "samosa": 15, "lassi": 40}
for item in prices:
    print(item)
for item, price in prices.items():
    print(f"{item:<8}Rs {price}")
print(sum(prices.values()))

Output

tea
samosa
lassi
tea     Rs 12
samosa  Rs 15
lassi   Rs 40
67

.keys() and .values() give just one side. A dictionary keeps its entries in the order you added them.

Counting with a dictionary

Counting how often each thing appears is a job dictionaries do well. get(key, 0) reads the current count, or 0 the first time.

votes = ["tea", "coffee", "tea", "lassi", "tea", "coffee"]
count = {}
for v in votes:
    count[v] = count.get(v, 0) + 1
print(count)

Output

{'tea': 3, 'coffee': 2, 'lassi': 1}

To find the winner, ask max to compare the keys by their counts.

count = {"tea": 3, "coffee": 2, "lassi": 1}
winner = max(count, key=count.get)
print(winner, count[winner])

Output

tea 3

Sorting a dictionary

sorted on a dictionary sorts its keys. To sort by value, sort the items with a key function.

marks = {"Ravi": 91, "Asha": 78, "Meena": 85}
for name in sorted(marks):
    print(name, marks[name])
by_marks = sorted(marks.items(), key=lambda pair: pair[1], reverse=True)
print(by_marks)

Output

Asha 78
Meena 85
Ravi 91
[('Ravi', 91), ('Meena', 85), ('Asha', 78)]

A list of dictionaries

Most real data looks like this: a list of records, where each record is a dictionary with the same keys. If you later read spreadsheets or data from websites, this is the shape you will get.

students = [
    {"name": "Asha", "marks": 78},
    {"name": "Ravi", "marks": 91},
    {"name": "Meena", "marks": 85},
]
for s in students:
    if s["marks"] >= 80:
        print(s["name"])
average = sum(s["marks"] for s in students) / len(students)
print(round(average, 1))

Output

Ravi
Meena
84.7

A dictionary inside a dictionary

school = {
    "Asha": {"class": 8, "house": "Blue"},
    "Ravi": {"class": 9, "house": "Red"},
}
print(school["Ravi"]["house"])

Output

Red

Read it left to right: first pick Ravi's record, then pick the house from it.

Common mistakes

Using square brackets to read a key that may be missing. Use get or check with in.
Forgetting the quotation marks around a string key: student[name] looks for a variable called name.
Expecting two entries with the same key. The later one wins.

Try it yourself

Exercise 1. Make a dictionary of three fruits and their prices per kg. Print the price of one fruit, then add a fourth fruit and print how many there are.

Show one answer

fruit = {"mango": 120, "banana": 50, "apple": 180}
print(fruit["banana"])
fruit["guava"] = 80
print(len(fruit))

Output

50
4

Exercise 2. Count how many times each letter appears in the word "banana".

Show one answer

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

Output

{'b': 1, 'a': 3, 'n': 2}

Exercise 3. A shop's stock is {"pen": 40, "book": 0, "bag": 7, "ink": 0}. Print the names of the items that are out of stock.

Show one answer

stock = {"pen": 40, "book": 0, "bag": 7, "ink": 0}
for item, qty in stock.items():
    if qty == 0:
        print(item)

Output

book
ink

Examples on this page were run with Python 3.14 on 12 November 2026. The output shown is the real output.

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