Task – Python Dictionary

1.Create a dictionary named student with the following keys and values. and print the same
"name": "Alice"
"age": 21
"major": "Computer Science"

CODE:

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

OUTPUT:

{'name': 'Alice', 'age': 21, 'major': 'Computer Science'}

EXPLANATION:

A dictionary stores data as key-value pairs

Keys: name, age, major

2.Using the student dictionary, print the values associated with the keys "name" and "major".

CODE:

print(student["name"])
print(student["major"])

OUTPUT:

Alice
Computer Science

EXPLANATION:

Use keys to access values
3.Add a new key-value pair to the student dictionary: "gpa": 3.8. Then update the "age" to 22.

CODE:

student["gpa"] = 3.8
student["age"] = 22
print(student)

OUTPUT:

{'name': 'Alice', 'age': 22, 'major': 'Computer Science', 'gpa': 3.8}

EXPLANATION:

New key added

Existing key updated

4.Remove the key "major" from the student dictionary using the del statement. Print the dictionary to confirm the removal.

CODE:

del student["major"]
print(student)

OUTPUT:

{'name': 'Alice', 'age': 22, 'gpa': 3.8}

EXPLANATION:

del removes a key-value pair
5.Check if the key "age" exists in the student dictionary. Print True or False based on the result.

CODE:

print("age" in student)

OUTPUT:

True

EXPLANATION:

in checks if key exists
6.Create a dictionary prices with three items, e.g., "apple": 0.5, "banana": 0.3, "orange": 0.7. Iterate over the dictionary and print each key-value pair.

CODE:

`prices = {"apple": 0.5, "banana": 0.3, "orange": 0.7}

for key, value in prices.items():
print(key, value)`

OUTPUT:

apple 0.5
banana 0.3
orange 0.7

EXPLANATION:

.items() gives key-value pairs
7.Use the len() function to find the number of key-value pairs in the prices dictionary. Print the result.

CODE:

print(len(prices))

OUTPUT:

3

EXPLANATION:

len() counts key-value pairs
8.Use the get() method to access the "gpa" in the student dictionary. Try to access a non-existing key, e.g., "graduation_year", with a default value of 2025.

CODE:

print(student.get("gpa"))
print(student.get("graduation_year", 2025))

OUTPUT:

3.8
2025

EXPLANATION:

get() avoids errors

Default value used if key missing

9.Create another dictionary extra_info with the following keys and values. Also merge extra_info into the student dictionary using the update() method.
"graduation_year": 2025
"hometown": "Springfield"

CODE:

`extra_info = {
"graduation_year": 2025,
"hometown": "Springfield"
}

student.update(extra_info)
print(student)`

OUTPUT:

{'name': 'Alice', 'age': 22, 'gpa': 3.8, 'graduation_year': 2025, 'hometown': 'Springfield'}

EXPLANATION:

update() merges dictionaries
10.Create a dictionary squares where the keys are numbers from 1 to 5 and the values are the squares of the keys. Use dictionary comprehension.

CODE:

squares = {x: x**2 for x in range(1, 6)}
print(squares)

OUTPUT:

{1: 1, 2: 4, 3: 9, 4: 16, 5: 25}

EXPLANATION:

Creates dictionary dynamically
11.Using the prices dictionary, print the keys and values as separate lists using the keys() and values() methods.

CODE:

print(list(prices.keys()))
print(list(prices.values()))

OUTPUT:

['apple', 'banana', 'orange']
[0.5, 0.3, 0.7]

EXPLANATION:

.keys() and .values() extract data
12.Create a dictionary school with two nested dictionaries. Access and print the age of "student2".
"student1": {"name": "Alice", "age": 21}
"student2": {"name": "Bob", "age": 22}

CODE:

`school = {
"student1": {"name": "Alice", "age": 21},
"student2": {"name": "Bob", "age": 22}
}

print(school["student2"]["age"])`

OUTPUT:

22

EXPLANATION:

Access nested dictionary using multiple keys
13.Use the setdefault() method to add a new key "advisor" with the value "Dr. Smith" to the student dictionary if it does not exist.

CODE:

student.setdefault("advisor", "Dr. Smith")
print(student)

OUTPUT:

{'name': 'Alice', 'age': 22, 'gpa': 3.8, 'graduation_year': 2025, 'hometown': 'Springfield', 'advisor': 'Dr. Smith'}

EXPLANATION:

Adds key only if it doesn’t exist