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
14.Use the pop() method to remove the "hometown" key from the student dictionary and store its value in a variable. Print the variable.
CODE:
value = student.pop("hometown")
print(value)
OUTPUT:
Springfield
EXPLANATION:
- pop() removes and returns value
15.Use the clear() method to remove all items from the prices dictionary. Print the dictionary to confirm it’s empty.
CODE:
prices.clear()
print(prices)
OUTPUT:
{}
EXPLANATION:
- Removes all items
16.Make a copy of the student dictionary using the copy() method. Modify the copy by changing "name" to "Charlie". Print both dictionaries to see the differences.
CODE:
`copy_student = student.copy()
copy_student["name"] = "Charlie"
print(student)
print(copy_student)`
OUTPUT:
{'name': 'Alice', ...}
{'name': 'Charlie', ...}
EXPLANATION:
- Copy creates separate dictionary
17.Create two lists: keys = ["name", "age", "major"] and values = ["Eve", 20, "Mathematics"]. Use the zip() function to create a dictionary from these lists.
CODE:
`keys = ["name", "age", "major"]
values = ["Eve", 20, "Mathematics"]
new_dict = dict(zip(keys, values))
print(new_dict)`
OUTPUT:
{'name': 'Eve', 'age': 20, 'major': 'Mathematics'}
EXPLANATION:
- zip() pairs elements
18.Use the items() method to iterate over the student dictionary and print each key-value pair.
CODE:
for k, v in student.items():
print(k, v)
OUTPUT:
name Alice
age 22
...
EXPLANATION:
- Iterates over key-value pairs
19.Given a list of fruits: ["apple", "banana", "apple", "orange", "banana", "banana"], create a dictionary fruit_count that counts the occurrences of each fruit.
CODE:
`fruits = ["apple", "banana", "apple", "orange", "banana", "banana"]
fruit_count = {}
for f in fruits:
fruit_count[f] = fruit_count.get(f, 0) + 1
print(fruit_count)`
OUTPUT:
{'apple': 2, 'banana': 3, 'orange': 1}
EXPLANATION:
- Counts occurrences using dictionary
20.Use collections.defaultdict to create a dictionary word_count that counts the number of occurrences of each word in a list: ["hello", "world", "hello", "python"].
CODE:
`from collections import defaultdict
words = ["hello", "world", "hello", "python"]
word_count = defaultdict(int)
for w in words:
word_count[w] += 1
print(dict(word_count))`
OUTPUT:
{'hello': 2, 'world': 1, 'python': 1}
EXPLANATION:
- defaultdict(int) sets default value to 0