This Python dictionary exercise aims to help Python developers to learn and practice dictionary operations. All questions are tested on Python 3.
Python dictionary is a mutable object, and it contains the data in the form of key-value pairs. Each key is separated from its value by a colon (:).
Dictionary is the most widely used data structure, and it is necessary to understand its methods and operations.
This Python dictionary exercise includes the following: –
- It contains 10 dictionary questions and solutions provided for each question.
- Practice different dictionary assignments, programs, and challenges.
It covers questions on the following topics:
- Dictionary operations and manipulations
- Dictionary functions
- Dictionary comprehension
Exercise 1: Convert two lists into a dictionary
There are the two lists. Write a Python program to convert them into a dictionary in a way that item from list1 is the key and item from list2 is the values.
Example
Input
keys = ['Ten', 'Twenty', 'Thirty'] values = [10, 20, 30]
Output
{'Ten': 10, 'Twenty': 20, 'Thirty': 30}
Hint:
- Use the
zip()function. This function takes two or more iterables (like list, dict, string), aggregates them in a tuple, and returns it. - Or, Iterate the list using a for loop and range() function. In each iteration, add a new key-value pair to a dict using the
update()method
Solution:
Solution 1: The zip() function and a dict() constructor
- Use the
zip(keys, values)to aggregate two lists. - Wrap the result of a
zip()function into adict()constructor.
keys = ['Ten', 'Twenty', 'Thirty'] values = [10, 20, 30] res_dict = dict(zip(keys, values)) print(res_dict)
Solution 2: Using a loop and update() method of a dictionary
keys = ['Ten', 'Twenty', 'Thirty']
values = [10, 20, 30]
# empty dictionary
res_dict = dict()
for i in range(len(keys)):
res_dict.update({keys[i]: values[i]})
print(res_dict)
Exercise 2: Merge two Python dictionaries into one
Example
Input
dict1 = {'Ten': 10, 'Twenty': 20, 'Thirty': 30}
dict2 = {'Thirty': 30, 'Fourty': 40, 'Fifty': 50}
Output
{'Ten': 10, 'Twenty': 20, 'Thirty': 30, 'Fourty': 40, 'Fifty': 50}
Solution:
#Python 3.5
dict1 = {'Ten': 10, 'Twenty': 20, 'Thirty': 30}
dict2 = {'Thirty': 30, 'Fourty': 40, 'Fifty': 50}
dict3 = {**dict1, **dict2}
print(dict3)
# Another version of Python:
dict1 = {'Ten': 10, 'Twenty': 20, 'Thirty': 30}
dict2 = {'Thirty': 30, 'Fourty': 40, 'Fifty': 50}
dict3 = dict1.copy()
dict3.update(dict2)
print(dict3)
Exercise 3: Print the value of key ‘history’ from the below dict
Example
Input
sampleDict = {
"class": {
"student": {
"name": "Mike",
"marks": {
"physics": 70,
"history": 80
}
}
}
}
Output
80
Hint:
- It is a nested dict. Use the correct chaining of keys to locate the specified key-value pair.
Solution:
sampleDict = {
"class": {
"student": {
"name": "Mike",
"marks": {
"physics": 70,
"history": 80
}
}
}
}
# understand how to located the nested key
# sampleDict['class'] = {'student': {'name': 'Mike', 'marks': {'physics': 70, 'history': 80}}}
# sampleDict['class']['student'] = {'name': 'Mike', 'marks': {'physics': 70, 'history': 80}}
# sampleDict['class']['student']['marks'] = {'physics': 70, 'history': 80}
# solution
print(sampleDict['class']['student']['marks']['history'])
Exercise 4: Initialize dictionary with default values
In Python, we can initialize the keys with the same values.
Example
Input
employees = ['Kelly', 'Emma']
defaults = {"designation": 'Developer', "salary": 8000}
Output
{'Kelly': {'designation': 'Developer', 'salary': 8000}, 'Emma': {'designation': 'Developer', 'salary': 8000}}
Hint:
- Use the
fromkeys()method of dict.
Solution:
The fromkeys() method returns a dictionary with the specified keys and the specified value.
employees = ['Kelly', 'Emma']
defaults = {"designation": 'Developer', "salary": 8000}
res = dict.fromkeys(employees, defaults)
print(res)
# Individual data
print(res["Kelly"])
Exercise 5: Create a dictionary by extracting the keys from a given dictionary
Write a Python program to create a new dictionary by extracting the mentioned keys from the below dictionary.
Example
Input
sample_dict = {
"name": "Kelly",
"age": 25,
"salary": 8000,
"city": "New york"}
# Keys to extract
keys = ["name", "salary"]
Output
{'name': 'Kelly', 'salary': 8000}
Hint:
- Iterate the mentioned keys using a loop
- Next, check if the current key is present in the dictionary, if it is present, add it to the new dictionary
Solution:
Solution 1: Dictionary Comprehension
sampleDict = {
"name": "Kelly",
"age":25,
"salary": 8000,
"city": "New york" }
keys = ["name", "salary"]
newDict = {k: sampleDict[k] for k in keys}
print(newDict)
Solution 2: Using the update() method and loop
sample_dict = {
"name": "Kelly",
"age": 25,
"salary": 8000,
"city": "New york"}
# keys to extract
keys = ["name", "salary"]
# new dict
res = dict()
for k in keys:
# add current key with its va;ue from sample_dict
res.update({k: sample_dict[k]})
print(res)
Exercise 6: Delete a list of keys from a dictionary
Example
Input
sample_dict = {
"name": "Kelly",
"age": 25,
"salary": 8000,
"city": "New york"
}
# Keys to remove
keys = ["name", "salary"]
Output
{'city': 'New york', 'age': 25}
Hint:
- Iterate the mentioned keys using a loop
- Next, check if the current key is present in the dictionary, if it is present, remove it from the dictionary
- To achieve the above result, we can use the dictionary comprehension or the
pop()method of a dictionary.
Solution:
Solution 1: Using the pop() method and loop
sample_dict = {
"name": "Kelly",
"age": 25,
"salary": 8000,
"city": "New york"
}
# Keys to remove
keys = ["name", "salary"]
for k in keys:
sample_dict.pop(k)
print(sample_dict)
Solution 2: Dictionary Comprehension
sample_dict = {
"name": "Kelly",
"age": 25,
"salary": 8000,
"city": "New york"
}
# Keys to remove
keys = ["name", "salary"]
sample_dict = {k: sample_dict[k] for k in sample_dict.keys() - keys}
print(sample_dict)
Exercise 7: Check if a value exists in a dictionary
We know how to check if the key exists in a dictionary. Sometimes it is required to check if the given value is present.
Write a Python program to check if value 200 exists in the following dictionary.
Example
Input
sample_dict = {'a': 100, 'b': 200, 'c': 300}
Output
200 present in a dict
Hint:
- Get all values of a dict in a list using the
values()method. - Next, use the if condition to check if 200 is present in the given list
Solution:
sample_dict = {'a': 100, 'b': 200, 'c': 300}
if 200 in sample_dict.values():
print('200 present in a dict')
Exercise 8: Rename key of a dictionary
Write a program to rename a key city to a location in the following dictionary.
Example
Input
sample_dict = {
"name": "Kelly",
"age":25,
"salary": 8000,
"city": "New york"
}
Output
{'name': 'Kelly', 'age': 25, 'salary': 8000, 'location': 'New york'}
Hint:
- Remove the city from a given dictionary
- Add a new key (location) into a dictionary with the same value
Solution:
sample_dict = {
"name": "Kelly",
"age": 25,
"salary": 8000,
"city": "New york"
}
sample_dict['location'] = sample_dict.pop('city')
print(sample_dict)
Exercise 9: Get the key of a minimum value from the following dictionary
Example
Input
sample_dict = {
'Physics': 82,
'Math': 65,
'history': 75
}
Output
Math
Hint:
- Use the built-in function
min()
Solution:
sample_dict = {
'Physics': 82,
'Math': 65,
'history': 75
}
print(min(sample_dict, key=sample_dict.get))
Exercise 10: Change value of a key in a nested dictionary
Write a Python program to change Brad’s salary to 8500 in the following dictionary.
Example
Input
sample_dict = {
'emp1': {'name': 'Jhon', 'salary': 7500},
'emp2': {'name': 'Emma', 'salary': 8000},
'emp3': {'name': 'Brad', 'salary': 500}
}
Output
{
'emp1': {'name': 'Jhon', 'salary': 7500},
'emp2': {'name': 'Emma', 'salary': 8000},
'emp3': {'name': 'Brad', 'salary': 8500}
}
Solution:
sample_dict = {
'emp1': {'name': 'Jhon', 'salary': 7500},
'emp2': {'name': 'Emma', 'salary': 8000},
'emp3': {'name': 'Brad', 'salary': 6500}
}
sample_dict['emp3']['salary'] = 8500
print(sample_dict)