Python Dictionary Exercise with Solutions

Exercises with Solutions for the Python Dictionary

The purpose of this Python dictionary exercise is to teach and practice dictionary operations for Python developers. Python 3 is used to test each question.

The data are presented as key-value pairs in the mutable Python dictionary object. A colon separates each key’s value from each key. ( : ).

The most used data structure is the dictionary, so it is important to comprehend how it works. The following are included in this Python dictionary exercise:

–

It contains 10 dictionary questions and solutions provided for each question.
Practice different dictionary assignments, programs, and challenges.

It addresses queries regarding the following

topics:

Dictionary operations and manipulations
Dictionary functions
Dictionary comprehension

Each question’s solution helps you become better acquainted with the Python dictionary. If you have any other suggestions, please let us know. It’ll assist others.

developers.

Use

Online Code Editor

to solve exercise questions

.
Read the

complete guide to

Python dictionaries

to solve this exercise

a table

contents

Exercise 1: Convert two lists into a dictionary
Exercise 2: Merge two Python dictionaries into one
Exercise 3: Print the value of key ‘history’ from the below dict
Exercise 4: Initialize dictionary with default values
Exercise 5: Create a dictionary by extracting the keys from a given dictionary
Exercise 6: Delete a list of keys from a dictionary
Exercise 7: Check if a value exists in a dictionary
Exercise 8: Rename key of a dictionary
Exercise 9: Get the key of a minimum value from the following dictionary
Exercise 10: Change value of a key in a nested dictionary

Exercise 1: Create a dictionary from two lists.

The two lists are listed below. Create a Python program to turn them into a dictionary where the first item from each list serves as the key and the second item serves as the definition.

value

keys = ['Ten', 'Twenty', 'Thirty']
values = [10, 20, 30]

Run Predicted result:

display@@@0 Show Hint

Utilize the zip() function. This function aggregates two or more iterables (such as a list, dict, or string) into a tuple.

, and gives it back.

Or, use a for loop to iterate the list.

, range()

function. Use the update command to add a new key-value pair to a dict each time.() method Demonstrate the solution

1: A dict and the zip() function()

function Object() { [native code] }

Use the
zip(keys, values)
to aggregate two lists.
Wrap the result of a
zip()
function into a
dict()
constructor.

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)

Run

Alternative 2

: Making use of the loop and update() methods of a

dictionary

dict1 = {'Ten': 10, 'Twenty': 20, 'Thirty': 30}
dict2 = {'Thirty': 30, 'Fourty': 40, 'Fifty': 50}
dict3 = {**dict1, **dict2}
print(dict3)

Run

Combine two Python dictionaries in exercise two.

one

sampleDict = {
    "class": {
        "student": {
            "name": "Mike",
            "marks": {
                "physics": 70,
                "history": 80
            }
        }
    }
}

Run Predicted result:

{'Ten': 10, 'Twenty': 20, 'Thirty': 30, 'Fourty': 40, 'Fifty': 50}

Python version 3.5+ Show Solution

employees = ['Kelly', 'Emma']
defaults = {"designation": 'Developer', "salary": 8000}

Run

Various Versions

sample_dict = {
    "name": "Kelly",
    "age": 25,
    "salary": 8000,
    "city": "New york"}
# Keys to extract
keys = ["name", "salary"]

Run

Print the value of the key “history” from the following in Exercise 3.

dict

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)

Run Predicted result:

80 Show Indicator

The dict is nested. Find the required key-value pair by using the proper key chaining. Display Solution

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)

Run

Exercise 4: Set default values for the dictionary’s initialization

The keys can be initialized in Python with the same values. Given

sample_dict = {
  "name": "Kelly",
  "age":25,
  "salary": 8000,
  "city": "New york"
}

Run Predicted result:

@@@2 output Show Hint

Use the dict’s fromkeys() method. Display Solution

A dictionary with the supplied keys and the specified values is returned by the fromkeys() method.

value.

sample_dict = {
  'Physics': 82,
  'Math': 65,
  'history': 75
}

Run

Exercise 5: Build a dictionary by removing the keys from one that is already there.

construct a Python program to take the keys from the dictionary below and extract them to construct a new dictionary. provided diction

sample_dict = {
    'emp1': {'name': 'Jhon', 'salary': 7500},
    'emp2': {'name': 'Emma', 'salary': 8000},
    'emp3': {'name': 'Brad', 'salary': 500}
}

Run Predicted result:

display@@@3 Show Hint

Solution 1:

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

Show Solution

Word reference

Comprehension

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)

Run

Alternative 2

: Making use of the update() method

loop

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)

Run

Remove a list of keys from a dictionary in exercise six

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)

Run Predicted result:

display@@@4 Show 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

We can use a dictionary’s pop() method or dictionary comprehension to get the aforementioned result. Solution 1 for Display

Description Employing the pop() technique and a loop

sample_dict equals Kelly, 25, makes $8,000 per year and lives in New York.

}

# Removed keys

[“Name,” “Salary”] in keys

keys: sample_dict.pop for k(k)

print(sample_dict) Alternate Solution

Word reference

Comprehension

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)

Run

Exercise 7: Verify whether a value is present in a dictionary.

We are aware of how to determine if a dictionary has the key. There are situations when it is necessary to verify that the given value is present.

To determine whether the value 200 is included in the following dictionary, write a Python program. Given :

sample_dict is “a”: 100, “b”: 200, and “c”: 300.” expected result

@@@5 output Show 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

Show Answer

dict1 = {'Ten': 10, 'Twenty': 20, 'Thirty': 30}
dict2 = {'Thirty': 30, 'Fourty': 40, 'Fifty': 50}

Run

Exercise 8: Rename the dictionary’s key

Create a program to change the name of a significant city to one from the following dictionaries. Given

dict1 = {'Ten': 10, 'Twenty': 20, 'Thirty': 30}
dict2 = {'Thirty': 30, 'Fourty': 40, 'Fifty': 50}

Run Predicted result:

display@@@6 Show Hint

Remove the city from a given dictionary
Add a new key (location) into a dictionary with the same value

Show Answer

dict1 = {'Ten': 10, 'Twenty': 20, 'Thirty': 30}
dict2 = {'Thirty': 30, 'Fourty': 40, 'Fifty': 50}

Run

Exercise 9: From the following, obtain the key with the lowest value.

dictionary

dict1 = {'Ten': 10, 'Twenty': 20, 'Thirty': 30}
dict2 = {'Thirty': 30, 'Fourty': 40, 'Fifty': 50}

Run Predicted result:

Math Show Suggestions

Utilize the native function min() Show Solution

dict1 = {'Ten': 10, 'Twenty': 20, 'Thirty': 30}
dict2 = {'Thirty': 30, 'Fourty': 40, 'Fifty': 50}

Run

Change the value of a key in a nested dictionary in exercise 10

To alter Brad’s pay to $8500 in the following dictionary, create a Python application. Given

dict1 = {'Ten': 10, 'Twenty': 20, 'Thirty': 30}
dict2 = {'Thirty': 30, 'Fourty': 40, 'Fifty': 50}
dict3 = {**dict1, **dict2}
print(dict3)

Run Predicted result:

@@@7 output: Show Solution

dict1 = {'Ten': 10, 'Twenty': 20, 'Thirty': 30}
dict2 = {'Thirty': 30, 'Fourty': 40, 'Fifty': 50}
dict3 = {**dict1, **dict2}
print(dict3)

Run

Python Dictionary Exercise with Solutions

Class method vs Static method in Python

Python Program to Count the Number of Matching Characters in a Pair of String

Coroutine in Python

Leave A Reply Cancel Reply

Python Dictionary Exercise with Solutions

Related Posts

Class method vs Static method in Python

Python Program to Count the Number of Matching Characters in a Pair of String

Coroutine in Python

Leave A Reply Cancel Reply