Practice Python Exercises and Challenges with Solutions

Steps to solve this question:

• Split the given string into a list of words using the split() method
• Use a list comprehension to create a new list by reversing each word from a list.
• Use the join() function to convert the new list into a string
• Display the resultant string

Solution:

# Exercise to reverse each word  of a string
def reverse_words(Sentence):
    # Split string on whitespace
    words = Sentence.split(" ")

    # iterate list and reverse each word using ::-1
    new_word_list = [word[::-1] for word in words]
    
    # Joining the new list of words
    res_str = " ".join(new_word_list)
    return res_str

# Given String
str1 = "My Name is Jessa"
print(reverse_words(str1))

Steps to solve this question: -

• First, open the file in a read mode
• Next, read all content from a file using the read() function and assign it to a variable.
• Next, use string replace() function to replace all newlines (\n) with space (' ').
• Display final string

with open('sample.txt', 'r') as file:
    data = file.read().replace('\n', ' ')
    print(data)

Solution 1: Using while loop

number_list = [10, 20, 30, 40, 50, 60, 70, 80, 90, 100]
i = 0
# get list's size
n = len(number_list)
# iterate list till i is smaller than n
while i < n:
    # check if number is greater than 50
    if number_list[i] > 50:
        # delete current index from list
        del number_list[i]
        # reduce the list size
        n = n - 1
    else:
        # move to next item
        i = i + 1
print(number_list)

Solution 2: Using for loop and range()

number_list = [10, 20, 30, 40, 50, 60, 70, 80, 90, 100]
for i in range(len(number_list) - 1, -1, -1):
    if number_list[i] > 50:
        del number_list[i]
print(number_list)

ascii_dict = {'A': 65, 'B': 66, 'C': 67, 'D': 68}
# Reverse mapping
new_dict = {value: key for key, value in ascii_dict.items()}
print(new_dict)

1. Use the counter() method of the collection module.
2. Create a dictionary that will maintain the count of each item of a list. Next, Fetch all keys whose value is greater than 2

Solution 1: - Using collections.Counter()

import collections

sample_list = [10, 20, 60, 30, 20, 40, 30, 60, 70, 80]

duplicates = []
for item, count in collections.Counter(sample_list).items():
    if count > 1:
        duplicates.append(item)
print(duplicates)

Solution 2: -

sample_list = [10, 20, 60, 30, 20, 40, 30, 60, 70, 80]
exist = {}
duplicates = []

for x in sample_list:
    if x not in exist:
        exist[x] = 1
    else:
        duplicates.append(x)
print(duplicates)

# Dictionary
d1 = {'A': 65, 'B': 66, 'C': 67, 'D': 68, 'E': 69, 'F': 70}
# Filter dict using following keys
l1 = ['A', 'C', 'F']
new_dict = {key: d1[key] for key in l1}
print(new_dict)

rows = 5
x = 0
# reverse for loop from 5 to 0
for i in range(rows, 0, -1):
    x += 1
    for j in range(1, i + 1):
        print(x, end=' ')
    print('\r')

def manipulate(x, y):
    # concatenate two strings
    def inner_fun(x, y):
        return x + y

    z = inner_fun(x, y)
    return z + 'Developers'

result = manipulate('Emma', 'Kelly')
print(result)

list1 = [5, [10, 15, [20, 25, [30, 35], 40], 45], 50]
# modify item
list1[1][2][2][1] = 3500
# print final result
print(list1)

# print(list1[1]) = [10, 15, [20, 25, [30, 400], 40], 45]
# print(list1[1][2]) = [20, 25, [30, 400], 40]
# print(list1[1][2][2]) = [30, 40]
# print(list1[1][2][2][1]) = 40

emp_dict = {
    "company": {
        "employee": {
            "name": "Jess",
            "payable": {
                "salary": 9000,
                "increment": 12
            }
        }
    }
}
print(emp_dict['company']['employee']['payable']['increment'])