Python Basic Exercise for Beginners (2024)

• Create a function that will take two numbers as parameters
• Next, Inside a function, multiply two numbers and save their product in a product variable
• Next, use the if condition to check if the product >1000. If yes, return the product
• Otherwise, use the else block to calculate the sum of two numbers and return it.

def multiplication_or_sum(num1, num2): # calculate product of two number product = num1 * num2 # check if product is less then 1000 if product <= 1000: return product else: # product is greater than 1000 calculate sum return num1 + num2# first conditionresult = multiplication_or_sum(20, 30)print("The result is", result)# Second conditionresult = multiplication_or_sum(40, 30)print("The result is", result)

• Create a variable called previous_num and assign it to 0
• Iterate the first 10 numbers one by one using for loop and range() function
• Next, display the current number (i), previous number, and the addition of both numbers in each iteration of the loop. Finally, change the value of the previous number to the current number ( previous_num = i).

print("Printing current and previous number and their sum in a range(10)")previous_num = 0# loop from 1 to 10for i in range(1, 11): x_sum = previous_num + i print("Current Number", i, "Previous Number ", previous_num, " Sum: ", x_sum) # modify previous number # set it to the current number previous_num = i

• Use the Python input() function to accept a string from a user.
• Calculate the length of the string using the len() function
• Next, iterate each character of a string using for loop and range() function.
• Use start = 0, stop = len(s)-1, and step =2. the step is 2 because we want only even index numbers
• In each iteration of a loop, use s[i] to print characters present at the current even index number

Solution 1:

# accept input string from a userword = input('Enter word ')print("Original String:", word)# get the length of a stringsize = len(word)# iterate a each character of a string# start: 0 to start with first character# stop: size-1 because index starts with 0# step: 2 to get the characters present at even index like 0, 2, 4print("Printing only even index chars")for i in range(0, size - 1, 2): print("index[", i, "]", word[i])

Solution 2: Using list slicing

# accept input string from a userword = input('Enter word ')print("Original String:", word)# using list slicing# convert string to list# pick only even index charsx = list(word)for i in x[0::2]: print(i)

def remove_chars(word, n): print('Original string:', word) x = word[n:] return xprint("Removing characters from a string")print(remove_chars("pynative", 4))print(remove_chars("pynative", 2))

def first_last_same(numberList): print("Given list:", numberList) first_num = numberList[0] last_num = numberList[-1] if first_num == last_num: return True else: return Falsenumbers_x = [10, 20, 30, 40, 10]print("result is", first_last_same(numbers_x))numbers_y = [75, 65, 35, 75, 30]print("result is", first_last_same(numbers_y))

num_list = [10, 20, 33, 46, 55]print("Given list:", num_list)print('Divisible by 5:')for num in num_list: if num % 5 == 0: print(num)

Solution 1: Use the count() method

str_x = "Emma is good developer. Emma is a writer"# use count method of a str classcnt = str_x.count("Emma")print(cnt)

Solution 2: Without string method

def count_emma(statement): print("Given String: ", statement) count = 0 for i in range(len(statement) - 1): count += statement[i: i + 4] == 'Emma' return countcount = count_emma("Emma is good developer. Emma is a writer")print("Emma appeared ", count, "times")

for num in range(10): for i in range(num): print (num, end=" ") #print number # new line after each row to display pattern correctly print("\n")

• Reverse the given number and save it in a different variable
• Use the if condition to check if the original and reverse numbers are identical. If yes, return True.

def palindrome(number): print("original number", number) original_num = number # reverse the given number reverse_num = 0 while number > 0: reminder = number % 10 reverse_num = (reverse_num * 10) + reminder number = number // 10 # check numbers if original_num == reverse_num: print("Given number palindrome") else: print("Given number is not palindrome")palindrome(121)palindrome(125)

• Create an empty list named result_list
• Iterate the first list using a for loop
• In each iteration, check if the current number is odd number using num % 2 != 0 formula. If the current number is an odd number, add it to the result list
• Now, Iterate the first list using a loop.
• In each iteration, check if the current number is odd number using num % 2 == 0 formula. If the current number is an even number, add it to the result list
• Print the result list

def merge_list(list1, list2): result_list = [] # iterate first list for num in list1: # check if current number is odd if num % 2 != 0: # add odd number to result list result_list.append(num) # iterate second list for num in list2: # check if current number is even if num % 2 == 0: # add even number to result list result_list.append(num) return result_listlist1 = [10, 20, 25, 30, 35]list2 = [40, 45, 60, 75, 90]print("result list:", merge_list(list1, list2))

Use while loop

number = 7536print("Given number", number)while number > 0: # get the last digit digit = number % 10 # remove the last digit and repeat the loop number = number // 10 print(digit, end=" ")

income = 45000tax_payable = 0print("Given income", income)if income <= 10000: tax_payable = 0elif income <= 20000: # no tax on first 10,000 x = income - 10000 # 10% tax tax_payable = x * 10 / 100else: # first 10,000 tax_payable = 0 # next 10,000 10% tax tax_payable = 10000 * 10 / 100 # remaining 20%tax tax_payable += (income - 20000) * 20 / 100print("Total tax to pay is", tax_payable)

• Create the outer for loop to iterate numbers from 1 to 10. So, the total number of iterations of the outer loop is 10.
• Create an inner loop to iterate 10 times.
• For each outer loop iteration, the inner loop will execute ten times.
• In the first iteration of the nested loop, the number is 1. In the next, it 2. and so on till 10.
• In each iteration of an inner loop, we calculated the multiplication of two numbers. (current outer number and current inner number)

for i in range(1, 11): for j in range(1, 11): print(i * j, end=" ") print("\t\t")

for i in range(6, 0, -1): for j in range(0, i - 1): print("*", end=' ') print(" ")

def exponent(base, exp): num = exp result = 1 while num > 0: result = result * base num = num - 1 print(base, "raises to the power of", exp, "is: ", result)exponent(5, 4)