Python OOPE Exam

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🔢 Function: Count Positive Integers (Ignore None) Let’s build an easy-to-use Python function! ✨ 🚩 Function Definition def count_positive_integers(lst): return sum(1 for x in lst if x is not None and isinstance(x, int) and x > 0) 📝 Step-by-Step Explanation Iterate through each element x in the list. Ignore None values (x is not None). Check if the value is an integer (isinstance(x, int)). Count only if the integer is positive (x > 0, so zero is ignored). Sum up the count (sum(1 for ...)). 🧪 Practice Questions Practice 1 lst = [4, -3, None, 7, 0, None, 10] print(count_positive_integers(lst)) # _______ Answer: 3 (Counts 4, 7, 10)

🔄 Swap and Reverse Lines Program in Python Below is a Python program that reads multiple lines of input, swaps every two consecutive lines, and then reverses every odd-numbered line in the resulting output sequence. This meets the problem’s requirements exactly. 🛠️ Complete Code def main(): n = int(input()) lines = [input() for _ in range(n)] # Swap every two consecutive lines for i in range(0, n - 1, 2): lines[i], lines[i + 1] = lines[i + 1], lines[i] # Reverse every odd-numbered line in the output (1-based index) for i in range(n): if (i + 1) % 2 == 1: # odd line number lines[i] = lines[i][::-1] # Print the transformed lines for line in lines: print(line) if __name__ == "__main__": main() 📝 Explanation Input Reading: Read the integer n for the number of lines, then read each line into a list. Swapping Consecutive Lines: Iterate through the list in steps of two, swapping the pair at indices i and i+1. Reversing Odd-Numbered Lines: After swapping, reverse every odd line (1st, 3rd, 5th, …) by checking (i+1) % 2 == 1. Output: Print each line after these transformations. 🌟 Example Run Input:

🧮 Function: is_positive_odd_or_negative_even Let’s build a Python function that checks if a number is either a positive odd or a negative even. 🎯 🚩 Function Definition def is_positive_odd_or_negative_even(n): # Check for positive odd OR negative even return (n > 0 and n % 2 == 1) or (n < 0 and n % 2 == 0) 📝 Step-by-Step Explanation Positive Odd Number: Condition: n > 0 and n % 2 == 1 Checks if the number is positive AND odd. Negative Even Number: Condition: n < 0 and n % 2 == 0 Checks if the number is negative AND even. Logical OR: The function returns True if any one of the above conditions is met. 🧪 Practice Questions Practice 1 print(is_positive_odd_or_negative_even(7)) # _______ 7 is positive and odd. Output: True Practice 2 print(is_positive_odd_or_negative_even(-4)) # _______ -4 is negative and even. Output: True Practice 3 print(is_positive_odd_or_negative_even(-3)) # _______ -3 is negative and odd. Output: False Practice 4 print(is_positive_odd_or_negative_even(8)) # _______ 8 is positive and even. Output: False Practice 5 print(is_positive_odd_or_negative_even(0)) # _______ 0 is neither positive nor negative (and it’s even). Output: False ✨ Key Points Odd number: Remainder when divided by 2 is 1 (n % 2 == 1). Even number: Remainder when divided by 2 is 0 (n % 2 == 0). Positive Odd or Negative Even: Only one of these is needed for True. Zero case: 0 is not positive or negative, so always returns False. Happy Coding! 🚀

def print_n_pattern(n): """ Prints an "N" shaped pattern with n rows. The pattern has vertical bars (|) on the left and right sides of each row, with a backslash (\) moving diagonally from the top-left to the bottom-right. There are no spaces to the right of the pattern. Args: n (int): The number of rows for the pattern (n > 0). """ for i in range(n): # Start each row with a vertical bar row = "|" # Add leading spaces before the diagonal backslash # The number of leading spaces is equal to the current row index 'i' row += " " * i # Add the diagonal backslash row += "\\" # Add trailing spaces after the diagonal backslash # The total width of the inner part (between vertical bars) is (n - 1) characters. # So, the number of trailing spaces is (n - 1) - i row += " " * (n - 1 - i) # End each row with a vertical bar row += "|" # Print the constructed row print(row) if __name__ == "__main__": try: # Read the integer input from the user n = int(input()) # Validate the input: n must be greater than 0 if n <= 0: print("Input 'n' must be a positive integer (n > 0).") else: # Call the function to print the pattern print_n_pattern(n) except ValueError: # Handle cases where the input is not a valid integer print("Invalid input. Please enter a single integer.")

🗓️ mm_dd_yy_to_yy_dd_mm Function Implementation Here’s how you can implement the mm_dd_yy_to_yy_dd_mm function to convert a date string from the format mm-dd-yy to yy-dd-mm. def mm_dd_yy_to_yy_dd_mm(date: str) -> str: """ Convert a date string from the format mm-dd-yy to yy-dd-mm. Args: date (str): The date in the format "mm-dd-yy". Returns: str: The date in the format "yy-dd-mm". Example: >>> mm_dd_yy_to_yy_dd_mm("12-25-21") "21-25-12" """ mm, dd, yy = date.split('-') return f"{yy}-{dd}-{mm}" 🐍 Step-by-Step Explanation Split the String: The split('-') method divides the input string into three parts: "mm", "dd", and "yy". Unpack and Rearrange: Assign these parts to mm, dd, and yy. Then, use an f-string to rearrange them into yy-dd-mm. Return the Result: The function returns the reformatted date string. 💡 Example Usage Input: "12-25-21" Output: "21-25-12" print(mm_dd_yy_to_yy_dd_mm("12-25-21")) # Output: 21-25-12 📚 Practice Questions What will mm_dd_yy_to_yy_dd_mm("04-01-99") return? Try mm_dd_yy_to_yy_dd_mm("11-30-00") What if the input is "07-04-76"? ✅ Solutions "99-01-04" "00-30-11" "76-04-07" 🚩 Note If the input is not in the expected "mm-dd-yy" format (for example, missing dashes or incorrect number of components), the function will raise a ValueError.

🔎 Function: Find Last Word Starting with Uppercase Here’s how you can implement the last_word_starts_with_upper_case function to find the last word in a sentence that starts with an uppercase letter. If no such word exists, it returns None. 🛠️ Function Definition def last_word_starts_with_upper_case(sentence: str): """ Find the last word in a sentence that starts with an uppercase letter. Args: sentence (str): The input sentence. Returns: str or None: The last word starting with an uppercase letter, or None if no such word exists. """ result = None for word in sentence.split(): if word and word[0].isupper(): result = word return result ✨ Step-by-Step Explanation Step 1: Split the sentence into individual words using split(). Step 2: Loop through each word: If the word starts with an uppercase letter (word.isupper()), save it as the current result. Step 3: After checking all words, return the last word found that met the criteria, or None if none did. 🚦 Example Usage print(last_word_starts_with_upper_case("This is a Test sentence")) # Output: "Test" print(last_word_starts_with_upper_case("no uppercase words here")) # Output: None 📝 Practice Questions What will last_word_starts_with_upper_case("Alice and Bob went to Paris") return? Try last_word_starts_with_upper_case("all lowercase words"). What is the result of last_word_starts_with_upper_case("Python Is Easy To Learn")? ✅ Solutions "Paris" None "Learn" With this function, you can quickly scan sentences for the last word that begins with a capital letter. Perfect for text analysis and simple parsing tasks! 🐍💡