# Noemi Cabrera # 2 December 2021 # In this lesson, I learned how to use and differentiate the pass, break, and continue statements. # These are efficient for when you're not ready to write code under certain places, continue with # code in the next line, or want to break a forever loop. The break and continue statements are # also useful to control loop iteration. Also, I learned how to use nested loops to iterate over the # elements of a table and how to employ compound conditional expressions in a loop structure. # All of these features allow the code to be more efficient and smooth. # I didn't have any major difficulties in this lesson.

# In this code, the pass statement is used inside the defined useless_function as its code so that an error is not displayed. It's # used as a placeholder for the actual code that can be put into the function. It displays nothing. # Does nothing, but syntactically valid def useless_function(): pass

# In this code, the pass statement is used inside a for loop to replace any code so that an error is not displayed. It's # used as a placeholder for the actual code that can be put into the loop. It displays nothing. # Does nothing, but syntactically valid for i in range(10): pass

# In this code, the pass statement is used inside an if statement to replace any code so that an error is not displayed. It's # used as a placeholder for the actual code that can be put into the statement. It displays nothing. # Does nothing, but syntactically valid if (5 < 6): pass

# In this code, a break statement is used in the same code defined first to iterate throughthe items in the 'lst' list until the the if statement evaluates # true. This is more efficient since you do not have to iterate through every single item in the list like the first code in this cell does. # Find the index of the first occurrence of 611 in an arbitrarily long list # Define a long list lst = [976, 618, 812, 898, 804, 689, 611, 630, 467, 411, 648, 931, 618, 425, 328, 196, 56, 615, 458, 166, 115, 118, 22, 585, 213, 855, 615, 478, 234, 360, 981, 174, 993, 627, 911, 385, 219, 577, 540, 49, 164, 109, 893, 727, 687, 709, 971, 781, 802, 701, 789, 421, 863, 44, 573, 91, 140, 338, 36, 102, 502, 723, 134, 336, 940, 969, 649, 598, 413, 307, 929, 951, 993, 436, 149, 280, 907, 733, 713, 268, 29, 946, 102, 277, 352, 449, 527, 201, 566, 643, 118, 587, 717, 358, 542, 223, 846, 244, 83, 268] # Find the index (without break) index = 0 iteration_count = 0 for num in lst: if (num == 611): found_at = index index = index + 1 iteration_count = iteration_count + 1 print("Without using a break, 611 was found at index:", found_at, "using", iteration_count, "iterations") # Find the index (with break) index = 0 iteration_count = 0 for num in lst: if (num == 611): found_at = index break # adding a break to improve efficiency index = index + 1 iteration_count = iteration_count + 1 print("Using a break, 611 was found at index:", found_at, "using", iteration_count, "iterations")

# In this code, a break statement is used under the if statement to break the infinite while loop that asks for user input. WHen # the user enters the correct answer, the loop breaks. # Prompt the user for a positive number while True: x = int(input("Enter a positive number: ")) if (x > 0): break #x is positive, break the loop

# In this code, a continue statement is used in a for loop that iterates through 'lst'. The continue statement indicates that when # the iteration is finished, the next line of code would be executed. All the odd numbers in 'lst' are displayed. # Print the odd elements in a list lst = [14, 6, 5, 10, 6, 9, 33, 103, 21, 55] for num in lst: # Skip all even numbers if (num % 2 == 0): continue print(num)

# In this program, a break statement is used in the for loop that iterates through num. The break is placed in # the if statement after the code in it has executed so that the number of iterations is reduced. # [ ] The following program tests if `num` is prime or not, use a `break` statement to improve its efficiency # and reduce the number of necessary iterations # Compare the number of necessary iterations with and without the `break` statement # Use the following numbers for the comparison: num = 45345 #num = 11579 #num = 948240 #num = 128093 #num = 519937 #num = 694394 prime = True # assume num is prime unless proven otherwise iteration_count = 0 for i in range(2, num): if num % i == 0: prime = False; break iteration_count = iteration_count + 1 # Display results if prime: print(num, "is prime") else: print(num, "is NOT prime") print("Total number of iterations:", iteration_count)

# In this code, the user is asked to enter an odd number. A while loop is used to ask for an odd number # until the user enters a correct answer. I f the number is odd, "Thank you" displays and the loop breaks. # [ ] Write a program to prompt the user for an odd number; use an infinite loop and a `break` statement. while True: odd_num = int(input("Enter an odd number: ")) if (odd_num % 2 != 0): print("Thank you") break

# In this code, the sqrt function is first imported from the math module. Then, with a for loop, numbers # 1 through 100 are iterated through. Then using a continue statement in the if statement, we check if the # number is not divisible by 7 and continue moving on until all the numbers have been iterated through. # The new line of code that prints the numbers divisible by 7 and their squate roots is executed because of the continue # statement. # [ ] Modify the following program to display numbers that are divisible by 7 along with their square roots. # HINT: Use a `continue` statement in the loop from math import sqrt for num in range(1, 100): #TODO if num % 7 != 0: continue print(num, "- square root:", sqrt(num))

# In this code, a nested for loop is used to iterate through each list found in the 'table' list. # The initial for loop iterates through table. Then, the nested loop iterates through each of the items in # table. A tab/space is added at the end of each item in each list. The items are displayed. # list of lists table = [[5, 2, 6], [4, 6, 0], [9, 1, 8], [7, 3, 8]] for row in table: for col in row: # print the value col followed by a tab print(col, end = "\t") # Print a new line print()

# In this code, a char_art function is defined and designed to display a staircase generated with [] # The initial for loop terates through the range specified when calling the function. Then, the second for loop # does the same thing, but an if statatement is added to compare the two numbers and when they match, a # [] is displayed. # Generate a staircase character art # Size controls the number of steps def char_art(steps): for row in range(steps): for col in range(steps): if(col <= row): print("[]", end = "") print() # Generate a staircase with 6 steps char_art(6)

# In this code, a nested for loop is used. First the initial for loop iterates through each list in the # "table" list. Then a nested for loop is used to iterate through each item in the lists in "table". # Each item found in one list is added together and then the final sum is displayed. This is done for # each of the list in "table". # [ ] Write a program to display the sum of each row in the table table = [[5, 2, 6], [4, 6, 0], [9, 1, 8], [7, 3, 8]] for row in table: sum=0 for col in row: sum+=col print("sum = ", sum) print()

# In this code, the generate_star function is defined. This function takes in 1 argument, size. # Then with an initial for loop, each number in the range, which is defined by the argument value entered. # Then a nested for loop and an if/else statement are used to display an asterick when the numbers # equal each other or when the numbers of the range equals the substraction of the size, a number in the range, and 1. # The final star is printed. # [ ] Complete the function `generate_star` so it displays a star of variable `size` using "*" # For size = 5 the star should look like: # * * # * * # * # * * # * * def generate_star(size): #TODO for row in range(size): for col in range(size): if(col == row): print("*", end = "") elif (row == size - col -1): print("*", end = "") else: print(" ", end = "") pass # Display star generate_star(5)

# In this code, the for loop iterates through each item in the lst list. Then, an if statement is used with # the and operator to check if the number is the largest and even in the list. # The largest even number is displayed. lst = [102, 34, 55, 166, 20, 67, 305] # Nesting a compound conditional in a for loop largest = 0 for num in lst: # test if num is even and greater than largest if((largest < num) and (num % 2 == 0)): largest = num print("Largest even number in the list is: ", largest)

# In this code, a for loop is used to iterate through the 'ages' list and then an if/else statement is # used to place the ages of 100 peolple into categories (children, teens, young adults, and adults) # This is done by using an and operator in the elif statements. # The final number of people in each category is printed. # Ages of 100 people ages = [86, 38, 30, 19, 29, 6, 95, 22, 23, 82, 39, 73, 30, 98, 5, 68, 57, 34, 35, 81, 54, 77, 29, 75, 83, 14, 88, 7, 8, 32, 93, 76, 42, 1, 32, 70, 70, 3, 34, 52, 44, 41, 7, 77, 73, 97, 34, 13, 33, 54, 8, 82, 21, 55, 72, 41, 34, 98, 72, 73, 24, 55, 50, 63, 38, 92, 43, 68, 52, 68, 69, 51, 19, 24, 35, 55, 74, 47, 8, 19, 69, 12, 96, 96, 11, 30, 97, 73, 22, 25, 19, 85, 37, 68, 39, 76, 73, 18, 45, 42] # Initial count children = 0 teens = 0 young_adults = 0 adults = 0 # Nesting compound conditionals within a for loop for age in ages: if(age <= 12): children = children + 1 elif((age >= 13) and (age <= 19)): teens = teens + 1 elif((age >= 20) and (age <= 30)): young_adults = young_adults + 1 elif(age > 30): adults = adults + 1 print("Number of children: ", children) print("Number of teens: ", teens) print("Number of young_adults: ", young_adults) print("Number of adults: ", adults)

# In this code, a while loop is created. Then, the user is asked to enter a number between 50 and 60 # until the number entered is valid. This is done with an 'and' operator and comparisons statements in the # loop. # Prompt the user for a number between 50 and 60 # Using infinite loop and break while True: x = int(input("Enter a number between 50 and 60: ")) if ((x >= 50) and (x <= 60)): print("Thank you!") break

# In this code, a for loop is used to iterate through the 'lst' list. Then an if/else statement is used to # to check if any number in the list is an even positive, an odd negative, or a zero. The and operator # is used in the elif statements to check if a number is both even and positive, or odd and negative. # The count of times each category is in 'lst' list is displayed. # [ ] Complete the following program to count the number of even positive numbers, odd negative numbers, and zeros in `lst` # then print the results lst = [9, 0, -2, -4, -5, 2, -15, 6, -65, -7] even_positives = 0 odd_negatives = 0 zeros = 0 #TODO: Count even_positives, odd_negatives, and zeros for num in lst: if(num == 0): zeros = zeros + 1 elif((num % 2 == 0) and (num > 0)): even_positives = even_positives + 1 elif((num % 2 != 0) and (num < 0)): odd_negatives = odd_negatives + 1 else: pass print("Number of even positive numbers: ", even_positives) print("Number of odd negative numbers: ", odd_negatives) print("Number of zeros: ", zeros)

# In this code, a for loop is used to iterate through the "s" list. The variable punc_m is set to # zero before the loop. Then using an if statement and the or operator, we check if any of the specified # punctuation marks are in s. If this evaluates to true, then the punc_m variable is added 1. # The total number of punctuation marks in 's' is displayed. # [ ] Write a program to count the number of punctuation marks (. , ? ! ' " : ;) in `s` s = "Once you eliminate the impossible, whatever remains, no matter how improbable, must be the truth." # Sherlock Holmes (by Sir Arthur Conan Doyle, 1859-1930) punc_m = 0 for c in s: if (c == ".") or (c ==',')or (c =='?') or (c =='!') or (c =="'") or (c =='"') or (c ==':') or (c ==';'): punc_m +=1 print("The number of punctuation marks is: ", punc_m)

# In this code, a infinite while loop with the True keyword is used. Then, the user is asked to # enter a odd positive number until the entered answer is correct, and the loop breaks. # [ ] Write a program to prompt the user for an odd positive number; use an infinite loop and a `break` statement. while True: odd_pos = int(input("Enter a odd positive number: ")) if (odd_pos % 2 != 0) and (odd_pos > 0): print("Thanks!") break