3.11. While Loops With Conditionals

So far the loops we have looked at have had simple blocks of code inside the loop, i.e. the blocks that were being repeated were quite simple. What you’ll see is that the code inside the loop can be as complicated as you’d like. For example, we can add conditionals inside the loop.

Here’s an example with an if statement.

i = 0

while i < 5:
    print(i)
    if i == 3:
        print("Third time's the charm!")
    i = i + 1

0
1
2
3
Third time's the charm!
4

What’s happening here?

We can see that the condition we have set for our while loop is i < 5.

At the first iteration i = 0. This means the loop condition is True. We execute the code block:

print(i)
if i == 3:
    print("Third time's the charm!")
i = i + 1

This means that we output:

0

Then since the condition i == 3 is False, we increment i and move onto the next iteration.

At the second iteration i = 1. This means the loop condition is True. Again, we execute the code block and we output:

1

Then since the condition i == 3 is False, we increment i and move onto the next iteration.

At the third iteration i = 2. This means the loop condition is True. Again, we execute the code block and we output:

2

Then since the condition i == 3 is False, we increment i and move onto the next iteration.

At the fourth iteration i = 3. This means the loop condition is True. Again, we execute the code block and we output:

3

This time the condition i == 3 is True, so we also output:

Third time's the charm!

We increment i and move onto the next iteration.

At the fifth iteration i = 4. This means that the loop condition is True. Again, we execute the code block and we output:

4

Then since the condition i == 3 is False, we increment i and move onto the next iteration.

At the sixth iteration i = 5. This means that the loop condition is False. We do not execute the code block.

This is how we can represent this code diagrammatically.

Note that there are two False routes running down the right hand side of the figure. This means that there should be two levels of indentation in the code.

Here’s an example of an if-else statement inside a loop using input().

n = input("Enter a number: ")

while n != "":
    if int(n) % 2 == 0:
        print("even")
    else:
        print("odd")
    n = input("Enter a number: ")

What you’ll notice with this loop is that this loop is going to keep asking the user for a number. With each number, the program will check whether int(n) % 2 == 0, i.e. checks whether the number is even or not. If the number is even the program will output:

even

otherwise it will output:

odd

This program will keep going until the user enters an empty string.

Again there are two False routes running down the right hand side of the figure, which corresponds to two levels of indentation in the code.

Question 1

Will the following two programs produce the same output?

Program 1

i = 0
while i < 100:
    print(i)
    i = i + 10

Program 2

i = 0
while i < 100:
    if i%10 == 0:
        print(i)
    i = i + 1

Solution

Let’s look at Program 1. This program is going start at i = 0 and then at each iteration i will increase by 10 since we update i with i = i + 10. Since we print each value of i, we get the output:

0
10
20
30
40
50
60
70
80
90

Note that we don’t print 100 because when i = 100 the while condition is False, the code block inside the loop doesn’t run and we exit the program.

Now let’s look at Program 2. This program is going start at i = 0 and then at each iteration i will increase by 1 since we update i with i = i + 1. This means we’ll run the loop 100 times (from i = 0, …, 99). Inside the loop we have the code block

if i%10 == 0:
    print(i)
i = i + 1

This means that at each iteration we check to see whether i is divisible by 10 and if it is, we print(i). This means we get the output:

0
10
20
30
40
50
60
70
80
90

Thus, the output of program 1 and program 2 are the same even though the code is different!

Note that program 1 has one False route running down the right hand side of the figure and program 1 only has one level of indentation. Program 2 has two False routes running down the right hand side of the figure and program 2 has two levels of indentation.

Question 2

What do you think the output of the following will be?

cars = ['red', 'white', 'blue', 'white', 'blue', 'green', 'red', 'blue']

count = 0

i = 0
while i < len(cars):
    if cars[i] == 'blue':
        count = count + 1
    i = i + 1

print(count)

Solution

Solution is locked

Question 3

What is wrong with the following code?

n = int(input())

while n != '':
    if n%3 == 0:
        print('divisible by 3')
    n = int(input())

1. The indentation on line 6 does not match the indentation on line 5.

2. n is not an integer, which means n%3 in line 4 results in a TypeError as you cannot take the modulus of a string.

3. To exit the loop the user needs to enter a blank lint, but this will result in a ValueError on line 1 or line 6 as you cannot convert an empty string to an integer.

4. The code will result in an infinite loop

Solution

Solution is locked

Question 4

We want to write a program that will read integers from a user and if the integer is divisible by 3 the program will output

divisible by 3

otherwise the program will do nothing.

The program will terminate only if the user doesn’t provide input, e.g. enters an empty string.

An example of how this program might run is shown below with user input shown in bold.

4
8
9
divisible by 3
12
divisible by 3

Our first attempt at the code did not work:

n = int(input())

while n != '':
    if n%3 == 0:
        print('divisible by 3')
    n = int(input())

Write a new program that fixes the error in our first attempt.

Solution

Solution is locked

Question 5

Implement the algorithm illustrated in the diagram below in Python

Solution

Solution is locked

Code challenge: Odd Numbers

You have been provided with a list called numbers.

numbers = [5, 12, 39, 45, 2, 61, 22, 49, 50, 73, 11, 84, 32, 40, 18, 39, 85, 16, 14, 49, 99, 68, 57, 35, 66]

Write a program that loops through the lists and sums all the odd numbers. Print your result.

Solution

Solution is locked