Section 4: Handling Errors With Try/Except

Exception Handling

Getting an error or exception in Python program, without any exception handling means entire program will crash.

In real world, this is not the desired behavior, and we want our program to detect errors, handle them, and then continue to run.

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def spam(divideBy):  
    return 42 / divideBy  
  
print(spam(2))  
print(spam(12))  
print(spam(0))  
print(spam(1))

When the program is run we will get ZeroDivisonError at line 6.

You can put the previous divide-by-zero code in a try clause and have an except clause contain code to handle what happens when this error occurs.

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def spam(divideBy):
    try:
        return 42 / divideBy
    except ZeroDivisionError:
        return ('Error: I cannot do that.')


print(spam(2))
print(spam(12))
print(spam(0))
print(spam(1))

When code in a try clause causes an error, the program execution immediately moves to the code in the except clause. After running that code, the execution continues as normal.

A Short Program: Zigzag

This program will create a back-and-forth, zigzag pattern until the user stops it by pressing the Mu editor’s Stop button or by pressing CTRL-C. When you run this program, the output will look something like this:

    ********  
   ********  
  ********  
 ********  
********  
 ********  
  ********  
   ********  
    ********

This is how I implemented:

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# An extra project from book's chapter 3
import sys
import time
def asterisks_pattern(startSpace, pattern):
    print(' ' * startSpace + pattern)
    time.sleep(0.1)
pattern = '******'

while True:
    try:
        for startSpace in range(10):
            asterisks_pattern(startSpace, pattern)

        for startSpace in range(10, 1, -1):
            asterisks_pattern(startSpace, pattern)
    except KeyboardInterrupt:
        print(' Quiting the animation pattern. Goodbye!')
        sys.exit()

Here is Al’s implementation

The Collatz Sequence

Write a function named collatz() that has one parameter named number. If number is even, then collatz() should print number // 2 and return this value. If number is odd, then collatz() should print and return 3 * number + 1.

Then write a program that lets the user type in an integer and that keeps calling collatz() on that number until the function returns the value 1. (Amazingly enough, this sequence actually works for any integer—sooner or later, using this sequence, you’ll arrive at 1! Even mathematicians aren’t sure why. Your program is exploring what’s called the Collatz sequence, sometimes called “the simplest impossible math problem.”)

Remember to convert the return value from input() to an integer with the int() function; otherwise, it will be a string value.

Hint: An integer number is even if number % 2 == 0, and it’s odd if number % 2 == 1.

The output of this program could look something like this:

Enter number:  
3  
10  
5  
16  
8  
4  
2  
1

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# Extra Project from book's chapter 3
def collatz(number):
    if number % 2 == 0:
        result = int(number / 2)
    else:
        result = int(3 * number + 1)
    print(result)
    return result


try:
    number = int(input("Enter your number:\n"))
    while number != 1:
        number = collatz(number)
except ValueError:
    print('Please enter a valid integer')