Non-Programmer's Tutorial for Python 3/Dealing with the imperfect

closing files with with

We use the "with" statement to open and close files.

with open("in_test.txt", "rt") as in_file:
    with open("out_test.txt", "wt") as out_file:
        text = in_file.read()
        data = parse(text)
        results = encode(data)
        out_file.write(results)
    print( "All done." )

If some sort of error happens anywhere in this code (one of the files is inaccessible, the parse() function chokes on corrupt data, etc.) the "with" statements guarantee that all the files will eventually be properly closed. Closing a file just means that the file is "cleaned up" and "released" by our program so that it can be used in another program.

catching errors with try

So you now have the perfect program, it runs flawlessly, except for one detail, it will crash on invalid user input. Have no fear, for Python has a special control structure for you. It's called try and it tries to do something. Here is an example of a program with a problem:

print("Type Control C or -1 to exit")
number = 1
while number != -1:
   number = int(input("Enter a number: "))
   print("You entered:", number)

Notice how when you enter @#& it outputs something like:

Traceback (most recent call last):
 File "try_less.py", line 4, in <module>
   number = int(input("Enter a number: "))
ValueError: invalid literal for int() with base 10: '\\@#&'

As you can see the int() function is unhappy with the number @#& (as well it should be). The last line shows what the problem is; Python found a ValueError. How can our program deal with this? What we do is first: put the place where errors may occur in a try block, and second: tell Python how we want ValueErrors handled. The following program does this:

print("Type Control C or -1 to exit")
number = 1
while number != -1:
    try:
        number = int(input("Enter a number: "))
        print("You entered:", number)
    except ValueError:
        print("That was not a number.")

Now when we run the new program and give it @#& it tells us "That was not a number." and continues with what it was doing before.

When your program keeps having some error that you know how to handle, put code in a try block, and put the way to handle the error in the except block.

Exercises

Update at least the phone numbers program (in section Dictionaries) so it doesn't crash if a user doesn't enter any data at the menu.

Solution

def print_menu():
	print('1. Print Phone Numbers')
	print('2. Add a Phone Number')
	print('3. Remove a Phone Number')
	print('4. Lookup a Phone Number')
	print('5. Quit')
	print()

numbers = {}
menu_choice = 0
print_menu()
while menu_choice != 5:
	try:
		menu_choice = int(input("Type in a number (1-5): "))
		if menu_choice == 1:
			print("Telephone Numbers:")
			for x in numbers.keys():
				print("Name: ", x, "\tNumber:", numbers[x])
			print()
		elif menu_choice == 2:
			print("Add Name and Number")
			name = input("Name: ")
			phone = input("Number: ")
			numbers[name] = phone
		elif menu_choice == 3:
			print("Remove Name and Number")
			name = input("Name: ")
			if name in numbers:
				del numbers[name]
			else:
				print(name, "was not found")
		elif menu_choice == 4:
			print("Lookup Number")
			name = input("Name: ")
			if name in numbers:
				print("The number is", numbers[name])
			else:
				print(name, "was not found")
		elif menu_choice != 5:
			print_menu()
	except ValueError:
		print("That was not a number.")