Published

2025-03-31

Open In Colab

Error Handling

Justin Post

We’ve already learned a lot about creating our own functions. One thing we haven’t touched on is how to have our functions fail in a good way. That is, when an error occurs, can we either avoid it and finish the function execution or can we at least provide a clear explanation of what error came up?

To do so we’ll talk about try/catch logic. Here we try some code and can plan to catch certain errors and do something about them.

Note: These types of webpages are built from Jupyter notebooks (.ipynb files). You can access your own versions of them by clicking here. It is highly recommended that you go through and run the notebooks yourself, modifying and rerunning things where you’d like!

Errors When Programming

Commonly you’ll have syntax errors and exceptions

  • A Syntax error is when you typed something in wrong (and it can’t be parsed by python)
for x in range(0,10) #missing colon
    print(x)
  File "<ipython-input-1-e60b965da02e>", line 1
    for x in range(0,10) #missing colon
                         ^
SyntaxError: expected ':'

SyntaxError: invalid syntax (<string>, line 1)

  • These aren’t things we can fix on the backend. Code needs to be typed in correctly!

  • An exception occurs when python can’t execute your code (during the execution something bad happens)

print("the number is " + 10)
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-2-8cddb0f2de90> in <cell line: 1>()
----> 1 print("the number is " + 10)

TypeError: can only concatenate str (not "int") to str

TypeError: can only concatenate str (not "int") to str

Dealing with Exceptions

Exceptions can be dealt with to some degree! Consider this function to print out strings passed by the user.

def print_strings(*x): #should all be strings!
    c = 0
    for i in x:
        print("The value in position " + str(c) + " is: " + i)
        c += 1

print_strings("cat", "dog", "bird")
The value in position 0 is: cat
The value in position 1 is: dog
The value in position 2 is: bird
  • If we pass a non-string, this will throw an exception.
def print_strings(*x): #should all be strings!
    c = 0
    for i in x:
        print("The value in position " + str(c) + " is: " + i)
        c += 1

print_strings("cat", 1, "bird")
The value in position 0 is: cat
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-4-15f25c060c0e> in <cell line: 7>()
      5         c += 1
      6 
----> 7 print_strings("cat", 1, "bird")

<ipython-input-4-15f25c060c0e> in print_strings(*x)
      2     c = 0
      3     for i in x:
----> 4         print("The value in position " + str(c) + " is: " + i)
      5         c += 1
      6 

TypeError: can only concatenate str (not "int") to str
  • Note that it says TypeError at the beginning of that big error message. What we can do is run a try block and set up what to do when we get certain exceptions via an except block.
  • The syntax for these is simlar to if and else blocks
def print_strings(*x):
    c = 0
    for i in x:
        try: #try the code that is indented below here
            print("The value in position " + str(c) + " is: " + i)
        except TypeError: #if we get a TypeError in the previous try block, do this and then continue execution!
            print("Oh no! Not a string")
        c += 1
print_strings("cat", "dog", 1, "bird")
The value in position 0 is: cat
The value in position 1 is: dog
Oh no! Not a string
The value in position 3 is: bird

  • Can have multiple except statements and an else block to account for many situations

  • Let’s create a quick function to printout information about a person given as key/value pairs

def print_stuff(**x): #now taking key value pairs (a dictionary within the function)
    print("Pay special attention to " + x.pop("Name") + "\nHis attributes are:") #we must have a Name argument
    for key in x: #now run through the other info given and print it out
        print("\t", key, " : ", str(x[key])) #\t is a tab

print_stuff(Name = "Jack London", Age = 41, Job = "Writer")
Pay special attention to Jack London
His attributes are:
     Age  :  41
     Job  :  Writer

If we don’t have a Name argument passed, we’d have an error!

print_stuff(Person = "Jack London", Age = 41, Job = "Writer")
---------------------------------------------------------------------------
KeyError                                  Traceback (most recent call last)
<ipython-input-16-66ebf0cac27f> in <cell line: 1>()
----> 1 print_stuff(Person = "Jack London", Age = 41, Job = "Writer")

<ipython-input-15-a21ff30dfb20> in print_stuff(**x)
      1 def print_stuff(**x): #now taking key value pairs (a dictionary within the function)
----> 2     print("Pay special attention to " + x.pop("Name") + "\nHis attributes are:") #we must have a Name argument
      3     for key in x: #now run through the other info given and print it out
      4         print("\t", key, " : ", str(x[key])) #\t is a tab
      5 

KeyError: 'Name'

In this case we get a KeyError (this is when we look for a certain key in a dictionary (or other similar object) and can’t find it) - We can look for this kind of error - Also look for other errors that might come up

def print_stuff(**x): #now taking key value pairs
    try:
        print("Pay special attention to " + x.pop("Name") + "\nHis attributes are:")
    except TypeError: #If x.pop("Name") doesn't give a string we'll get a TypeError
        print("Oh no! 'Name' is not a string")
    except KeyError: #If the user didn't give a "Name" argument we'll get a KeyError
        print("You didn't supply a 'Name'!")
    for key in x: #Now print out the rest of the info about the person
        print("\t", key, " : ", str(x[key]))
print_stuff(Name = "Jack London", Age = 41, Job = "Writer")
Pay special attention to Jack London
His attributes are:
     Age  :  41
     Job  :  Writer
print_stuff(Name = 11, Age = 41, Job = "Writer")
Oh no! 'Name' is not a string
     Age  :  41
     Job  :  Writer
print_stuff(Person = "Jack London", Age = 41, Job = "Writer")
You didn't supply a 'Name'!
     Person  :  Jack London
     Age  :  41
     Job  :  Writer

An else block can be given that is similar to what we use with if, elif logic - This specifies what to do if things aren’t accounted for above

def print_stuff(**x): #now taking key value pairs
    try:
        print("Pay special attention to " + x.pop("Name") + "\nHis attributes are:")
    except TypeError:
        print("Oh no! 'Name' is not a string")
    except KeyError:
        print("You didn't supply a 'Name'!")
    else: #if no errors occurred
        print("(Valid name by the way - you rule)")
    for key in x:
        print("\t", key, " : ", str(x[key]))

print_stuff(Name = "Jack London", Age = 41, Job = "Writer")
Pay special attention to Jack London
His attributes are:
(Valid name by the way - you rule)
     Age  :  41
     Job  :  Writer
  • A finally clause can be given to always execute at end regardless
  • Like an else block but it always runs
def print_stuff(**x): #now taking key value pairs
    print(x)
    try:
        print("Pay special attention to " + x.pop("Name") + "\nHis attributes are:")
    except TypeError:
        print("Oh no! 'Name' is not a string")
    except KeyError:
        print("You didn't supply a 'Name'!")
    else:
        print("(Valid name by the way)")
    finally:
        print("This string prints no matter what")
    for key in x:
        print("\t", key, " : ", str(x[key]))

print_stuff(name = "Jack London", Age = 41, Job = "Writer")
{'name': 'Jack London', 'Age': 41, 'Job': 'Writer'}
You didn't supply a 'Name'!
This string prints no matter what
     name  :  Jack London
     Age  :  41
     Job  :  Writer

Raising an Exception Yourself

  • You can define your own exceptions to be more descriptive!

  • If we try to divide by 0 we get a ZeroDivisionError exception

3/0
---------------------------------------------------------------------------
ZeroDivisionError                         Traceback (most recent call last)
<ipython-input-25-f6cc6d14333b> in <cell line: 1>()
----> 1 3/0

ZeroDivisionError: division by zero
  • Instead we can raise our own exception
def my_divide(x, y):
    if y == 0:
        raise Exception("Can't divide by 0...")
    return x/y
my_divide(3, 9) #works
0.3333333333333333
my_divide(3,0) #raises our custom exception and provides that note!
---------------------------------------------------------------------------
Exception                                 Traceback (most recent call last)
<ipython-input-28-c7c11902dc96> in <cell line: 1>()
----> 1 my_divide(3,0) #raises our custom exception and provides that note!

<ipython-input-26-418f7e821d7e> in my_divide(x, y)
      1 def my_divide(x, y):
      2     if y == 0:
----> 3         raise Exception("Can't divide by 0...")
      4     return x/y

Exception: Can't divide by 0...

Quick Video

This video shows an example of using error control! Remember to pop the video out into the full player.

The notebook written in the video is available here.

from IPython.display import IFrame
IFrame(src="https://ncsu.hosted.panopto.com/Panopto/Pages/Embed.aspx?id=cd799e52-f4f1-4531-8174-b10301708125&autoplay=false&offerviewer=true&showtitle=true&showbrand=true&captions=false&interactivity=all", height="405", width="720")

Recap

  • Syntax errors and exceptions

  • When an exception is raised:

    • try, except, else, and finally can help!

That wraps up our week 4 content! Head out our Moodle site and start on homework 4.

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If you are on Google Colab, head back to our course website for our next lesson!