Fix: KeyError: exception in Python

Example:

u2_album = {
    'title': 'The Joshua Tree',
    'year': 1987,
    'genre': 'Rock',
    'tracks': ['Where the Streets Have No Name', 'I Still Haven\'t Found What I\'m Looking For', 'With or Without You', 'Bullet the Blue Sky', 'Running to Stand Still', 'Red Hill Mining Town', 'In God\'s Country', 'Trip Through Your Wires', 'One Tree Hill', 'Exit', 'Mothers of the Disappeared']
}

print(u2_album['title'])
print(u2_album['year']) 
print(u2_album['track'])

If you try to run the above Python code, you will get a KeyError error.

The Joshua Tree
1987
---------------------------------------------------------------------------
KeyError                                  Traceback (most recent call last)
<ipython-input-25-be282156e277> in <cell line: 10>()
      8 print(u2_album['title'])
      9 print(u2_album['year'])
---> 10 print(u2_album['track'])

KeyError: 'track'

Why KeyError exception?

Python will throw a KeyError exception when a dictionary key is not found in the set of existing keys.

Fix:

If you look closely at the line where the error is thrown, we have made use of track which does not exist, its tracks.

Should be:

print(u2_album['tracks'])

Reference: https://docs.python.org/3/library/exceptions.html#KeyError

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Author Info:

Rakesh is a seasoned developer with over 10 years of experience in web and app development, and a deep knowledge of operating systems. Author of insightful How-To articles for Code2care.

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