I'm Getting Key Error in Python

I'm getting Key error in python

A KeyError generally means the key doesn't exist. So, are you sure the path key exists?

From the official python docs:

exception KeyError

Raised when a mapping (dictionary) key is not found in the set of
existing keys.

For example:

>>> mydict = {'a':'1','b':'2'}
>>> mydict['a']
'1'
>>> mydict['c']
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
KeyError: 'c'
>>>

So, try to print the content of meta_entry and check whether path exists or not.

>>> mydict = {'a':'1','b':'2'}
>>> print mydict
{'a': '1', 'b': '2'}

Or, you can do:

>>> 'a' in mydict
True
>>> 'c' in mydict
False

I keep on getting a KeyError in Python

Key error means the dictionary you are trying to access does not have the key you are using to get a value from it. It looks like you are trying to access the row dictionaries "Email" key. Your csv file does not have an "Email" column in some rows and thus is giving you this error. To solve, you can do row.get("Email","") which will just return an empty string if there is no email.

You can also just do a check before you append so that you arent adding empy items to your list by doing

  for row in user_reader:
    email = row.get("Email")
    if email is None: continue
    list_of_email_addresses.append(email)

Can insertion in dictionary cause KeyError?

Insertion will never raise a KeyError, but it can raise TypeError if you supply a bad key, and that generally means you're trying to use a mutable object as a key. Python integers are immutable, so they won't raise a TypeError.

Note that if you try to add too many items to your dict you will get a MemoryError due to lack of available RAM, or an OverflowError because you've attempted to exceed the maximum size of a collection (which can be read from sys.maxsize). But there's not much point bothering to catch those. :)

I'm getting KeyError: ' ' in my code and I don't see why

Like I said in the comment, the problem seems to be from trying to get the value from the dict returned from self.get_eln() and self.get_esr() when the self.combo3 or self.combo4 has no value.

I guess you could either check if both have values before trying to do so like this:

if self.combo3.get() and self.combo4.get():
    var2 = self.get_eln()[self.combo3.get()] + "-" + self.get_esr([self.combo4.get()]
else:
    var2 = '' # not sure which value you want for this var

and this way you ensure you don't get a KeyError.

Otherwise, if you wish to follow another approach with try ... except you could always do something like

try:
    var2 = self.get_eln()[self.combo3.get()] + "-" + self.get_esr([self.combo4.get()]
except KeyError:
    # make sure you don't want to 
    var2 = ''

I don't want to give you a simple copy-paste answer, so if you don't understand something, just comment and I'll try to help!

How to I get the missing key from a KeyError in Python?

Alredy answered here Link

map1 = {}
try:
    x = map1['key2']
except KeyError as e:    
    print(e.args[0])

Why am I getting KeyError: 0

it looks like you store cells locations as a tuple of two integers.
so you should address cell_dict with that instead of an integer.

Use cell_dict[(i, j)] instead of cell_dict[i][j].

Why am I getting this KeyError?

You have the keys and values backward. You are using the scores as keys in your dictionary and strings of letters as the values. In addition you are not using strip() on the input value, so you are processing incidental whitespace instead of alphabetical characters alone.

How to resolve Key Error Exception when accessing dictionary key inside an if statement

From a quick look at your code I can determine that you Never initialize the key 'user_res' in your dictionary unless an error occurs, so in some way nothing is going wrong and that is causing your crash.

Your django code tries to build the context dictionary and is getting a KeyError exception because the 'user_res' key is simply not there!, a simple solution would be to initialize this key in all your dictionaries, for example like:

my_dict = {
            'odd1': home_odd,
            'odd2': away_odd,
            'wager_amount': wager_amount,
            'ev': round(ev, 2),
            'back_odd': back_odd,
            'user_res' : 'everything went fine!'
        }

this would be sufficient in order for your application to work!

P.S, I must also add that it would be a slightly more elegant solution if you had your user simply pick the bet by entering a number (1 or 2 depending on the bet) rather than entering a long floating point value and being frustrated if he has a typo.

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