You can do this:

d.pop("", None)
d.pop(None, None)

Pops dictionary with a default value that you ignore.

input on Python 3 returns a string, and '1001' != 1001. If your key is the int 1001, you will need to convert the input to an int:

self._answer= int(input("Which appointment would you like to delete?"))

As an aside, the name self._useless indicates that you might not care about the result of the pop operation. If you just want to remove the entry from the dict, you can use del:

del self._book[self._answer]

or you could ignore the return value of pop and not assign it to anything:

self._book.pop(self._answer)

Just check if self.members is not empty:

if self.members:
    self.members.pop()

or, catch KeyError via try/except:

try:
    self.members.pop()
except KeyError:
    # do smth

dict.pop(key, default) will never raise a KeyError, if the key doesn't exist it just returns the default value. So your try:except: is not useful.

That aside using pop in such a way is fine, especially if you expect that case to be pretty common.

The issue with your second note is that the calling convention of your function is inconsistent, sometimes it returns two results and sometimes it returns none, so it's difficult to use.

• common Python idiom would recommend that your function raise an exception in case of an invalid email (whether missing or improper), that way the "happy path" always gets two values returned and the "unhappy path" is an exception handler, just replace your bare return by raising a suitable exception
• if you don't want to raise an exception for some reason, then you need to either change the result to always be a single value (e.g. a dataclass), or you need to change your error cases to return something like None, None (which is itself somewhat risky in a different manner than the current issue, as if some_function() will always pass).

Your dict size decreases with pop while i increases to the (original) length of the list, at some point you are going to be referencing an key which is higher than the size of the list, resulting in a KeyError.

What you want would be something like:

for i, k in enumerate(nums_d.keys()):
    # code stuff

This way you have i as the range to perform your indexing within nums, and you have k which refers to each key within nums_d, in order to use pop and avoid using an element twice.

This is a problem with the name. The error message is clearly stating that 'Factor_2' is not valid. So you need to figure out what the actual column name is.

Use df.columns to get the column names.

Check for leading and trailing spaces. Is the comma part of the name?

Alternatively you could rename the columns.

eval_data.columns = ["Factor_1", "Factor_2", "Sum"]

Then you should be able to pop your column.

read the documentation for how pop works:

In [65]: dict.pop?
Docstring:
D.pop(k[,d]) -> v, remove specified key and return the corresponding value.
If key is not found, d is returned if given, otherwise KeyError is raised
Type:      method_descriptor

now check out your stack trace. it looks like django is trying to instantiate the object somehow without passing in the requisite keyword-args. so it's best to just add a default to pop