python how to convert Nonetype to int or string

Question

Hi, I've got an Nonetype value x, it's generally a number, but could be None. I want to divide it by a number, but python says

TypeError: int() argument must be a string or a number, not 'NoneType'

How could I solve that

Answer 1

That TypeError only appears when you try to pass int() None (which is the only NoneType value, as far as I know). I would say that your real goal should not be to convert NoneType to int, but to figure out where/why you're getting None instead of a number as expected, and either fix it or handle the None properly.

Answer 2

You should check to make sure the value is not None before trying to perform any calculations on it:

my_value = None
if my_value is not None:
    print int(my_value) / 2

Note: my_value was intentionally set to None to prove the code works and that the check is being performed.

Answer 3

int(value or 0)

This will use 0 in the case when you provide any value that Python considers False, such as None, 0, [], "", etc. Since 0 is False, you should only use 0 as the alternative value (otherwise you will find your 0s turning into that value).

int(0 if value is None else value)

This replaces only None with 0. Since we are testing for None specifically, you can use some other value.

Answer 4

This can happen if you forget to return a value from a function: it then returns None. Look at all places where you are assigning to that variable, and see if one of them is a function call where the function lacks a return statement.

Answer 5

In one of the comments, you say:

Somehow I got an Nonetype value, it supposed to be an int, but it's now a Nonetype object

Well, if it's your code, figure out how you're getting None when you expect a number and stop that from happening.

If it's someone else's code, find out the conditions under which it gives None and determine a sensible value to use for that.

Answer 6

A common "Pythonic" way to handle this kind of situation is known as EAFP for "It's easier to ask forgiveness than permission". Which usually means writing code that assumes everything is fine, but then wrapping it with a try..except block just in case to handle things when it's not.

Here's that coding style applied to your problem:

try:
    my_value = int(my_value)
except TypeError:
    my_value = 0  # or whatever you want to do

answer = my_value / divisor

Or perhaps the even simpler and slightly faster:

try:
    answer = my_value / divisor
except TypeError:
    answer = 0

The inverse and more traditional approach is known as LBYL which stands for "Look before you leap" is what @Soviut and some of the others have suggested. For additional coverage of this topic see my answer and associated comments to the question "Determine whether a key is present in a Python dict" elsewhere on this site.

One potential problem with EAFP is that it can hide the fact that something is wrong with some other part of your code or third-party module you're using, especially when the exceptions frequently occur (and therefore aren't really "exceptional" cases at all).