How to check if a variable is None in Python using an if statement?

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#1

I have a method that sometimes returns a NoneType value. How can I check if a variable is of NoneType in Python? For example, I want to use an if statement to handle cases where the variable might be None. Here’s what I’m trying to do:

if not new: new = ‘#’

I realize that this might not be the correct approach, and I hope you understand what I’m aiming to achieve with this. How can I properly handle a NoneType value in Python?

#2

Hello Asheenaraghunuhan,

To check if a variable is of NoneType, use the is operator, like this:

if variable is None: This approach works because None is the sole singleton object of NoneType in Python. The is operator checks for object identity, meaning it verifies whether two variables point to the same object.

As stated in the documentation: The operators is and is not test for object identity: x is y is true if and only if x and y are the same object. x is not y yields the inverse truth value. Since there is only one instance of None, using is is the preferred and reliable way to check for None.

#3

Hey Asheenaraghununan,

I hope this example will be helpful:

print(type(None)) # NoneType You can check the type of a variable against NoneType like this:

Example

name = 12 # name = None

if type(name) is type(None): print(“Can’t find name”) else: print(name)

In this example, if name is None, it will print “Can’t find name”. Otherwise, it will print the value of name.

#4

Hey Asheenaraghununan,

You can explicitly test for NoneType by importing it:

from types import NoneType

Alternatively, without importing:

NoneType = type(None)

This is useful when you have a variable that is optionally set to either its type or None. You can validate its type using a single isinstance call:

from types import NoneType foo = ‘bar’ assert isinstance(foo, (str, NoneType))

This is equivalent to: And also:

assert isinstance(foo, str) or (foo is None)