NameError: name 'X' is not defined in Python

NameError: name 'X' is not defined in Python #

The Python "NameError: name is not defined" occurs when we try to access a variable or function that is not defined or before it is defined. To solve the error, make sure you haven't misspelled the variable's name and access it after it has been declared.

Here is an example of how the error occurs.

Copied!
employee = {
    'name': 'Alice',
    'age': 30,
}

# ⛔️ NameError: name 'Employee' is not defined. Did you mean: 'employee'?
print(Employee) # 👈️ misspelled variable's name

The issue is that we have misspelled the variable's name. Note that the names of variables, functions and classes are case-sensitive.

To solve the error in this scenario, we have to spell the variable's name correctly.

Copied!
employee = {
    'name': 'Alice',
    'age': 30,
}

print(employee)

The Python "NameError: name is not defined" occurs for multiple reasons:

1. Accessing a variable that doesn't exist.
2. Accessing a variable, function or class before it is declared.
3. Misspelling the name of a variable, a function or a class (names are case-sensitive).
4. Not wrapping a string in quotes, e.g. print(hello).
5. Not wrapping a key of a dictionary in quotes.
6. Using built-in modules without importing them first.
7. Accessing a scoped variable from outside. For example, declaring a variable in a function and trying to access it from outside.

The first thing you need to make sure is that you aren't accessing a variable that doesn't exist or has not yet been defined.

Copied!
# ⛔️ NameError: name 'do_math' is not defined
print(do_math(15, 15))


def do_math(a, b):
    return a + b

The issue in the example above is that we are trying to call the do_math function before it has been declared.

To solve the error, we have to move the line that calls the function or accesses the variable after it has been declared.

Copied!
def do_math(a, b):
    return a + b


print(do_math(15, 15))  # 👉️ 30

Another cause of the error is forgetting to wrap a string in single or double quotes.

Copied!
def greet(name):
    return 'Hello ' + name


# ⛔️ NameError: name 'Alice' is not defined. Did you mean: 'slice'?
greet(Alice) # 👈️ forgot to wrap string in quotes

The greet function expects to get called with a string but we forgot to wrap the string in quotes, so the name 'X' is not defined error occurred.

To solve the error, wrap the string in quotes.

Copied!
def greet(name):
    return 'Hello ' + name

greet('Alice')

The error is also caused if you have a dictionary and forget to wrap its keys in quotes.

Copied!
employee = {
    'name': 'Alice',
    # ⛔️ NameError: name 'age' is not defined
    age: 30 # 👈️ dictionary key not wrapped in quotes
}

Unless you have numeric keys in the dictionary, make sure to wrap them in single or double quotes.

Copied!
employee = {
    'name': 'Alice',
    'age': 30
}

The "NameError: name is not defined" is also caused if you use a built-in module without importing it.

Copied!
# ⛔️ NameError: name 'math' is not defined
print(math.floor(15.5))

To solve the error, make sure to import any modules you are using.

Copied!
import math

print(math.floor(15.5))  # 👉️ 15

The error also occurs if you try to access a scoped variable from outside.

Copied!
def get_message():
    message = 'hello world' # 👈️ variable declared in function
    return message


get_message()

# ⛔️ NameError: name 'message' is not defined
print(message)

The message variable is declared in the get_message function, so it isn't accessible from the outer scope.

The best way to solve the error is to declare the variable in the outer scope if you have to access it from outside.

Copied!
# 👇️ declare variable in outer scope
message = 'hello world'

def get_message():
    return message


get_message()

print(message)  # 👉️ "hello world"

An alternative in this scenario would also be to return the value from the function and store it in a variable.

Copied!
def get_message():
    message = 'hello world'
    return message


result = get_message()

print(result)  # 👉️ "hello world"

Another alternative would be to mark the variable as global.

Copied!
def get_message():
    # 👇️ mark message as global
    global message

    # 👇️ change its value
    message = 'hello world'

    return message


get_message()

print(message)  # 👉️ "hello world"

If you are trying to access a variable that is declared in a nested function from an outer function, you can mark the variable as nonlocal.

Copied!

def outer():
    def inner():
        message = 'hello world'
        print(message)

    inner()

    # ⛔️ NameError: name 'message' is not defined
    print(message)


outer()

The inner function declares a variable named message but we try to access the variable from the outer function and get the "name message is not defined" error.

To get around this, we can mark the message variable as nonlocal.

Copied!
def outer():
    # 👇️ initialize message variable
    message = ''

    def inner():
        # 👇️ Mark message as nonlocal
        nonlocal message
        message = 'hello world'
        print(message)

    inner()

    print(message)  # 👉️ "hello world"


outer()

The nonlocal keyword allows us to work with the local variables of enclosing functions.

Had we not used the nonlocal statement, the call to the print() function would have returned an empty string.

Copied!

def outer():
    # 👇️ initialize message variable
    message = ''

    def inner():
        # 👇️ declares message in inner's scope
        message = 'hello world'
        print(message)

    inner()

    print(message)  # 👉️ ""

outer()

Conclusion #

The Python "NameError: name is not defined" occurs when we try to access a variable or function that is not defined or before it is defined. To solve the error, make sure you haven't misspelled the variable's name and access it after it has been declared.