UnboundLocalError: cannot access local variable 'X' where it is not associated with a value

# UnboundLocalError: cannot access local variable 'X' where it is not associated with a value

The Python "UnboundLocalError: cannot access local variable 'X' where it is not associated with a value" occurs for multiple reasons:

• Referencing a local variable before assigning a value to it in a function.
• Only assigning a value to a variable if a certain condition is met.
• Only assigning a value to a variable in a try/except statement.

To solve the error, mark the variable as global before referencing it in the function or make sure to assign a value to the variable regardless if a condition is met.

Here is an example of how the error occurs.

Copied!
country = 'Austria'


def my_func():
    # ⛔️ UnboundLocalError: cannot access local variable 'X' where it is not associated with a value
    print(country)

    country = 'Germany'


my_func()

We reference the country variable in the function before assigning a value to it.

The local country variable shadows the country variable that is declared in the global scope.

# Mark the variable as global before accessing it within the function

To solve the error, mark the variable as global before accessing it within your function.

Copied!
country = 'Austria'


def my_func():
    # 👇️ Mark as global first
    global country

    print(country)

    country = 'Germany'


# 👇️ Call function to reassign the variable
my_func()  # 👉️ Austria

# 👇️ Now the `country` variable is set to `Germany`

my_func()  # 👉️ Germany

When we mark the variable as global, we tell to the interpreter to look for it in the global scope.

The first call to the function reassigns the global country variable, setting it to "Germany".

Once we call the function the second time and print the country variable, it's already set to "Germany".

# Local variables shadow global ones with the same name

You can reference the country global variable from within the function, however, if you assign a value to the variable, the local variable shadows the global one.

Copied!
country = 'Austria'


def my_func():
    # 👇️ Can access global variable here
    print(country)


my_func()  # 👉️ Austria

If you assign a value to the variable within the function without marking it as global, the assignment only applies to the function's local scope.

Copied!
country = 'Austria'


def my_func():
    # 👇️ This only applies within the function
    country = 'Germany'

    print(country)


print(country)  # 👉️ Austria

my_func()  # 👉️ Germany

print(country)  # 👉️ Austria

The country variable in the function is set to 'Germany' and the country variable outside the function has a value of 'Austria'.

This behavior is very confusing and should generally be avoided.

It is much better to pick different names and avoid clashes.

# Assigning values to local variables from an outer function

If you have a nested function and are trying to assign a value to the local variables from the outer function, use the nonlocal keyword.

Copied!
def outer():
    # 👇️ Initialize country variable
    country = ''

    def inner():
        # 👇️ Mark country as nonlocal
        nonlocal country
        country = 'Germany'
        print(country)

    # 👇️ Invoke inner function
    inner()

    print(country)  # 👉️ "Germany"


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 country variable
    country = ''

    def inner():
        # 👇️ Declares country in inner function's scope
        country = 'Germany'
        print(country)

    inner()

    # 👇️ Refers to outer's country variable
    print(repr(country))  # 👉️ ""


outer()

If you get the error "Local variable referenced before assignment", check out the following article.

# Returning a value that is used to reassign the global variable

An alternative solution to using the global keyword is to return a value from the function and use the value to reassign the global variable.

Copied!
country = 'Austria'


def my_func():
    print(country)  # 👉️ Austria

    new_country = 'Germany'
    return new_country


result = my_func()
print(result)  # 👉️ Germany

country = result
print(country)  # 👉️ Germany

We simply return the value that we eventually assign to the country global variable.

# Passing the global variable as an argument to the function

You should also consider passing the global variable as an argument to the function.

Copied!
country = 'Austria'


def my_func(ctry):
    location = 'Linz, ' + ctry

    return location


result = my_func(country)
print(result)  # 👉️ Linz, Austria

We passed the country global variable as an argument to the function.

If we assign a value to a variable in a function, the variable is assumed to be local unless explicitly declared as global.

# Only assigning a value to a variable is a condition is met

Another common cause of the error is assigning a value to a variable only if a condition is met and trying to access it after.

Copied!
def my_func():
    if 5 > 10:
        country = 'Germany'
        print(country)

    # ⛔️ UnboundLocalError: cannot access local variable 'country' where it is not associated with a value
    return f'Country: {country}'


my_func()

The if statement checks if 5 is greater than 10.

The condition is never met, so the country variable never gets set.

When we try to access the variable after the if block, we get the error because the variable was never associated with a value.

To solve the error, set the variable to a default value before the if block.

Copied!
def my_func():
    country = None

    if 5 > 10:
        country = 'Germany'
        print(country)

    return f'Country: {country}'


print(my_func())

We initialized the country variable to None outside the if block.

We used a default value of None, but you can also use an empty string "" or 0 as the default value, depending on your use case.

The variable is set even if the if condition isn't met, so the error is never raised.

# Only assigning a value to a variable in a try/except statement

Another cause of the error is only assigning a value to a variable in a try/except statement.

Copied!
def my_func():
    try:
        raise ValueError('Something went wrong')

        country = 'Germany'
        print(country)
    except ValueError:
        pass

    # ⛔️ UnboundLocalError: cannot access local variable 'country' where it is not associated with a value
    return f'Country: {country}'


print(my_func())

Notice that an error is raised before the country variable is declared in the try block.

The variable is never given a value, so when we try to access it in the return statement, an UnboundLocalError occurs.

To solve the error, set the variable to an initial value before the try statement.

Copied!
def my_func():
    country = None

    try:
        raise ValueError('Something went wrong')

        country = 'Germany'
        print(country)
    except ValueError:
        pass

    return f'Country: {country}'


print(my_func())  # 👉️ Country: None

We initialized the country variable to None before the try block.

We used None as the default value, but you can also set the variable to an empty string "" or 0 depending on your use case.

Even if an error is raised in the try block, the variable is still associated with a value because we've initialized it prior.

# Additional Resources

You can learn more about the related topics by checking out the following tutorials:

• Python: Convert string with comma separator and dot to float
• Format number with comma as thousands separator in Python
• Not all arguments converted during string formatting (Python)
• Python: locale.Error: unsupported locale setting [Solved]