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UnboundLocalError: local variable 'batch_outputs' referenced before assignment #66
Spartan859 commented Aug 19, 2021
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Local variable referenced before assignment in Python
Last updated: Apr 8, 2024 Reading time · 4 min
# Local variable referenced before assignment in Python
The Python "UnboundLocalError: Local variable referenced before assignment" occurs when we reference a local variable before assigning a value to it in a function.
To solve the error, mark the variable as global in the function definition, e.g. global my_var .
Here is an example of how the error occurs.
We assign a value to the name variable in the function.
# Mark the variable as global to solve the error
To solve the error, mark the variable as global in your function definition.
If a variable is assigned a value in a function's body, it is a local variable unless explicitly declared as global .
# Local variables shadow global ones with the same name
You could reference the global name variable from inside the function but if you assign a value to the variable in the function's body, the local variable shadows the global one.
Accessing the name variable in the function is perfectly fine.
On the other hand, variables declared in a function cannot be accessed from the global scope.
The name variable is declared in the function, so trying to access it from outside causes an error.
Make sure you don't try to access the variable before using the global keyword, otherwise, you'd get the SyntaxError: name 'X' is used prior to global declaration error.
# Returning a value from the function instead
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.
We simply return the value that we eventually use to assign to the name 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.
We passed the name 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 .
# Assigning a value to a local variable from an outer scope
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.
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.
Printing the message variable on the last line of the function shows an empty string because the inner() function has its own scope.
Changing the value of the variable in the inner scope is not possible unless we use the nonlocal keyword.
Instead, the message variable in the inner function simply shadows the variable with the same name from the outer scope.
# Discussion
As shown in this section of the documentation, when you assign a value to a variable inside a function, the variable:
- Becomes local to the scope.
- Shadows any variables from the outer scope that have the same name.
The last line in the example function assigns a value to the name variable, marking it as a local variable and shadowing the name variable from the outer scope.
At the time the print(name) line runs, the name variable is not yet initialized, which causes the error.
The most intuitive way to solve the error is to use the global keyword.
The global keyword is used to indicate to Python that we are actually modifying the value of the name variable from the outer scope.
- If a variable is only referenced inside a function, it is implicitly global.
- If a variable is assigned a value inside a function's body, it is assumed to be local, unless explicitly marked as global .
If you want to read more about why this error occurs, check out [this section] ( this section ) of the docs.
# Additional Resources
You can learn more about the related topics by checking out the following tutorials:
- SyntaxError: name 'X' is used prior to global declaration
Borislav Hadzhiev
Web Developer
Copyright © 2024 Borislav Hadzhiev
How to Fix Local Variable Referenced Before Assignment Error in Python
Table of Contents
Fixing local variable referenced before assignment error.
In Python , when you try to reference a variable that hasn't yet been given a value (assigned), it will throw an error.
That error will look like this:
In this post, we'll see examples of what causes this and how to fix it.
Let's begin by looking at an example of this error:
If you run this code, you'll get
The issue is that in this line:
We are defining a local variable called value and then trying to use it before it has been assigned a value, instead of using the variable that we defined in the first line.
If we want to refer the variable that was defined in the first line, we can make use of the global keyword.
The global keyword is used to refer to a variable that is defined outside of a function.
Let's look at how using global can fix our issue here:
Global variables have global scope, so you can referenced them anywhere in your code, thus avoiding the error.
If you run this code, you'll get this output:
In this post, we learned at how to avoid the local variable referenced before assignment error in Python.
The error stems from trying to refer to a variable without an assigned value, so either make use of a global variable using the global keyword, or assign the variable a value before using it.
Thanks for reading!
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