unboundlocalerror local variable 'values' referenced before assignment

Local variable referenced before assignment in Python

Last updated: Feb 17, 2023 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

Fixing Python UnboundLocalError: Local Variable ‘x’ Accessed Before Assignment

Understanding unboundlocalerror.

The UnboundLocalError in Python occurs when a function tries to access a local variable before it has been assigned a value. Variables in Python have scope that defines their level of visibility throughout the code: global scope, local scope, and nonlocal (in nested functions) scope. This error typically surfaces when using a variable that has not been initialized in the current function’s scope or when an attempt is made to modify a global variable without proper declaration.

Solutions for the Problem

To fix an UnboundLocalError, you need to identify the scope of the problematic variable and ensure it is correctly used within that scope.

Method 1: Initializing the Variable

Make sure to initialize the variable within the function before using it. This is often the simplest fix.

Method 2: Using Global Variables

If you intend to use a global variable and modify its value within a function, you must declare it as global before you use it.

Method 3: Using Nonlocal Variables

If the variable is defined in an outer function and you want to modify it within a nested function, use the nonlocal keyword.

That’s it. Happy coding!

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[SOLVED] Local Variable Referenced Before Assignment

Python treats variables referenced only inside a function as global variables. Any variable assigned to a function’s body is assumed to be a local variable unless explicitly declared as global.

Why Does This Error Occur?

Unboundlocalerror: local variable referenced before assignment occurs when a variable is used before its created. Python does not have the concept of variable declarations. Hence it searches for the variable whenever used. When not found, it throws the error.

Before we hop into the solutions, let’s have a look at what is the global and local variables.

Local Variable Declarations vs. Global Variable Declarations

Local Variable Referenced Before Assignment Error with Explanation

Try these examples yourself using our Online Compiler.

Let’s look at the following function:

Explanation

The variable myVar has been assigned a value twice. Once before the declaration of myFunction and within myFunction itself.

Using Global Variables

Passing the variable as global allows the function to recognize the variable outside the function.

Create Functions that Take in Parameters

Instead of initializing myVar as a global or local variable, it can be passed to the function as a parameter. This removes the need to create a variable in memory.

UnboundLocalError: local variable ‘DISTRO_NAME’

This error may occur when trying to launch the Anaconda Navigator in Linux Systems.

Upon launching Anaconda Navigator, the opening screen freezes and doesn’t proceed to load.

Try and update your Anaconda Navigator with the following command.

If solution one doesn’t work, you have to edit a file located at

After finding and opening the Python file, make the following changes:

In the function on line 159, simply add the line:

DISTRO_NAME = None

Save the file and re-launch Anaconda Navigator.

DJANGO – Local Variable Referenced Before Assignment [Form]

The program takes information from a form filled out by a user. Accordingly, an email is sent using the information.

Upon running you get the following error:

We have created a class myForm that creates instances of Django forms. It extracts the user’s name, email, and message to be sent.

A function GetContact is created to use the information from the Django form and produce an email. It takes one request parameter. Prior to sending the email, the function verifies the validity of the form. Upon True , .get() function is passed to fetch the name, email, and message. Finally, the email sent via the send_mail function

Why does the error occur?

We are initializing form under the if request.method == “POST” condition statement. Using the GET request, our variable form doesn’t get defined.

Local variable Referenced before assignment but it is global

This is a common error that happens when we don’t provide a value to a variable and reference it. This can happen with local variables. Global variables can’t be assigned.

This error message is raised when a variable is referenced before it has been assigned a value within the local scope of a function, even though it is a global variable.

Here’s an example to help illustrate the problem:

In this example, x is a global variable that is defined outside of the function my_func(). However, when we try to print the value of x inside the function, we get a UnboundLocalError with the message “local variable ‘x’ referenced before assignment”.

This is because the += operator implicitly creates a local variable within the function’s scope, which shadows the global variable of the same name. Since we’re trying to access the value of x before it’s been assigned a value within the local scope, the interpreter raises an error.

To fix this, you can use the global keyword to explicitly refer to the global variable within the function’s scope:

However, in the above example, the global keyword tells Python that we want to modify the value of the global variable x, rather than creating a new local variable. This allows us to access and modify the global variable within the function’s scope, without causing any errors.

Local variable ‘version’ referenced before assignment ubuntu-drivers

This error occurs with Ubuntu version drivers. To solve this error, you can re-specify the version information and give a split as 2 –

Here, p_name means package name.

With the help of the threading module, you can avoid using global variables in multi-threading. Make sure you lock and release your threads correctly to avoid the race condition.

When a variable that is created locally is called before assigning, it results in Unbound Local Error in Python. The interpreter can’t track the variable.

Therefore, we have examined the local variable referenced before the assignment Exception in Python. The differences between a local and global variable declaration have been explained, and multiple solutions regarding the issue have been provided.

Trending Python Articles

How to fix UnboundLocalError: local variable 'x' referenced before assignment in Python

by Nathan Sebhastian

Posted on May 26, 2023

Reading time: 2 minutes

One error you might encounter when running Python code is:

This error commonly occurs when you reference a variable inside a function without first assigning it a value.

You could also see this error when you forget to pass the variable as an argument to your function.

Let me show you an example that causes this error and how I fix it in practice.

How to reproduce this error

Suppose you have a variable called name declared in your Python code as follows:

Next, you created a function that uses the name variable as shown below:

When you execute the code above, you’ll get this error:

This error occurs because you both assign and reference a variable called name inside the function.

Python thinks you’re trying to assign the local variable name to name , which is not the case here because the original name variable we declared is a global variable.

How to fix this error

To resolve this error, you can change the variable’s name inside the function to something else. For example, name_with_title should work:

As an alternative, you can specify a name parameter in the greet() function to indicate that you require a variable to be passed to the function.

When calling the function, you need to pass a variable as follows:

This code allows Python to know that you intend to use the name variable which is passed as an argument to the function as part of the newly declared name variable.

Still, I would say that you need to use a different name when declaring a variable inside the function. Using the same name might confuse you in the future.

Here’s the best solution to the error:

Now it’s clear that we’re using the name variable given to the function as part of the value assigned to name_with_title . Way to go!

The UnboundLocalError: local variable 'x' referenced before assignment occurs when you reference a variable inside a function before declaring that variable.

To resolve this error, you need to use a different variable name when referencing the existing variable, or you can also specify a parameter for the function.

I hope this tutorial is useful. See you in other tutorials.

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Python local variable referenced before assignment Solution

When you start introducing functions into your code, you’re bound to encounter an UnboundLocalError at some point. This error is raised when you try to use a variable before it has been assigned in the local context .

In this guide, we talk about what this error means and why it is raised. We walk through an example of this error in action to help you understand how you can solve it.

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What is unboundlocalerror: local variable referenced before assignment.

Trying to assign a value to a variable that does not have local scope can result in this error:

Python has a simple rule to determine the scope of a variable. If a variable is assigned in a function , that variable is local. This is because it is assumed that when you define a variable inside a function you only need to access it inside that function.

There are two variable scopes in Python: local and global. Global variables are accessible throughout an entire program; local variables are only accessible within the function in which they are originally defined.

Let’s take a look at how to solve this error.

An Example Scenario

We’re going to write a program that calculates the grade a student has earned in class.

We start by declaring two variables:

These variables store the numerical and letter grades a student has earned, respectively. By default, the value of “letter” is “F”. Next, we write a function that calculates a student’s letter grade based on their numerical grade using an “if” statement :

Finally, we call our function:

This line of code prints out the value returned by the calculate_grade() function to the console. We pass through one parameter into our function: numerical. This is the numerical value of the grade a student has earned.

Let’s run our code and see what happens:

An error has been raised.

The Solution

Our code returns an error because we reference “letter” before we assign it.

We have set the value of “numerical” to 42. Our if statement does not set a value for any grade over 50. This means that when we call our calculate_grade() function, our return statement does not know the value to which we are referring.

We do define “letter” at the start of our program. However, we define it in the global context. Python treats “return letter” as trying to return a local variable called “letter”, not a global variable.

We solve this problem in two ways. First, we can add an else statement to our code. This ensures we declare “letter” before we try to return it:

Let’s try to run our code again:

Our code successfully prints out the student’s grade.

If you are using an “if” statement where you declare a variable, you should make sure there is an “else” statement in place. This will make sure that even if none of your if statements evaluate to True, you can still set a value for the variable with which you are going to work.

Alternatively, we could use the “global” keyword to make our global keyword available in the local context in our calculate_grade() function. However, this approach is likely to lead to more confusing code and other issues. In general, variables should not be declared using “global” unless absolutely necessary . Your first, and main, port of call should always be to make sure that a variable is correctly defined.

In the example above, for instance, we did not check that the variable “letter” was defined in all use cases.

That’s it! We have fixed the local variable error in our code.

The UnboundLocalError: local variable referenced before assignment error is raised when you try to assign a value to a local variable before it has been declared. You can solve this error by ensuring that a local variable is declared before you assign it a value.

Now you’re ready to solve UnboundLocalError Python errors like a professional developer !

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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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How to Fix – UnboundLocalError: Local variable Referenced Before Assignment in Python

Developers often encounter the  UnboundLocalError Local Variable Referenced Before Assignment error in Python. In this article, we will see what is local variable referenced before assignment error in Python and how to fix it by using different approaches.

What is UnboundLocalError: Local variable Referenced Before Assignment?

This error occurs when a local variable is referenced before it has been assigned a value within a function or method. This error typically surfaces when utilizing try-except blocks to handle exceptions, creating a puzzle for developers trying to comprehend its origins and find a solution.

Below, are the reasons by which UnboundLocalError: Local variable Referenced Before Assignment error occurs in  Python :

Nested Function Variable Access

Global variable modification.

In this code, the outer_function defines a variable ‘x’ and a nested inner_function attempts to access it, but encounters an UnboundLocalError due to a local ‘x’ being defined later in the inner_function.

In this code, the function example_function tries to increment the global variable ‘x’, but encounters an UnboundLocalError since it’s treated as a local variable due to the assignment operation within the function.

Solution for Local variable Referenced Before Assignment in Python

Below, are the approaches to solve “Local variable Referenced Before Assignment”.

In this code, example_function successfully modifies the global variable ‘x’ by declaring it as global within the function, incrementing its value by 1, and then printing the updated value.

In this code, the outer_function defines a local variable ‘x’, and the inner_function accesses and modifies it as a nonlocal variable, allowing changes to the outer function’s scope from within the inner function.

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UnboundLocalError: local variable 'values' referenced before assignment in lr_scheduler

This my code:

how to solve this error?

Could you post a code snippet to reproduce this issue, please? This dummy example runs fine:

Need any other snippet code? I’m not sure how much complete code you need

A minimal and executable code snippet would be great. Could you try to remove unnecessary functions and use some random inputs, so that we can reproduce this issue locally?

I’m sorry for my slow response。This code might suit your needs。

You can see that the first call to scheduler. Step should have been fault-free because it printed the following print statement, as well as eval for the model. But an error should have occurred on the second call

I forgot the code that came out and this is the following code

Now it turns out that if you use annotated code, you get an error, while unannotated code works fine

I really don’t know why, is it the data problem that caused the error

Thanks for the code so far. Could you also post the code you are using to initialize the model, optimizer, and scheduler? Also, could you try to run the code on the CPU only and check, if you see the same error? If not, could you rerun the GPU code using CUDA_LAUNCH_BLOCKING=1 python script.py args and post the stack trace again?

This is my code:

It was very embarrassing that I could not debug with CPU because of the machine, but if I used GPU to debug, the error result did not change

Could you call scheudler.get_lr() before the error is thrown and check the return value, please?

I am very sorry for replying to you a few days later, because I am a sophomore student in university and I have a lot of things to do recently, so I didn’t deal with this problem for a few days. As you requested, I added this line of code. The problem is that it returned the value successfully without any problems during the first epoch. But after the second epoch, it reported an error

Are you recreating or manipulating the scheduler or optimizer in each epoch somehow?

29/5000 Thank you for answering my question so patiently. My code should not have this problem。 If I use following code,the error will not appear( it will appear when using the annotating code):

Here is my complete training code(When you put scheduler. Step () into each batch iteration,error will apear)：

I had the same error that I think has been fixed.

In the end it seems like the number of epochs you had mentioned in your scheduler was less than the number of epochs you tried training for. I went into %debug in the notebook and tried calling self.get_lr() as suggested. I got this message: *** ValueError: Tried to step 3752 times. The specified number of total steps is 3750

Then with some basic math and a lot of code search I realised that I had specified 5 epochs in my scheduler but called for 10 epochs in my fit function.

Hope this helps.

There is an error with total_steps.i am also getting same error but i rectified it

Python has lexical scoping by default, which means that although an enclosed scope can access values in its enclosing scope, it cannot modify them (unless they’re declared global with the global keyword). A closure binds values in the enclosing environment to names in the local environment. The local environment can then use the bound value, and even reassign that name to something else, but it can’t modify the binding in the enclosing environment. UnboundLocalError happend because when python sees an assignment inside a function then it considers that variable as local variable and will not fetch its value from enclosing or global scope when we execute the function. To modify a global variable inside a function, you must use the global keyword.

Hello,I had the same error that I can’t solve it. I want to ask you some question about it.Thank you. What is the ‘The specified number of total steps is 3750’? How to change the number of steps? Thank you.

Hi make sure that your dataloader and the scheduler have the same number of iterations. If I remember correctly I got this error when using the OneCycle LR scheduler which needs you to specify the max number of steps as init parameter. Hope this helps! If this isn’t the error you have, then please provide code and try to see what your scheduler.get_lr() method returns.

Local variable referenced before assignment in Python

The “local variable referenced before assignment” error occurs in Python when you try to use a local variable before it has been assigned a value.

This error typically arises in situations where you declare a variable within a function but then try to access or modify it before actually assigning a value to it.

Here’s an example to illustrate this error:

In this example, you would encounter the “local variable ‘x’ referenced before assignment” error because you’re trying to print the value of x before it has been assigned a value. To fix this, you should assign a value to x before attempting to access it:

In the corrected version, the local variable x is assigned a value before it’s used, preventing the error.

Keep in mind that Python treats variables inside functions as local unless explicitly stated otherwise using the global keyword (for global variables) or the nonlocal keyword (for variables in nested functions).

If you encounter this error and you’re sure that the variable should have been assigned a value before its use, double-check your code for any logical errors or typos that might be causing the variable to not be assigned properly.

Using the global keyword

If you have a global variable named letter and you try to modify it inside a function without declaring it as global, you will get error.

This is because Python assumes that any variable that is assigned a value inside a function is a local variable, unless you explicitly tell it otherwise.

To fix this error, you can use the global keyword to indicate that you want to use the global variable:

Using nonlocal keyword

The nonlocal keyword is used to work with variables inside nested functions, where the variable should not belong to the inner function. It allows you to modify the value of a non-local variable in the outer scope.

For example, if you have a function outer that defines a variable x , and another function inner inside outer that tries to change the value of x , you need to use the nonlocal keyword to tell Python that you are referring to the x defined in outer , not a new local variable in inner .

Here is an example of how to use the nonlocal keyword:

If you don’t use the nonlocal keyword, Python will create a new local variable x in inner , and the value of x in outer will not be changed:

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What is UnboundLocalError: local variable referenced before assignment?

Trying to assign a value to a variable that does not have local scope can result in this error:

Python has a simple rule to determine the scope of a variable. To clarify, a variable is assigned in a function, that variable is local. Because it is assumed that when you define a variable inside a function, you only need to access it inside that function.

There are two variable scopes in Python: local and global. Global variables are accessible throughout an entire program. Whereas, local variables are only accessible within the function in which they are originally defined.

An example of Local variable referenced before assignment

We’re going to write a program that calculates the grade a student has earned in class.

Firstly, we start by declaring two variables:

These variables store the numerical and letter grades a student has earned, respectively. By default, the value of “letter” is “F”. Then, we write a function that calculates a student’s letter grade based on their numerical grade using an “if” statement:

Finally, we call our function:

This line of code prints out the value returned by the  calculate_grade()  function to the console. We pass through one parameter into our function: numerical. This is the numerical value of the grade a student has earned.

Let’s run our code of Local variable referenced before assignment and see what happens:

Here is an error!

The Solution of Local variable referenced before assignment

The code returns an error: Unboundlocalerror local variable referenced before assignment because we reference “letter” before we assign it.

We have set the value of “numerical” to 42. Our  if  statement does not set a value for any grade over 50. This means that when we call our  calculate_grade()  function, our return statement does not know the value to which we are referring.

Moreover, we do define “letter” at the start of our program. However, we define it in the global context. Because Python treats “return letter” as trying to return a local variable called “letter”, not a global variable.

Therefore, this problem of variable referenced before assignment could be solved in two ways. Firstly, we can add an  else  statement to our code. This ensures we declare “letter” before we try to return it:

Let’s try to run our code again:

Our code successfully prints out the student’s grade. This approach is good because it lets us keep “letter” in the local context. To clarify, we could even remove the “letter = “F”” statement from the top of our code because we do not use it in the global context.

Alternatively, we could use the “global” keyword to make our global keyword available in the local context in our  calculate_grade()  function:

We use the “global” keyword at the start of our function.

This keyword changes the scope of our variable to a global variable. This means the “return” statement will no longer treat “letter” like a local variable. Let’s run our code. Our code returns: F.

The code works successfully! Let’s try it using a different grade number by setting the value of “numerical” to a new number:

Our code returns: B.

Finally, we have fixed the local variable referenced before assignment error in the code.

To sum up, as you can see, the UnboundLocalError: local variable referenced before assignment error is raised when you try to assign a value to a local variable before it has been declared. Then, you can solve this error by ensuring that a local variable is declared before you assign it a value. Moreover, if a variable is declared globally that you want to access in a function, you can use the “global” keyword to change its value. In case you have any inquiry, let’s CONTACT US . With a lot of experience in Mobile app development services , we will surely solve it for you instantly.

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• fix python error , Local variable referenced before assignment , python , python dictionary , python error , python learning , UnboundLocalError , UnboundLocalError in Python

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UnboundLocalError: local variable 'new_value' referenced before assignment #82

kannes commented Feb 16, 2019

Sorry, something went wrong.

mdonahoe commented Apr 16, 2019

• 😕 4 reactions

kannes commented Apr 17, 2019

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