unboundlocalerror local variable 'xxx' referenced before assignment

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UnboundLocalError Local variable Referenced Before Assignment in Python

Handling errors is an integral part of writing robust and reliable Python code. One common stumbling block that developers often encounter is the “UnboundLocalError” raised within a try-except block. This error can be perplexing for those unfamiliar with its nuances but fear not – in this article, we will delve into the intricacies of the UnboundLocalError and provide a comprehensive guide on how to effectively use try-except statements to resolve it.

What is UnboundLocalError Local variable Referenced Before Assignment in Python?

The UnboundLocalError 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.

Why does UnboundLocalError: Local variable Referenced Before Assignment Occur?

below, are the reasons of occurring “Unboundlocalerror: Try Except Statements” in Python :

Variable Assignment Inside Try Block

Reassigning a global variable inside except block.

• Accessing a Variable Defined Inside an If Block

In the below code, example_function attempts to execute some_operation within a try-except block. If an exception occurs, it prints an error message. However, if no exception occurs, it prints the value of the variable result outside the try block, leading to an UnboundLocalError since result might not be defined if an exception was caught.

In below code , modify_global function attempts to increment the global variable global_var within a try block, but it raises an UnboundLocalError. This error occurs because the function treats global_var as a local variable due to the assignment operation within the try block.

Solution for UnboundLocalError Local variable Referenced Before Assignment

Below, are the approaches to solve “Unboundlocalerror: Try Except Statements”.

Initialize Variables Outside the Try Block

Avoid reassignment of global variables.

In modification to the example_function is correct. Initializing the variable result before the try block ensures that it exists even if an exception occurs within the try block. This helps prevent UnboundLocalError when trying to access result in the print statement outside the try block.

Below, code calculates a new value ( local_var ) based on the global variable and then prints both the local and global variables separately. It demonstrates that the global variable is accessed directly without being reassigned within the function.

In conclusion , To fix “UnboundLocalError” related to try-except statements, ensure that variables used within the try block are initialized before the try block starts. This can be achieved by declaring the variables with default values or assigning them None outside the try block. Additionally, when modifying global variables within a try block, use the `global` keyword to explicitly declare them.

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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 UnboundLocalError: local variable 'x' referenced before assignment in Python

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

How to reproduce this error

How to fix this error.

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

Take your skills to the next level ⚡️

Python UnboundLocalError: local variable referenced before assignment

by Suf | Programming , Python , Tips

If you try to reference a local variable before assigning a value to it within the body of a function, you will encounter the UnboundLocalError: local variable referenced before assignment.

The preferable way to solve this error is to pass parameters to your function, for example:

Alternatively, you can declare the variable as global to access it while inside a function. For example,

This tutorial will go through the error in detail and how to solve it with code examples .

Table of contents

What is scope in python, unboundlocalerror: local variable referenced before assignment, solution #1: passing parameters to the function, solution #2: use global keyword, solution #1: include else statement, solution #2: use global keyword.

Scope refers to a variable being only available inside the region where it was created. A variable created inside a function belongs to the local scope of that function, and we can only use that variable inside that function.

A variable created in the main body of the Python code is a global variable and belongs to the global scope. Global variables are available within any scope, global and local.

UnboundLocalError occurs when we try to modify a variable defined as local before creating it. If we only need to read a variable within a function, we can do so without using the global keyword. Consider the following example that demonstrates a variable var created with global scope and accessed from test_func :

If we try to assign a value to var within test_func , the Python interpreter will raise the UnboundLocalError:

This error occurs because when we make an assignment to a variable in a scope, that variable becomes local to that scope and overrides any variable with the same name in the global or outer scope.

var +=1 is similar to var = var + 1 , therefore the Python interpreter should first read var , perform the addition and assign the value back to var .

var is a variable local to test_func , so the variable is read or referenced before we have assigned it. As a result, the Python interpreter raises the UnboundLocalError.

Example #1: Accessing a Local Variable

Let’s look at an example where we define a global variable number. We will use the increment_func to increase the numerical value of number by 1.

Let’s run the code to see what happens:

The error occurs because we tried to read a local variable before assigning a value to it.

We can solve this error by passing a parameter to increment_func . This solution is the preferred approach. Typically Python developers avoid declaring global variables unless they are necessary. Let’s look at the revised code:

We have assigned a value to number and passed it to the increment_func , which will resolve the UnboundLocalError. Let’s run the code to see the result:

We successfully printed the value to the console.

We also can solve this error by using the global keyword. The global statement tells the Python interpreter that inside increment_func , the variable number is a global variable even if we assign to it in increment_func . Let’s look at the revised code:

Let’s run the code to see the result:

Example #2: Function with if-elif statements

Let’s look at an example where we collect a score from a player of a game to rank their level of expertise. The variable we will use is called score and the calculate_level function takes in score as a parameter and returns a string containing the player’s level .

In the above code, we have a series of if-elif statements for assigning a string to the level variable. Let’s run the code to see what happens:

The error occurs because we input a score equal to 40 . The conditional statements in the function do not account for a value below 55 , therefore when we call the calculate_level function, Python will attempt to return level without any value assigned to it.

We can solve this error by completing the set of conditions with an else statement. The else statement will provide an assignment to level for all scores lower than 55 . Let’s look at the revised code:

In the above code, all scores below 55 are given the beginner level. Let’s run the code to see what happens:

We can also create a global variable level and then use the global keyword inside calculate_level . Using the global keyword will ensure that the variable is available in the local scope of the calculate_level function. Let’s look at the revised code.

In the above code, we put the global statement inside the function and at the beginning. Note that the “default” value of level is beginner and we do not include the else statement in the function. Let’s run the code to see the result:

Congratulations on reading to the end of this tutorial! The UnboundLocalError: local variable referenced before assignment occurs when you try to reference a local variable before assigning a value to it. Preferably, you can solve this error by passing parameters to your function. Alternatively, you can use the global keyword.

If you have if-elif statements in your code where you assign a value to a local variable and do not account for all outcomes, you may encounter this error. In which case, you must include an else statement to account for the missing outcome.

For further reading on Python code blocks and structure, go to the article: How to Solve Python IndentationError: unindent does not match any outer indentation level .

Go to the  online courses page on Python  to learn more about Python for data science and machine learning.

Have fun and happy researching!

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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.

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Python 3: UnboundLocalError: local variable referenced before assignment

This error occurs when you are trying to access a variable before it has been assigned a value. Here is an example of a code snippet that would raise this error:

Watch a video course Python - The Practical Guide

The error message will be:

In this example, the variable x is being accessed before it is assigned a value, which is causing the error. To fix this, you can either move the assignment of the variable x before the print statement, or give it an initial value before the print statement.

Both will work without any error.

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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.

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UnboundLocalError: local variable referenced before assignment

I have following simple function to get percent values for different cover types from a raster. It gives me following error: UnboundLocalError: local variable 'a' referenced before assignment

which isn't clear to me. Any suggestions?

• arcgis-10.1
• unboundlocalerror

• 1 Because if row.getValue("Value") == 1 might be false and so a never gets assigned. –  Nathan W Commented May 20, 2013 at 2:39
• It has value and do gets assigned. I checked it in arcmap interactive python window but can't get it to work in a stand alone script. –  Ibe Commented May 20, 2013 at 2:44
• 1 your loop will also only give you the values of the last loop iteration as you are returning out of the loop and not doing anything with each value. –  Nathan W Commented May 20, 2013 at 2:49
• You could use 3 x elif and an else to see if any values other than 1-4 are encountered. –  PolyGeo ♦ Commented May 20, 2013 at 3:44
• I tried that way as well but still hung up with error. –  Ibe Commented May 20, 2013 at 4:15

This error is pretty much explained here and it helped me to get assignments and return values for all variables.

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having a problem with error code: UnboundLocalError: local variable 'player' referenced before assignment

hi hoping someone here can actually help me. I asked this question on stack overflow and they closed it and redirected to similar previously asked questions. none of which actually helped me to understand how to correct the error code.

I am following a video tutorial series for learning to code with python/pygame. and i have checked my coding aginst the source material and they are the same, because i typed it as was told to in the video.

here's the full tracebackback including the error:

File "/home/dev/PycharmProjects/game7_meteor_game/game7_meteor_game.py", line 160, in <module>

File "/home/dev/PycharmProjects/game7_meteor_game/game7_meteor_game.py", line 152, in main

GameInterface(num_player=1, screen=screen)

File "/home/dev/PycharmProjects/game7_meteor_game/game7_meteor_game.py", line 82, in GameInterface

if player.cooling_time > 0:

UnboundLocalError: local variable 'player' referenced before assignment.

here's the actual code:

if player.cooling_time > 0: player.cooling_time -= 1

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UnboundLocalError: local variable 'boxprops' referenced before assignment #3647

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UnboundLocalError in IMDiffusion: Handling Local Variable References

Abstract: In this article, we will discuss how to handle UnboundLocalError exceptions in IMDiffusion, a Python-based diffusion model.

UnboundLocalError: Handling Local Variable References in Python

In Python, a UnboundLocalError is a type of exception that is raised when a local variable is referenced before it is assigned. This error typically occurs when a variable is used within a function before it has been given a value.

Understanding Local Variables

In Python, a variable is considered local to a function if it is assigned a value within the function. Local variables are created when the function is called and are destroyed when the function returns. This means that local variables are only accessible within the function where they are defined.

For example, consider the following code:

In this example, the variable result is a local variable because it is assigned a value within the function. It is only accessible within the add\_numbers function and cannot be accessed from outside the function.

The UnboundLocalError Exception

An UnboundLocalError exception is raised when a local variable is referenced before it is assigned. This can occur if a variable is used within a function before it has been given a value.

In this example, the variable result is used in the print statement before it is assigned a value. This will result in an UnboundLocalError exception being raised:

Handling the UnboundLocalError Exception

The UnboundLocalError exception can be handled using a try-except block. For example:

In this example, the try-except block is used to handle the UnboundLocalError exception. If the exception is raised, the code within the except block will be executed.

Avoiding the UnboundLocalError Exception

The best way to avoid the UnboundLocalError exception is to ensure that local variables are only used after they have been assigned a value. This can be done by assigning a value to the variable at the beginning of the function or by using a default value for the variable.

For example:

In this example, the variable result is assigned a default value of 0 at the beginning of the function. This ensures that the variable is always defined before it is used.

Python: Errors and Exceptions

Python: Try-Except

Code Example

In this example, the function compute\_one\_subset\_one\_strategy takes four arguments: dataset\_name , subset\_name , compute\_sum , and compute\_abs . The function uses a for loop to iterate over the files in the specified directory and calculates the residuals for each file. The residuals are then returned as a list.

The function uses a default value of 0 for the residual variable, which ensures that the variable is always defined before it is used. This helps to avoid the UnboundLocalError exception.

The function also uses a try-except block to handle any exceptions that may be raised. If an exception is raised, the function will return a message indicating that an error has occurred.

Learn how to properly assign and handle local variables in IMDiffusion to avoid UnboundLocalError exceptions.

Understanding undefined behavior in c++: directly removing elements under range filter view.

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Tags: :  Python Error Handling IMDiffusion

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Seaborn Error: local variable 'boxprops' referenced before assignment [closed]

I am trying to plot using seaborn boxplot, however, I get the following error:

These codes were executing fine last week. I have tried updating seaborn and pyfolio, yet I still get the same error. Does anyone know how to fix this?

**Attempted Solutions: **

• Updating seaborn and pyfolio to the latest versions.

Does anyone have suggestions on how to resolve this issue?

• Can you test the code as shown? Does it still generate the error? Do you get other warnings? –  JohanC Commented Jun 24 at 13:58
• @JohanC It works ! thank you! Do you know what the exact problem was? –  Rhea Groenenberg Commented Jun 24 at 14:05
• I think there was something wrong with the input data. Maybe the problem is "solved" in this test, but not with your real data. In that case, you could try to print out information about filtered_df and create test data similar to that one. Or maybe your update of pyfolio finally had effect (I suppose pyfolio changed something to the input they are sending to seaborn). –  JohanC Commented Jun 24 at 14:10
• I'm voting to close this issue as not reproducible. –  Trenton McKinney Commented Jun 25 at 1:05

Browse other questions tagged python matplotlib seaborn pyfolio or ask your own question .

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