unboundlocalerror local variable 'value_is_number' referenced before assignment

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

[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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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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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 'value_is_number' referenced before assignment #331

peepeepeepoopoopoo commented Sep 5, 2023 • edited Loading

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【Python】成功解决Python报错 UnboundLocalError: local variable ‘xxx‘ referenced before assignment问题

😎 作者介绍：我是程序员洲洲，一个热爱写作的非著名程序员。CSDN全栈优质领域创作者、华为云博客社区云享专家、阿里云博客社区专家博主。 🤓 同时欢迎大家关注其他专栏，我将分享Web前后端开发、人工智能、机器学习、深度学习从0到1系列文章。

在Python编程中，UnboundLocalError是一个运行时错误，它发生在尝试访问一个在当前作用域内未被绑定（即未被赋值）的局部变量时。 错误信息UnboundLocalError: local variable ‘xxx’ referenced before assignment指出变量xxx在赋值之前就被引用了。 这种情况通常发生在函数内部，尤其是在使用循环或条件语句时，变量的赋值逻辑可能因为某些条件未满足而未能执行，导致在后续的代码中访问了未初始化的变量。

我们来看看粉丝跟我说的具体的报错情况：

运行后会显示报错：UnboundLocalError: local variable ‘xxx’ referenced before assignment

把变量声明称global，global sum_score。

• 条件语句中未初始化变量
• 循环中变量初始化位置错误
• 循环的退出条件导致变量未初始化
• 确保变量在使用前被初始化
• 调整循环中变量的作用域
• 检查循环退出条件，确保变量被初始化
• 明确变量作用域：理解Python中变量的作用域，确保在变量的作用域内使用前已经初始化。
• 使用初始化值：为变量提供一个初始值，特别是在不确定变量是否会被赋值的情况下。
• 条件语句的使用：在条件语句中使用变量前，确保变量已经在所有分支中被初始化。
• 循环逻辑检查：在循环中使用变量前，确保循环的逻辑允许变量被正确初始化。
• 代码审查：定期进行代码审查，检查变量的使用是否符合预期，特别是变量初始化的逻辑。
• 编写测试：编写单元测试来验证函数或方法在所有预期的使用情况下都能正确处理变量初始化。

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UnboundLocalError: local variable 'b' referenced before assignment

I keep getting this error but I have defined the local variable inside function#Get data checkpoint size. Initially I thought that this error might be because of indentation but that's also not working for me.

What if data_incomingbytes is empty? Then the loop won't run and b won't be assigned to. The possibility of that happening is what Python is complaining about. You need to assign to b (and all the other variables) whether or not the loop runs.

• you are right I did not think about this possibility.. Thanks a bundle –  Imo Commented Aug 27, 2015 at 14:43

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