unboundlocalerror local variable 'ip' referenced before assignment

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.

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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

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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UnboundLocalError Local variable Referenced Before Assignment in Python

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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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[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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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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UnboundLocalError: tokenizer_one referenced before assignment in train_dreambooth_lora_sd3.py #8594

C0nsumption commented Jun 16, 2024

xhinker commented Jun 17, 2024

Sorry, something went wrong.

satani99 commented Jun 17, 2024

• 👍 4 reactions
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Forainest commented Jun 20, 2024

Successfully merging a pull request may close this issue.

UndboundLocalError: local variable referenced before assignment

Hello all, I’m using PsychoPy 2023.2.3 Win 10 x64bits

What I’m trying to do? The experiment will show in the middle of the screen an abstracted stimuli (B1 or B2), and after valid click on it, the stimulus will remain on the middle of the screen and three more stimuli will appear in the cornor of the screen.

I’m having this erro (attached above), a simple error, but I can not see where the error is. Also the experiment isn’t working proberly and is the old version (I don’t know but someone are having troubles with this version of PscyhoPy)? ba_training_block.xlsx (13.8 KB) SMTS.psyexp (91.6 KB) stimuli, instructions and parameters.xlsx (12.8 KB)

You have a routine called sample but you also use that name for your image file in sample_box .

I changed the name of the routine for ‘stimulus_sample’ and manteined the image file in sample_box as ‘sample’. But, the error still remain. But it do not happen all the time, this is very interesting…

Can u give it a look again? (I made some minor changes here)

Here the exp file ba_training_block.xlsx (13.7 KB) SMTS.psyexp (89.7 KB) stimuli, instructions and parameters.xlsx (12.8 KB)

Thanks again

Please could you confirm/show the new error message? Is it definitely still related to sample?

I think you have blank rows in your spreadsheet. The loop claims that there are 19 conditions but I think you only want 12. Without a value for sample_category sample doesn’t get set. With random presentation this will happen at a random point.

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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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【Python】已解决UnboundLocalError: local variable ‘xxx‘ referenced before assignment的报错解决方案

【Python】已解决UnboundLocalError: local variable ‘xxx’ referenced before assignment的报错解决方案

😎 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。 🌼 同时洲洲已经建立了程序员技术交流群，如果您感兴趣，可以私信我加入社群，可以直接vx联系（文末有名片）v：bdizztt 🖥 随时欢迎您跟我沟通，一起交流，一起成长、进步！ 点此也可获得联系方式~

今天有粉丝问我，遇到了这个报错该怎么办：

其实很简单，我们先来看看两种最简单的情况：

在子程序中对全局变量的操作代码：

这个时候我们只需要用global关键字来进行说明该变量是全局变量

如果是局部变量，但仍然报出unboundLocal Error问题，比如下面的代码示例：

错误提示：UnboundLocalError: local variable ‘bbb2’ referenced before assignment。

其实一下就知道了,报错的原因是python认为bbb2不一定能被赋值。

UnboundLocalError是一种常见的错误，发生在尝试访问一个在当前作用域内未被赋值的局部变量时。

Python的作用域规则决定了变量的可见性和生命周期，错误的使用可能会导致此类错误。

在变量声明后直接使用，而没有进行赋值。

错误代码示例：

在条件语句中对变量赋值，但在某些分支下变量未被赋值。

在循环中对变量赋值，但循环未执行或未达到赋值条件。

函数参数未提供默认值，调用时未传入参数。

在使用变量之前，确保已经对其进行了赋值。

正确代码示例：

在条件语句之外为变量提供默认值。

在循环后检查变量的状态，确保在所有分支中变量都被赋值。

为函数参数提供默认值，确保即使调用时未传入参数，变量也有一个初始值。

• 理解Python的作用域规则，避免在局部作用域内引用未赋值的变量。
• 在函数或代码块的开始处为变量赋默认值，可以减少未赋值的错误。
• 使用None或其他合适的默认值作为变量的初始状态。
• 在编写条件语句或循环时，考虑所有可能的执行路径，确保变量在所有路径中都被赋值。

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How do i fix UnboundLocalError: local variable 'command' referenced before assignment

I'm trying to make a virtual assistant and the only problem left is this one. I don't know how to fix it but here is the code:

• If an exception is raised in take_command() , then command isn't assigned a value and the return command will cause an error. Set command to something before the try: block. –  Craig Commented Feb 12, 2021 at 15:12
• @Craig he defines the variable in run_lora he just has to globalize it –  Yash Commented Feb 12, 2021 at 15:21

3 Answers 3

Because you defined command in run_lora it is a private variable you just need to add global command .

Change the word command to cmnd in take_command() function. Your command variable is referenced before it was actually declared

I think my problem was that I needed pyadio but the pyaudio version is way to old for python 3.8

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