unboundlocalerror local variable 'output_list' referenced before assignment

[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

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

成功解决python报错：UnboundLocalError: local variable ‘xxx’ referenced before assignment。在Python中， UnboundLocalError 是一种特定的 NameError ，它会在尝试引用一个还未被赋值的局部变量时发生。Python解释器需要知道变量的类型和作用域，因此，在局部作用域内引用一个未被赋值的变量时，就会抛出这个错误。

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【Python】解决Python报错：

1. 什么是unboundlocalerror？, 2. 常见的场景和原因, 方法一：全局变量, 方法二：函数参数, 方法三：局部变量初始化, 方法四：结合条件语句.

在Python编程中， UnboundLocalError: local variable 'xxx' referenced before assignment 是一个常见的错误，尤其是在写函数时可能会遇到。这篇技术博客将详细介绍 UnboundLocalError ，为什么会发生，以及如何解决这个错误。

在Python中， UnboundLocalError 是一种特定的 NameError ，它会在尝试引用一个还未被赋值的局部变量时发生。Python解释器需要知道变量的类型和作用域，因此，在局部作用域内引用一个未被赋值的变量时，就会抛出这个错误。

这是一个简单的代码示例来说明这个错误：

运行以上代码会抛出以下错误：

在这个例子中，Python解释器看到 print(x) 时，寻找局部作用域中的变量 x ，但这个变量在局部作用域内尚未被赋值（虽然在后面有赋值，解释器是从上到下执行代码的）。

理解错误的原因后，可以通过以下几种方式来解决 UnboundLocalError ：

如果变量希望在函数内和函数外都使用，可以将其声明为全局变量：

通过在函数内使用 global 关键字，将 x 声明为全局变量，这样即使在函数内也能访问全局变量 x 。

通过将变量作为参数传递给函数，使得函数内可以访问并使用这个变量：

在这种情况下， x 是函数 my_function 的一个参数，无需在函数内部声明。

在使用变量之前，先初始化该局部变量：

确保在函数内部引用变量之前，该变量已经被赋值。

在复杂的逻辑中，特别是在涉及条件语句时，可以先在函数开始部分初始化变量，确保无论哪条路径都可以正确访问该变量：

在这个例子中，我们确保了变量 x 在函数内部任何地方都能被适当地引用。

• 命名冲突 ：在全局变量和局部变量重名情况下，优先使用局部变量。如果不小心混用，容易引发错误。
• 提前规划变量作用域 ：代码设计时，可以提前规划好变量应该属于哪个作用域，以减少变量冲突和未定义变量的情况。

UnboundLocalError: local variable 'xxx' referenced before assignment 错误是一个常见的初学者错误，但只要理解了Python的变量作用域规则和执行顺序，就可以轻松避开。通过合适的解决方法，如使用全局变量、函数参数、局部变量初始化或结合条件语句，可以高效且清晰地管理变量的使用。

希望这篇文章能帮助你理解和解决这个错误。如果有任何问题或其他建议，欢迎在评论中与我们讨论。Happy coding!

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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!

Next Article: Fixing Python TypeError: Descriptor 'lower' for 'str' Objects Doesn't Apply to 'dict' Object

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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 ⚡️

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Python Tutorials & Tips

Fixing ‘UnboundLocalError’ in Python: A Simple Guide with Code Samples

Python is a popular programming language that is widely used for developing various applications. However, like any other programming language, it is not free from errors. One of the common errors that Python developers encounter is the ‘UnboundLocalError’. This error occurs when a local variable is referenced before it is assigned a value. In this article, we will discuss in detail what ‘UnboundLocalError’ is, why it occurs, and how to fix it.

When a variable is defined inside a function, it is considered a local variable. If the function tries to access this variable before it is assigned a value, it results in an ‘UnboundLocalError’. This error can be frustrating for developers, especially when they are working on a large project. However, it is not difficult to fix this error. One of the ways to fix it is by using the ‘global’ keyword to declare the variable as a global variable.

In conclusion, understanding ‘UnboundLocalError’ in Python is crucial for developers who want to avoid errors in their code. By following best practices and using the right techniques, developers can easily fix this error and ensure that their code runs smoothly. In the next section, we will explore in detail how to fix ‘UnboundLocalError’ using code examples.

Understanding UnboundLocalError

UnboundLocalError is a common error in Python that occurs when a local variable is referenced before it has been assigned a value. This error can be confusing for beginners because it is not always clear why it occurs or how to fix it. In this section, we will explore what UnboundLocalError is, why it occurs, and how to identify it.

What is UnboundLocalError?

UnboundLocalError is an exception that occurs when a local variable is referenced before it has been assigned a value. In Python, variables can have either a local or global scope. Local variables are defined within a function and are only accessible within that function. Global variables, on the other hand, are defined outside of a function and can be accessed by any function within the module.

Why Does UnboundLocalError Occur?

UnboundLocalError occurs when a local variable is referenced before it has been assigned a value. This can happen if the variable is defined within a function but is not assigned a value before it is referenced. It can also happen if the variable is defined as a global variable but is not explicitly declared as such using the global statement.

How to Identify UnboundLocalError

UnboundLocalError can be identified by the traceback message that is generated when the error occurs. The traceback message will indicate the line number where the error occurred and provide information about the variable that caused the error.

To fix UnboundLocalError, you need to ensure that all local variables are assigned a value before they are referenced. You can also use the global statement to explicitly declare a variable as a global variable, allowing it to be accessed by any function within the module.

In conclusion, UnboundLocalError is a common error in Python that occurs when a local variable is referenced before it has been assigned a value. To fix this error, you need to ensure that all local variables are assigned a value before they are referenced and use the global statement to declare global variables. By understanding UnboundLocalError and how to fix it, you can write more robust and error-free Python code.

Fixing UnboundLocalError

If you are a Python developer, you may have encountered the UnboundLocalError error while working with local variables or functions. This error occurs when a local variable is referenced before it is assigned a value within a function. In this section, we will discuss how to fix UnboundLocalError in Python.

Solutions for UnboundLocalError

There are several ways to fix UnboundLocalError in Python. One solution is to explicitly declare the variable as global using the global keyword. This will make the variable a global variable instead of a local variable. Here is an example:

In this example, we declared num as a global variable inside the test() function using the global keyword. This allowed us to access and modify the value of num inside the function without raising an UnboundLocalError .

Another solution is to use the int() function to initialize the variable with a value of 0. This will ensure that the variable has a value before it is referenced. Here is an example:

In this example, we used the int() function to initialize num with a value of 0. This prevented the UnboundLocalError from being raised when we referenced num before assigning it a value.

How to Avoid UnboundLocalError

To avoid UnboundLocalError , it is important to understand the concept of local scope and local names in Python. Local scope refers to the area of a program where a variable is defined and can be accessed. Local names refer to the variables defined within a function.

To prevent UnboundLocalError , you should always make sure to assign a value to a local variable before referencing it within a function. You should also avoid using the same name for both global and local variables, as this can cause confusion and lead to errors.

Another way to avoid UnboundLocalError is to use lexical scoping. This means defining a function within another function, which allows the inner function to access the variables of the outer function. This can help prevent UnboundLocalError by ensuring that all variables are defined and assigned a value before they are referenced.

In conclusion, UnboundLocalError is a common error in Python that can be fixed by explicitly declaring variables as global or initializing them with a value using the int() function. To avoid UnboundLocalError , it is important to understand the concept of local scope and local names, and to assign values to local variables before referencing them within a function.

In conclusion, understanding the ‘UnboundLocalError’ in Python is essential for any programmer. This error occurs when a local variable is referenced before it has been assigned a value within a function. It can be frustrating to deal with, but fortunately, there are several ways to fix it.

One common solution is to use the global keyword to declare the variable as global within the function. This allows the function to access the variable outside of its scope. Another solution is to use default arguments in the function definition to initialize the variable with a default value.

It is important to note that this error is a runtime error and can only be detected when the code is executed. Therefore, it is crucial to test your code thoroughly to catch any ‘UnboundLocalError’ before deploying it.

Python is a versatile programming language that is widely used in various fields. Understanding the ‘UnboundLocalError’ and how to fix it is a crucial aspect of programming in Python. By following the tips and tricks outlined in this article, you can avoid this error and write efficient and effective code.

In summary, this guide has covered the basics of the ‘UnboundLocalError’ in Python, including its causes and solutions. We have seen how to use the global keyword and default arguments to fix this error. Hopefully, this article has been helpful in your programming journey, and you can now write better code in Python.

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

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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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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:

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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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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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Local variable referenced before assignment in Python

The “local variable referenced before assignment” error occurs when you try to use a local variable before it has been assigned a value. This is a general programming concept describing the situation 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.

In Python, the compiler might throw the exact error: “UnboundLocalError: cannot access local variable ‘x’ where it is not associated with a value”

Here’s an example to illustrate this error:

In this example, you would encounter the above 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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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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UnboundLocalError: Local Variable Referenced Before Assignment

Updated Feb 09, 2023

The “local variable referenced before assignment” error occurs when you give reference of a local variable without assigning any value.

Explanation:

In the above example, we have given the value of variable “v1” in two places.

• Outside the function “myfunction()” .
• And at the end of the function “myfunction()” .

If we assign a value of a variable in the function it becomes local variable to that function, but in the above example we have assigned the value to “v1” variable at the end of the function and we are referring this variable before assigning.

And the variable “v1” which we have assigned at the beginning of the code block is not declared as a global variable.

To avoid an error like “UnboundLocalError: local variable referenced before assignment” to occur, we have to:

• Declare GLOBAL variable
• Pass parameters with the function

Declare Global Variable

Code example with global variable:

As we know if we declare any variable as global then its scope becomes global.

Pass function with Parameters

Code example passing parameters with function:

In the above example, as you can see, we are not using a global variable but passing the value of variable “v1” as a parameter with the function “myfunction()”.

Example 2.1

In the "example2 ", we have called a function “dayweek()” with parameter value “10” which gives the error but the same function with value “1” which runs properly in “example 2.1” and returns the output as “Weekday”.

Because in the above function we are assigning the value to variable “wd” if the value of variable " day " is the range from (0 to 7) . If the value of variable " day " greater than "7" or lower then " 0" we are not assigning any value to variable " wd " That's why, whenever the parameter is greater than 7 or less than 0, python compiler throws the error “ UnboundLocalError: local variable 'wd' referenced before assignment ”

To avoid such type of error you need assign the function variable which lies within the range or we need to assign some value like " Invalid Value " to variable " wd " if the value of variable " day " is not in range from ( 0 to 7 )

Correct Example with Exception

How To Resolve UnboundLocalError On Local Variable When Reassigned After The First Use?

Summary: To resolve an UnboundLocalError when the local variable is reassigned after the first use, you can either use the global keyword or the nonlocal keyword. The global keyword allows you to modify the values of a global variable from within a function’s local scope while the nonlocal keyword provides similar functionality in case of nested functions.

Problem: Given a local variable. How to resolve an UnboundLocalError when the local variable is reassigned after the first use?

Before we deal with the solutions to our problem, we must understand the the root cause behind UnboundLocalError .

Root Cause Of UnboundLocalErrors:

When a variable is assigned within a function, it is treated as a local variable by default in Python. If a local variable is referenced before a value has been assigned/bound to it, an UnboundLocalError is raised. In the above example when the variable 'val' is read by the Python interpreter inside the func() function, it assumes that 'val' is a local variable. However, it soon realizes that the local variable has been referenced before any value has been assigned to it within the function. Thus it throws an UnboundLocalError .

In other words, we can only access a global variable inside a function but cannot modify it from within the function (unless you force a global or nonlocal assignment using the global or nonlocal keywords ).

Note: The parent class of all Python exceptions is BaseException . The following diagram represents the exception hierarchy of the UnboundLocalError .

Now that we know why and when an UnboundLocalError is raised, let us have a look at the probable solutions to overcome this error.

Method 1: Using The global Keyword

We can use the global keyword to access and modify a global variable from the local scope of a function. Any variable created outside a function is global by default while any variable created within the function is local by default. Thus, to modify a global variable from within the function and avoid an UnboundLocalError we can use the global keyword.

Let us have a look at the following code which simplifies the above concept:

Method 2: Using nonlocal Keyword

The nonlocal keyword is used when we have a nested function. In such a situation local scope of a variable might not be defined, that is, the scope of a variable is neither local nor global.

Let us have a look at the following example to understand the usage of the nonlocal keyword to deal with our problem.

In the above program, the nonlocal keyword is used with the variable val inside the nested function foo() to create a nonlocal variable. If the value of the nonlocal variable changes, the changes are reflected in the local variable as well.

Disclaimer: the nonlocal keyword works only in Python 3 and above .

Method 3: Using A Mutable Object

Since nonlocal does not work in versions of Python before Python 3, you can use a mutable object like a python dictionary , to store the values if you do not want to create a global variable.

Let us have a look at the following program to understand the usage of a mutable object in order to overcome the UnboundLocalError .

Method 4: Passing Parameters To The Function

This is the traditional way of solving our problem. Instead of using a global variable, we can pass parameters to the defined function to avoid an UnboundLocalError . Let us have a look at the following code to see how this can be done:

Global Keyword vs Nonlocal Keyword

Before concluding this article, let us have a look at the key differences between a global and nonlocal variable/keywords.

• Unlike the global keyword, the nonlocal keyword works only in Python 3 and above .
• The global keyword can be used with pre-existing global variables or new variables whereas the nonlocal keyword must be defined with a pre-existing variable.

From the above discussion we learned that to deal with UnboundLocalError in our code we can use one of the following methods:

• Use the global keyword.
• Use the nonlocal keyword in case of nested functions.
• Use a mutable object.
• Pass parameters to a function.

I hope that you found this article helpful and after reading it you can deal with UnboundLocalError with ease. Please stay tuned for more interesting articles.

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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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Getting error "local variable referenced before assignment - how to fix?

I'm making a gambling program (i know this shouldn't be incredibly hard), and want to have multiple games which will be in subroutines. However, python seems to think my variables are being assigned in strange places.

I am semi-new to subroutines, and still have some issues here and there. Here is what I am working with:

And here is how it is called:

I expect it to just go through, add a tally into win or loss, then update the money. However, I am getting this error: UnboundLocalError: local variable 'losses' referenced before assignment If I win, it says the same thing with local variable 'wins' .

As shown all variables are assigned at the top, then referenced below in subroutines. I am completely unsure on how python thinks I referenced it before assignment?

I would appreciate any help, thank you in advance!

• Can you post the whole code? –  Manuel Fedele Commented Mar 23, 2019 at 14:55
• pastebin.com/WkrPRPfb - check this pastebin for the whole code –  xupaii Commented Mar 23, 2019 at 14:57

2 Answers 2

The reason is that losses is defined as a global variable. Within functions (local scope), you can, loosely speaking, read from global variables but not modify them.

This will work:

This won't:

You should assign to your variables within your function body if you want them to have local scope. If you explicitly want to modify global variables, you need to declare them with, for example, global losses in your function body.

The variables wins , money and losses were declared outside the scope of the fiftyfifty() function, so you cannot update them from inside the function unless you explicitly declare them as global variables like this:

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