unboundlocalerror local variable 'iter' 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

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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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

Previous Article: Python TypeError: write() argument must be str, not bytes

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

by Nathan Sebhastian

Posted on May 26, 2023

Reading time: 2 minutes

One error you might encounter when running Python code is:

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

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

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

How to reproduce this error

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

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

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

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

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

How to fix this error

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

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

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

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

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

Here’s the best solution to the error:

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

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

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

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

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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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[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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What is UnboundLocalError: local variable referenced before assignment?

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

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

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

An example of Local variable referenced before assignment

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

Firstly, we start by declaring two variables:

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

Finally, we call our function:

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

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

Here is an error!

The Solution of Local variable referenced before assignment

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

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

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

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

Let’s try to run our code again:

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

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

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

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

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

Our code returns: B.

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

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

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UnboundLocalError: local variable referenced before assignment

The unboundlocalerror: local variable referenced before assignment is raised when you try to use a variable before it has been assigned in the local context. Python doesn't have variable declarations , so it has to figure out the scope of variables itself. It does so by a simple rule: If there is an assignment to a variable inside a function, that variable is considered local .

If you set the value of a variable inside the function , python understands it as creating a local variable with that name. This local variable masks the global variable .

Here, the line

implicitly makes "counter" local to increment(). Trying to execute this line, though, will try to read the value of the local variable "counter" before it is assigned, resulting in an UnboundLocalError . To solve this problem, you can explicitly say it's a global by putting global declaration in you function.

The global statement does not have to be at the beginning of the function definition, but that is where it is usually placed. Wherever it is placed, the global declaration makes a variable to global variable everywhere in the function. If increment() is a local function and counter a local variable, you can use "nonlocal" in Python 3.x. This keyword is useful when we need to assign any value to nested scope variable.

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). All variable assignments in a function store the value in the local symbol table; whereas variable references first look in the local symbol table, then in the global symbol table, and then in the table of built-in names. Thus, global variables cannot be directly assigned a value within a function (unless named in a global statement), although they may be referenced.

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Why do I get UnboundLocalError: local variable 'batch_idx' referenced before assignment when using interleaved data sets with Hugging Face (HF)?

I get the following error:

it happens when I interleave my data set:

why is this happening?

ref: machine learning - Why do I get UnboundLocalError: local variable 'batch_idx' referenced before assignment when using interleaved data sets with Hugging Face (HF)? - Stack Overflow ref: Why do I get UnboundLocalError: local variable 'batch_idx' referenced before assignment when using interleaved data sets with Hugging Face (HF)? ref: Discord

The issue happened because llama2 does not have a max sequence length set. So it defaults to max int 1000000000000000019884624838656 . You can set it manually if you google the max seq len for your model e.g., for llama2-7b:

That should remove the issue in the code because

will not be zero when trying to concatenate all 1000 texts in the batch when forming a batch all of which have a length of block size (assumign block size is max seq len or some value you chose).

ref: [Tokenizers]What this max_length number?

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UnboundLocalError: local variable 'pdf' referenced before assignment #172

awalker4 commented Aug 1, 2023 • edited

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UnboundLocalError: local variable 'iou' referenced before assignment

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这个 UnboundLocalError 通常发生在Python中，当你尝试引用一个在当前作用域内未被初始化的局部变量时。在这个 train.py 文件中，错误出现在 loss.py 的 __call__ 方法中，具体是检查 iou 是否为tuple的部分。

为了修复这个问题，你需要确保 iou 已经在 compute_loss 函数内部被正确地定义并赋值。我猜测 iou 可能是在计算某个指标（比如IOU）的过程中产生的，但在你引用它之前，它还没有被计算出来。

首先，让我们检查 compute_loss 函数的代码，看看 iou 是否在计算过程中被定义。如果 iou 是一个依赖于 pred 和 targets 的值，那么可能需要在计算前先检查它们或者确保它们已经被计算了。

如果 iou 是通过之前的计算得到的，但没有在适当的地方返回，你可以尝试在计算后添加一个返回语句，例如：

确保 calculate_iou 函数在返回 iou 时已经执行了必要的计算，并将其存储在 iou 变量中。这样，当 __call__ 方法试图访问 iou 时，它就不会引发 UnboundLocalError 了。

如果你能提供 calculate_iou 函数的代码，我可以更准确地帮助你定位问题。同时，检查 loss.py 中 __call__ 方法的上下文，确认 iou 是否在预期的地方被使用。

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