unboundlocalerror local variable referenced before assignment in python

• Free Python 3 Tutorial
• Control Flow
• Exception Handling
• Python Programs
• Python Projects
• Python Interview Questions
• Python Database
• Data Science With Python
• Machine Learning with Python
• Python Program That Prints Out the Decimal Equivalents of 1/2, 1/3, 1/4, . . . ,1/10
• How to Convert PIL Image into pygame surface image?
• Python program to print the binary value of the numbers from 1 to N
• Python Program for Sum the digits of a given number
• Python program to print number of bits to store an integer and also the number in Binary format
• Python program to get all pairwise combinations from a list
• Python program to repeat M characters of a string N times
• How to format a string using a dictionary in Python
• Python program to convert binary to ASCII
• Python program to convert unix timestamp string to readable date
• Python program to print an array of bytes representing an integer
• Python Program to Multiply Two Binary Numbers
• Python Program to Find Sum of First and Last Digit
• Python Program to Sort Words in Alphabetical Order
• Python program to convert integer to roman
• Python - Lambda Function to Check if value is in a List
• Python - Get first element from a List of tuples
• Get Your System Information - Using Python Script
• Python - Get the object with the max attribute value in a list of objects

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.

Please Login to comment...

Similar reads.

• Python Errors
• Google Introduces New AI-powered Vids App
• Dolly Chaiwala: The Microsoft Windows 12 Brand Ambassador
• 10 Best Free Remote Desktop apps for Android in 2024
• 10 Best Free Internet Speed Test apps for Android in 2024
• 30 OOPs Interview Questions and Answers (2024)

Improve your Coding Skills with Practice

What kind of Experience do you want to share?

Fix "local variable referenced before assignment" in Python

Introduction

If you're a Python developer, you've probably come across a variety of errors, like the "local variable referenced before assignment" error. This error can be a bit puzzling, especially for beginners and when it involves local/global variables.

Today, we'll explain this error, understand why it occurs, and see how you can fix it.

The "local variable referenced before assignment" Error

The "local variable referenced before assignment" error in Python is a common error that occurs when a local variable is referenced before it has been assigned a value. This error is a type of UnboundLocalError , which is raised when a local variable is referenced before it has been assigned in the local scope.

Here's a simple example:

Running this code will throw the "local variable 'x' referenced before assignment" error. This is because the variable x is referenced in the print(x) statement before it is assigned a value in the local scope of the foo function.

Even more confusing is when it involves global variables. For example, the following code also produces the error:

But wait, why does this also produce the error? Isn't x assigned before it's used in the say_hello function? The problem here is that x is a global variable when assigned "Hello ". However, in the say_hello function, it's a different local variable, which has not yet been assigned.

We'll see later in this Byte how you can fix these cases as well.

Fixing the Error: Initialization

One way to fix this error is to initialize the variable before using it. This ensures that the variable exists in the local scope before it is referenced.

Let's correct the error from our first example:

In this revised code, we initialize x with a value of 1 before printing it. Now, when you run the function, it will print 1 without any errors.

Fixing the Error: Global Keyword

Another way to fix this error, depending on your specific scenario, is by using the global keyword. This is especially useful when you want to use a global variable inside a function.

No spam ever. Unsubscribe anytime. Read our Privacy Policy.

Here's how:

In this snippet, we declare x as a global variable inside the function foo . This tells Python to look for x in the global scope, not the local one . Now, when you run the function, it will increment the global x by 1 and print 1 .

Similar Error: NameError

An error that's similar to the "local variable referenced before assignment" error is the NameError . This is raised when you try to use a variable or a function name that has not been defined yet.

Running this code will result in a NameError :

In this case, we're trying to print the value of y , but y has not been defined anywhere in the code. Hence, Python raises a NameError . This is similar in that we are trying to use an uninitialized/undefined variable, but the main difference is that we didn't try to initialize y anywhere else in our code.

Variable Scope in Python

Understanding the concept of variable scope can help avoid many common errors in Python, including the main error of interest in this Byte. But what exactly is variable scope?

In Python, variables have two types of scope - global and local. A variable declared inside a function is known as a local variable, while a variable declared outside a function is a global variable.

Consider this example:

In this code, x is a global variable, and y is a local variable. x can be accessed anywhere in the code, but y can only be accessed within my_function . Confusion surrounding this is one of the most common causes for the "variable referenced before assignment" error.

In this Byte, we've taken a look at the "local variable referenced before assignment" error and another similar error, NameError . We also delved into the concept of variable scope in Python, which is an important concept to understand to avoid these errors. If you're seeing one of these errors, check the scope of your variables and make sure they're being assigned before they're being used.

Building Your First Convolutional Neural Network With Keras

Most resources start with pristine datasets, start at importing and finish at validation. There's much more to know. Why was a class predicted? Where was...

© 2013- 2024 Stack Abuse. All rights reserved.

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

Series: Common Errors in Python and How to Fix Them

Related Articles

• Python Warning: Secure coding is not enabled for restorable state
• 4 ways to install Python modules on Windows without admin rights
• Python TypeError: object of type ‘NoneType’ has no len()
• Python: How to access command-line arguments (3 approaches)
• Understanding ‘Never’ type in Python 3.11+ (5 examples)
• Python: 3 Ways to Retrieve City/Country from IP Address
• Using Type Aliases in Python: A Practical Guide (with Examples)
• Python: Defining distinct types using NewType class
• Using Optional Type in Python (explained with examples)
• Python: How to Override Methods in Classes
• Python: Define Generic Types for Lists of Nested Dictionaries
• Python: Defining type for a list that can contain both numbers and strings

Search tutorials, examples, and resources

• PHP programming
• Symfony & Doctrine
• Laravel & Eloquent
• Tailwind CSS
• Sequelize.js
• Mongoose.js

Local variable referenced before assignment in Python

Last updated: Apr 8, 2024 Reading time · 4 min

# Local variable referenced before assignment in Python

The Python "UnboundLocalError: Local variable referenced before assignment" occurs when we reference a local variable before assigning a value to it in a function.

To solve the error, mark the variable as global in the function definition, e.g. global my_var .

Here is an example of how the error occurs.

We assign a value to the name variable in the function.

# Mark the variable as global to solve the error

To solve the error, mark the variable as global in your function definition.

If a variable is assigned a value in a function's body, it is a local variable unless explicitly declared as global .

# Local variables shadow global ones with the same name

You could reference the global name variable from inside the function but if you assign a value to the variable in the function's body, the local variable shadows the global one.

Accessing the name variable in the function is perfectly fine.

On the other hand, variables declared in a function cannot be accessed from the global scope.

The name variable is declared in the function, so trying to access it from outside causes an error.

Make sure you don't try to access the variable before using the global keyword, otherwise, you'd get the SyntaxError: name 'X' is used prior to global declaration error.

# Returning a value from the function instead

An alternative solution to using the global keyword is to return a value from the function and use the value to reassign the global variable.

We simply return the value that we eventually use to assign to the name global variable.

# Passing the global variable as an argument to the function

You should also consider passing the global variable as an argument to the function.

We passed the name global variable as an argument to the function.

If we assign a value to a variable in a function, the variable is assumed to be local unless explicitly declared as global .

# Assigning a value to a local variable from an outer scope

If you have a nested function and are trying to assign a value to the local variables from the outer function, use the nonlocal keyword.

The nonlocal keyword allows us to work with the local variables of enclosing functions.

Had we not used the nonlocal statement, the call to the print() function would have returned an empty string.

Printing the message variable on the last line of the function shows an empty string because the inner() function has its own scope.

Changing the value of the variable in the inner scope is not possible unless we use the nonlocal keyword.

Instead, the message variable in the inner function simply shadows the variable with the same name from the outer scope.

# Discussion

As shown in this section of the documentation, when you assign a value to a variable inside a function, the variable:

• Becomes local to the scope.
• Shadows any variables from the outer scope that have the same name.

The last line in the example function assigns a value to the name variable, marking it as a local variable and shadowing the name variable from the outer scope.

At the time the print(name) line runs, the name variable is not yet initialized, which causes the error.

The most intuitive way to solve the error is to use the global keyword.

The global keyword is used to indicate to Python that we are actually modifying the value of the name variable from the outer scope.

• If a variable is only referenced inside a function, it is implicitly global.
• If a variable is assigned a value inside a function's body, it is assumed to be local, unless explicitly marked as global .

If you want to read more about why this error occurs, check out [this section] ( this section ) of the docs.

# Additional Resources

You can learn more about the related topics by checking out the following tutorials:

• SyntaxError: name 'X' is used prior to global declaration

Borislav Hadzhiev

Web Developer

Copyright © 2024 Borislav Hadzhiev

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

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.

Take your skills to the next level ⚡️

I'm sending out an occasional email with the latest tutorials on programming, web development, and statistics. Drop your email in the box below and I'll send new stuff straight into your inbox!

Hello! This website is dedicated to help you learn tech and data science skills with its step-by-step, beginner-friendly tutorials. Learn statistics, JavaScript and other programming languages using clear examples written for people.

Learn more about this website

Connect with me on Twitter

Or LinkedIn

Type the keyword below and hit enter

Click to see all tutorials tagged with:

Explore your training options in 10 minutes Get Started

• Graduate Stories
• Partner Spotlights
• Bootcamp Prep
• Bootcamp Admissions
• University Bootcamps
• Coding Tools
• Software Engineering
• Web Development
• Data Science
• Tech Guides
• Tech Resources
• Career Advice
• Online Learning
• Internships
• Apprenticeships
• Tech Salaries
• Associate Degree
• Bachelor's Degree
• Master's Degree
• University Admissions
• Best Schools
• Certifications
• Bootcamp Financing
• Higher Ed Financing
• Scholarships
• Financial Aid
• Best Coding Bootcamps
• Best Online Bootcamps
• Best Web Design Bootcamps
• Best Data Science Bootcamps
• Best Technology Sales Bootcamps
• Best Data Analytics Bootcamps
• Best Cybersecurity Bootcamps
• Best Digital Marketing Bootcamps
• Los Angeles
• San Francisco
• Browse All Locations
• Digital Marketing
• Machine Learning
• See All Subjects
• Bootcamps 101
• Full-Stack Development
• Career Changes
• View all Career Discussions
• Mobile App Development
• Cybersecurity
• Product Management
• UX/UI Design
• What is a Coding Bootcamp?
• Are Coding Bootcamps Worth It?
• How to Choose a Coding Bootcamp
• Best Online Coding Bootcamps and Courses
• Best Free Bootcamps and Coding Training
• Coding Bootcamp vs. Community College
• Coding Bootcamp vs. Self-Learning
• Bootcamps vs. Certifications: Compared
• What Is a Coding Bootcamp Job Guarantee?
• How to Pay for Coding Bootcamp
• Ultimate Guide to Coding Bootcamp Loans
• Best Coding Bootcamp Scholarships and Grants
• Education Stipends for Coding Bootcamps
• Get Your Coding Bootcamp Sponsored by Your Employer
• GI Bill and Coding Bootcamps
• Tech Intevriews
• Our Enterprise Solution
• Connect With Us
• Publication
• Reskill America
• Partner With Us

• Resource Center
• Bachelor’s Degree
• Master’s Degree

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.

Find your bootcamp match

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 !

About us: Career Karma is a platform designed to help job seekers find, research, and connect with job training programs to advance their careers. Learn about the CK publication .

What's Next?

Get matched with top bootcamps

Ask a question to our community, take our careers quiz.

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Decode Python

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.

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!

Share this:

• Click to share on Facebook (Opens in new window)
• Click to share on LinkedIn (Opens in new window)
• Click to share on Reddit (Opens in new window)
• Click to share on Pinterest (Opens in new window)
• Click to share on Telegram (Opens in new window)
• Click to share on WhatsApp (Opens in new window)
• Click to share on Twitter (Opens in new window)
• Click to share on Tumblr (Opens in new window)

[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

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!

• Privacy Policy
• Terms of Service

Local variable referenced before assignment in Python

The “local variable referenced before assignment” error occurs in Python when you try to use a local variable before it has been assigned a value.

This error 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.

Here’s an example to illustrate this error:

In this example, you would encounter the “local variable ‘x’ referenced before assignment” 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:

• Fix: "syntax error: unexpected eof" while parsing Python input
• Do streamers use VPNs?
• Is a vpn worth it for torrenting?
• Converting Uppercase to Lowercase in Python
• Converting a Comma-Separated String to a List in Python - Multiple Approaches
• Counting the Occurrences of Unique Values in a Python List: Multiple Approaches
• Remove Special Characters from a String in Python Using Regex
• Get File Size in Human Readable Format in Python
• How to Get Current Time in Milliseconds, Seconds(epoch) Using Various Methods in Python
• Find and Replace First Occurrence Of a String in python
• Converting Python Strings to Datetime Objects with Timezone
• Fixing the 'NoneType object is not iterable' Error in Python
• Scope rules in C Programming
• Find The Most Frequent Element in an Array
• Program to find sum of n natural numbers in C++ [3 Methods]
• Finding the Maximum and Minimum Elements of an Array using C++
• The Difference Between int main( ), void main( ) and int main (void)
• Convering a string into upper or lower case in C++
• Exceptions in C++
• How to Improve Technical Skills in Programming
• Next smaller element to the right
• Next Smaller Element to Left in an array
• Next greater to left
• Next Larger Element
• Check for subtree in a Binary Tree

How To Fix Unbound Local Errors In Python

Rahul Kumar Yadav

The beginner users of python may get confused when they see an Unbound Local Error problem in Python. To save them from this headache we have provided the cause, and the fixes of this error, which often occurs while coding in Python.

How UnboundLocalError: local variable referenced before assignment occurs

UnboundLocalError: local variable referenced before assignment , error occurs when we are trying to reference the variable inside a function before assigning any value to it.

This error occurs only when the local variable which is in focus is being assigned some value in the function, and we have referenced it before the assignment operation.

Let us take the help of the code to see what I am trying to say:-

The above example will show an error in line 3, where it says UnboundLocalError: local_variable 'string' referenced before assignment .

The above error occurs because there is both assignment and reference of the variable string, and thus, program creates a local variable named as string for the function example, but it is referenced before any value has been assigned(in the line 3, as we are trying to use the value of string to update the value of string) and since, till then there is no value for local variable string in the function, it shows unbound local error.

Had the code been, as given below:-

It would have simply displayed the output:-

As in this case, there is no assignment of the value to the string variable, and thus, no local variable string is created, and thus, it uses the value in the global scope to display.

The problem of UnboundLocalError occurs only when there is an assignment operation of the variable having the same name as outside the function and inside the function, and the variable is referenced before the assignment operation has taken place.

How to Fix UnboundLocalError error

Only way to avoid this error is to assign the variable before referencing it, or to modify it inside the function with a different name and then return that modified value and assign it to the variable outside the function.

Various Ways of Fixing the Above Error

Sometimes we need to use the value of the variable outside the function, and thus it may require referencing before assignment.

There are some ways by which we can bypass this problem, where we need to reference before assignment.

• Using global keyword:-

The output of the above code is:-

Using the global keyword we can easily circumvent the problem of no referencing before assignment, as then it would be able to use the global variable for further operations inside the function.

• We can pass the value of variable outside the function as a parameter to the function:-

Upon passing the value, we will be able to use the variable’s value without declaring it as a global variable, as sometimes it is beneficial to not expose the value to every function.

• Sometimes we want to modify the value of global variable without declaring it as a global variable, we can simply use some other variable, and modify it accordingly and then return its value and assign it to variable whose value we wanted to modify, like in the given example:-

The output of the above code will be :-

In the above code, we have modified the variable string without declaring it as a global variable, by returning the modified value from the function.

• If we are using a nested function and we want to assign a value to the local variables from the outer function, then we can use the “nonlocal” keyword to access that variable inside the nested function.

The output of the above function will be:-

We can see in the output that we have modified the string present in the outer_example() by accessing it inside the inner_example() .

We can also use the methods numbered 2 and 3 along with this one to modify the variable present in the outer function from the inner function.

We just saw the reasons behind the occurrence of Unbound Local Error in Python, along with various fixes which we can use in our function to bypass the problem along with getting our task accomplished.

• C Tutorials
• Java Tutorials
• C++ Tutorials
• SQL Tutorials

Sometimes you may suddenly start getting UnboundLocalError in your code which was working perfectly just minutes ago. And what you did is just added an assignment statement.

Lets say your code is as below, which is working fine.

Now you added an assignment statement inside printx function where you are increasing the value of global variable x . You will start getting UnboundLocalError .

What caused UnboundLocalError?

Lets understand few things first.

In python all the variables inside a function are global if they are not assigned any value to them. That means if a variable is only referenced inside a function, it is global. However if we assign any value to a variable inside a function, its scope becomes local to that unless explicitly declared global. 

In first example, we are not assigning any value to x variable inside function and just referencing it to print. Hence x is global. 

In second example, we assigned the value 7 to x, hence x's scope is local to function and we tried to print it before assigning any value to x.

Similarly, you will get UnboundLocalError in scenario similar to below example.

UnboundLocalError can be solved by changing the scope of the variable which is complaining. You need to explicitly declare the variable global.

Variable x's scope in function printx is global. You can verify the same by printing the value of x in terminal and it will be 6.

Remember that same thing can be done using nonlocal keyword. But mind it that nonlocal binds the variable to nearest enclosing scope except global. Lets understand this with an example.

Output will be:

So take away is:

1. Use global keyword with a variable if only you are assigning value to that variable inside a function and you want to overwrite the value of variable declared in global scope.

2. Use nonlocal keyword in nested scopes.

Host your Django App for free (or in only $5 with custom domain name)

Search code, repositories, users, issues, pull requests...

Provide feedback.

We read every piece of feedback, and take your input very seriously.

Saved searches

Use saved searches to filter your results more quickly.

To see all available qualifiers, see our documentation .

• Notifications

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement . We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

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

awalker4 commented Aug 1, 2023 • edited

Successfully merging a pull request may close this issue.

码龄 粉丝数 原力等级 --

UnboundLocalError: local variable 'iou' referenced before assignment

• 编辑 收藏 删除 结题

¥ 0 (可追加 ¥500）

提供问题酬金的用户不参与问题酬金结算和分配

支付即为同意 《付费问题酬金结算规则》

4 条回答 默认 最新

引自免费微信小程序： 皆我百晓生

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。

• 没有解决我的问题, 去提问

• ¥15 java根据模板，生成word文档，需要带目录
• ¥15 广告联盟的兜底广告是什么意思
• ¥15 如何证明高斯噪声的包络公式
• ¥150 寻找王者荣耀开发作者，合作或者解答
• ¥15 乳腺癌数据集 相关矩阵 特征选择
• ¥15 我的游戏账号被盗取，请问我该怎么做
• ¥15 通关usb3.0.push文件，导致usb频繁断连
• ¥15 有没有能解决微信公众号，只能实时拍照，没有选择相册上传功能，我不懂任何技术，，有没有给我发个软件就能搞定的方法
• ¥15 Pythontxt文本可视化
• ¥15 如何基于Ryu环境下使用scapy包进行数据包构造