python global variable unboundlocalerror local variable referenced before assignment

• Python Basics
• Interview Questions
• Python Quiz
• Popular Packages
• Python Projects
• Practice Python
• AI With Python
• Learn Python3
• Python Automation
• Python Web Dev
• DSA with Python
• Python OOPs
• Dictionaries

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
• Python Programs

Improve your Coding Skills with Practice

What kind of Experience do you want to share?

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

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 ⚡️

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)

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 *

[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

Consultancy

• Technology Consulting
• Customer Experience Consulting
• Solution Architect Consulting

Software Development Services

• Ecommerce Development
• Web App Development
• Mobile App Development
• SAAS Product Development
• Content Management System
• System Integration & Data Migration
• Cloud Computing
• Computer Vision

Dedicated Development Team

• Full Stack Developers For Hire
• Offshore Development Center

Marketing & Creative Design

• UX/UI Design
• Customer Experience Optimization
• Digital Marketing
• Devops Services
• Service Level Management
• Security Services
• Odoo gold partner

By Industry

• Retail & Ecommerce
• Manufacturing
• Import & Distribution
• Financical & Banking
• Technology For Startups

Business Model

• MARKETPLACE ECOMMERCE

Our realized projects

MB Securities - A Premier Brokerage

iONAH - A Pioneer in Consumer Electronics Industry

Emers Group - An Official Nike Distributing Agent

Academy Xi - An Australian-based EdTech Startup

• Market insight

• Ohio Digital
• Onnet Consoulting

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.

>>> Read more

• Average python length: How to find it with examples
• List assignment index out of range: Python indexerror solution you should know
• Spyder vs Pycharm: The detailed comparison to get best option for Python programming
• fix python error , Local variable referenced before assignment , python , python dictionary , python error , python learning , UnboundLocalError , UnboundLocalError in Python

Our Other Services

• E-commerce Development
• Web Apps Development
• Web CMS Development
• Mobile Apps Development
• Software Consultant & Development
• System Integration & Data Migration
• Dedicated Developers & Testers For Hire
• Remote Working Team
• Saas Products Development
• Web/Mobile App Development
• Outsourcing
• Hiring Developers
• Digital Transformation
• Advanced SEO Tips

Lastest News

Challenges and Advices When Using AWS Cloud Managed Services

Comprehensive Guide about AWS Well Architected Framework

AWS Cloud Migration: Guide-to-Guide from A to Z

Uncover The Treasures Of Cloud Computing For Healthcare 

A Synopsis of Cloud Computing in Financial Services 

Discover Cutting-Edge Cloud Computing Applications To Optimize Business Resources

Tailor your experience

• Success Stories

Copyright ©2007 – 2021 by AHT TECH JSC. All Rights Reserved.

Thank you for your message. It has been sent.

• Password Generator
• HTML Editor
• HTML Encoder
• JSON Beautifier
• CSS Beautifier
• Markdown Convertor
• Find the Closest Tailwind CSS Color
• Phrase encrypt / decrypt
• Browser Feature Detection
• Number convertor
• CSS Maker text shadow
• CSS Maker Text Rotation
• CSS Maker Out Line
• CSS Maker RGB Shadow
• CSS Maker Transform
• CSS Maker Font Face
• Color Picker
• Colors CMYK
• Color mixer
• Color Converter
• Color Contrast Analyzer
• Color Gradient
• String Length Calculator
• MD5 Hash Generator
• Sha256 Hash Generator
• String Reverse
• URL Encoder
• URL Decoder
• Base 64 Encoder
• Base 64 Decoder
• Extra Spaces Remover
• String to Lowercase
• String to Uppercase
• Word Count Calculator
• Empty Lines Remover
• HTML Tags Remover
• Binary to Hex
• Hex to Binary
• Rot13 Transform on a String
• String to Binary
• Duplicate Lines Remover

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:

Watch a video course Python - The Practical Guide

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.

Related Resources

• Using global variables in a function
• "Least Astonishment" and the Mutable Default Argument
• Why is "1000000000000000 in range(1000000000000001)" so fast in Python 3?
• HTML Basics
• Javascript Basics
• TypeScript Basics
• React Basics
• Angular Basics
• Sass Basics
• Vue.js Basics
• Python Basics
• Java Basics
• NodeJS Basics

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)

Stack Exchange Network

Stack Exchange network consists of 183 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Q&A for work

Connect and share knowledge within a single location that is structured and easy to search.

UnboundLocalError: local variable referenced before assignment

I have following simple function to get percent values for different cover types from a raster. It gives me following error: UnboundLocalError: local variable 'a' referenced before assignment

which isn't clear to me. Any suggestions?

• arcgis-10.1
• unboundlocalerror

• 1 Because if row.getValue("Value") == 1 might be false and so a never gets assigned. –  Nathan W Commented May 20, 2013 at 2:39
• It has value and do gets assigned. I checked it in arcmap interactive python window but can't get it to work in a stand alone script. –  Ibe Commented May 20, 2013 at 2:44
• 1 your loop will also only give you the values of the last loop iteration as you are returning out of the loop and not doing anything with each value. –  Nathan W Commented May 20, 2013 at 2:49
• You could use 3 x elif and an else to see if any values other than 1-4 are encountered. –  PolyGeo ♦ Commented May 20, 2013 at 3:44
• I tried that way as well but still hung up with error. –  Ibe Commented May 20, 2013 at 4:15

This error is pretty much explained here and it helped me to get assignments and return values for all variables.

Your Answer

Sign up or log in, post as a guest.

Required, but never shown

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy .

Not the answer you're looking for? Browse other questions tagged arcpy arcgis-10.1 unboundlocalerror or ask your own question .

• Featured on Meta
• Upcoming sign-up experiments related to tags

Hot Network Questions

• Would a spaceport on Ceres make sense?
• What to fill in the document number while applying for Schengen visa "C"?
• Roll-adjustment definition for swaps schedule generation
• Which numbers are sums of finite numbers of reciprocal squares?
• Drawing waves using tikz in latex
• How to model an optimization problem with mutual exclusivity of two variables, without introducing integer variables?
• what is the difference between prayer (προσευχῇ) and prayer also translated as supplication (δέησις) in Philippians 4:6?
• Why can Ethernet NICs bridge to VirtualBox and most Wi-Fi NICs don't?
• What does ‘a grade-hog’ mean?
• Visit USA via land border for French citizen
• Collaborators write their departments for my (undergraduate) affiliation
• Did James Madison say or write that the 10 Commandments are critical to the US nation?
• Were there engineers in blimp nacelles, and why were they there?
• Tombs of Ancients
• How to produce this table: Probability datatable with multirow
• Tubeless tape width?
• In By His Bootstraps (Heinlein) why is Hitler's name Schickelgruber?
• Less ridiculous way to prove that an Ascii character compares equal with itself in Coq
• How do you say "living being" in Classical Latin?
• Weird behavior by car insurance - is this legit?
• Power of multiplication of powers
• Huygens' principle and the laws of reflection/refraction
• Why was William Tyndale executed but nothing happened to Miles Coverdale?
• Is it consistent with ZFC that the real line is approachable by sets with no accumulation points?

UnboundLocalError: local variable 'fig' referenced before assignment

I am getting the above error while plotting the bar graphs and appending them to results. Is there any solution.

UnboundLocalError: local variable ‘fig’ referenced before assignment Traceback (most recent call last): File “C:\Users\Local\Temp\ipykernel_20424\180331076.py”, line 16, in update_graphs File “C:\Users\Lib\site-packages\plotly\express_chart_types.py”, line 368, in bar return make_figure( File “C:\Users\Lib\site-packages\plotly\express_core.py”, line 2182, in make_figure fig = init_figure( File “C:\Users\Lib\site-packages\plotly\express_core.py”, line 2327, in init_figure for annot in fig.layout.annotations: UnboundLocalError: local variable ‘fig’ referenced before assignment

fig =px.bar(x='country',y='population')

I assume x and y are columns of a DataFrame, but you never specify the DataFrame to use.

could also be a scope issue when fig=px.bar() was declared in a different function or not in the function where result.append() was declared.

@dashapp , is result.append() right below fig = px.bar()?

@adamschroeder I updated it as a new topic

hi @dashapp I don’t think this is a solution yet, but heads up that you have a spelling mistake with ‘var’. You probably meant val.

What is “output.append”? What is output coming from?

@adamschroeder Sorry that was a typo. so when I click a value from the dropdown output should return a bar plot and empty container. Suppose I select two columns output should be two bar graphs and empty container which can take hoverdata and click data. My code is working properly and only error is with fig. Even if I remove fig variable and add the graph in figure variable in dcc.graph, the error persists. It has something to do with plotly. Can you check this.

An UnboundLocalError is raised when a local variable is referenced before it has been assigned. In most cases this will occur when trying to modify a local variable before it is actually assigned within the local scope. 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.

Python has lexical scoping by default, which means that although an enclosed scope can access values in its enclosing scope, it cannot modify them (unless they’re declared global with the global keyword). A closure binds values in the enclosing environment to names in the local environment. The local environment can then use the bound value, and even reassign that name to something else, but it can’t modify the binding in the enclosing environment. UnboundLocalError happend because when python sees an assignment inside a function then it considers that variable as local variable and will not fetch its value from enclosing or global scope when we execute the function. However, to modify a global variable inside a function, you must use the global keyword.

In my case the error is UnboundLocalError: local variable ‘fig’ referenced before assignment

Traceback (most recent call last): File “C:\Users\Local\Temp\ipykernel_20424\180331076.py”, line 16, in update_graphs

File “C:\Users\Lib\site-packages\plotly\express_chart_types.py”, line 368, in bar return make_figure( File “C:\Users\Lib\site-packages\plotly\express_core.py”, line 2182, in make_figure fig = init_figure( File “C:\Users\Lib\site-packages\plotly\express_core.py”, line 2327, in init_figure for annot in fig.layout.annotations:

I haven’t used fig variable at all. Its just related to plotly fig.

I get the error too. But I found my d1 is empty dataframe. when I fix the filt operation, it works

Related Topics

• View Active Threads
• View Today's Posts
• View New Posts
• My Discussions
• Unanswered Posts
• Unread Posts
• Active Threads
• Mark all forums read
• Member List
• Interpreter

UnboundLocalError: local variable referenced before assignment

• Python Forum
• Python Coding
• General Coding Help
• 0 Vote(s) - 0 Average

• View a Printable Version

User Panel Messages

Announcements.

Login to Python Forum

【已解决】Python报错：UnboundLocalError：local variable ‘xxx‘ referenced before assignment

Python编程实战：深入解析与解决UnboundLocalError的策略

在Python编程过程中，开发者可能会遭遇 UnboundLocalError 这一常见错误，其错误信息通常表现为“local variable ‘xxx’ referenced before assignment”，意味着尝试访问一个在当前作用域内未被事先赋值的局部变量。本文将通过具体场景分析，揭示错误背后的原理，并提供一系列实用解决方案，助你轻松跨越这一编程挑战。

运行上述代码时，由于在累加成绩前未对 total 进行初始化，程序将抛出 UnboundLocalError 。

• 确保变量初始化 ：在使用任何变量前，务必确保已经赋予了初始值。对于上述案例，应在循环前正确初始化 total 变量。

修正后的代码示例：

明确作用域管理 ：如果变量需要在多个作用域内共享，考虑其定义的位置，或使用全局变量（谨慎使用，以免造成不必要的副作用）。

循环和条件语句中的变量处理 ：在循环或条件判断中，确保变量在所有可能的执行路径上都得到初始化。

利用默认值 ：在函数参数中提供默认值，可以避免因参数未传入而导致的未初始化错误。

UnboundLocalError 的产生，实质上是对Python作用域规则理解和应用上的疏漏。通过上述策略的应用，开发者不仅能够有效避免此类错误，还能进一步加深对Python语言特性的理解。记住，良好的编程习惯，如及时初始化变量、清晰地界定作用域，是编写高质量代码的基石。在编程之旅上，每一步的谨慎与反思，都是通往卓越的必经之路。

祝大家学习顺利~ 如有任何错误，恳请批评指正~~ 以上是我通过各种方式得出的经验和方法，欢迎大家评论区留言讨论呀，如果文章对你们产生了帮助，也欢迎点赞收藏，我会继续努力分享更多干货~

🎈关注我的公众号AI Sun可以获取Chatgpt最新发展报告以及腾讯字节等众多大厂面经。 😎也欢迎大家和我交流，相互学习，提升技术，风里雨里，我在等你~

“相关推荐”对你有帮助么？

请填写红包祝福语或标题

你的鼓励将是我创作的最大动力

您的余额不足，请更换扫码支付或 充值

1.余额是钱包充值的虚拟货币，按照1:1的比例进行支付金额的抵扣。 2.余额无法直接购买下载，可以购买VIP、付费专栏及课程。

• Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers
• Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand
• OverflowAI GenAI features for Teams
• OverflowAPI Train & fine-tune LLMs
• Labs The future of collective knowledge sharing
• About the company Visit the blog

Collectives™ on Stack Overflow

Find centralized, trusted content and collaborate around the technologies you use most.

Q&A for work

Connect and share knowledge within a single location that is structured and easy to search.

Get early access and see previews of new features.

Seaborn Error: local variable 'boxprops' referenced before assignment [closed]

I am trying to plot using seaborn boxplot, however, I get the following error:

These codes were executing fine last week. I have tried updating seaborn and pyfolio, yet I still get the same error. Does anyone know how to fix this?

**Attempted Solutions: **

• Updating seaborn and pyfolio to the latest versions.

Does anyone have suggestions on how to resolve this issue?

• Can you test the code as shown? Does it still generate the error? Do you get other warnings? –  JohanC Commented Jun 24 at 13:58
• @JohanC It works ! thank you! Do you know what the exact problem was? –  Rhea Groenenberg Commented Jun 24 at 14:05
• I think there was something wrong with the input data. Maybe the problem is "solved" in this test, but not with your real data. In that case, you could try to print out information about filtered_df and create test data similar to that one. Or maybe your update of pyfolio finally had effect (I suppose pyfolio changed something to the input they are sending to seaborn). –  JohanC Commented Jun 24 at 14:10
• I'm voting to close this issue as not reproducible. –  Trenton McKinney Commented Jun 25 at 1:05

Browse other questions tagged python matplotlib seaborn pyfolio or ask your own question .

• Featured on Meta
• Upcoming sign-up experiments related to tags
• Should we burninate the [lib] tag?
• Policy: Generative AI (e.g., ChatGPT) is banned
• What makes a homepage useful for logged-in users

Hot Network Questions

• Algorithm to evaluate "connectedness" of a binary matrix
• Stiff differential equation
• Synthesis of racemic nicotine
• Why did Geordi have his visor replaced with ocular implants between Generations and First Contact?
• How to make D&D easier for kids?
• What could explain that small planes near an airport are perceived as harassing homeowners?
• Can I tell a MILP solver to prefer solutions with fewer fractions?
• Can a planet have a warm, tropical climate both at the poles and at the equator?
• DSP Puzzle: Advanced Signal Forensics
• How do you say "living being" in Classical Latin?
• Weird behavior by car insurance - is this legit?
• Co-authors with little contribution
• How are "pursed" and "rounded" synonymous?
• Does Not(A and not-A) = Not(A nand A) in intuitionistic logic?
• What is the relationship between gravitation, centripetal and centrifugal force on the Earth?
• Is there a way to non-destructively test whether an Ethernet cable is pure copper or copper-clad aluminum (CCA)?
• Are positions with different physical pieces on two squares but same kind and same color, considered the same?
• How to Control StringContainsQ
• Would a spaceport on Ceres make sense?
• Why are there no Goldstone modes in superconductor?
• Could a transparent frequency-altering material be possible?
• Why can Ethernet NICs bridge to VirtualBox and most Wi-Fi NICs don't?
• Do IDE data lines need pull-up resistors?
• Is there any instance of Sati practice described in Itihasa and Purana?