local variable 'version' referenced before assignment

[SOLVED] Local Variable Referenced Before Assignment

Python treats variables referenced only inside a function as global variables. Any variable assigned to a function’s body is assumed to be a local variable unless explicitly declared as global.

Why Does This Error Occur?

Unboundlocalerror: local variable referenced before assignment occurs when a variable is used before its created. Python does not have the concept of variable declarations. Hence it searches for the variable whenever used. When not found, it throws the error.

Before we hop into the solutions, let’s have a look at what is the global and local variables.

Local Variable Declarations vs. Global Variable Declarations

Local Variable Referenced Before Assignment Error with Explanation

Try these examples yourself using our Online Compiler.

Let’s look at the following function:

Explanation

The variable myVar has been assigned a value twice. Once before the declaration of myFunction and within myFunction itself.

Using Global Variables

Passing the variable as global allows the function to recognize the variable outside the function.

Create Functions that Take in Parameters

Instead of initializing myVar as a global or local variable, it can be passed to the function as a parameter. This removes the need to create a variable in memory.

UnboundLocalError: local variable ‘DISTRO_NAME’

This error may occur when trying to launch the Anaconda Navigator in Linux Systems.

Upon launching Anaconda Navigator, the opening screen freezes and doesn’t proceed to load.

Try and update your Anaconda Navigator with the following command.

If solution one doesn’t work, you have to edit a file located at

After finding and opening the Python file, make the following changes:

In the function on line 159, simply add the line:

DISTRO_NAME = None

Save the file and re-launch Anaconda Navigator.

DJANGO – Local Variable Referenced Before Assignment [Form]

The program takes information from a form filled out by a user. Accordingly, an email is sent using the information.

Upon running you get the following error:

We have created a class myForm that creates instances of Django forms. It extracts the user’s name, email, and message to be sent.

A function GetContact is created to use the information from the Django form and produce an email. It takes one request parameter. Prior to sending the email, the function verifies the validity of the form. Upon True , .get() function is passed to fetch the name, email, and message. Finally, the email sent via the send_mail function

Why does the error occur?

We are initializing form under the if request.method == “POST” condition statement. Using the GET request, our variable form doesn’t get defined.

Local variable Referenced before assignment but it is global

This is a common error that happens when we don’t provide a value to a variable and reference it. This can happen with local variables. Global variables can’t be assigned.

This error message is raised when a variable is referenced before it has been assigned a value within the local scope of a function, even though it is a global variable.

Here’s an example to help illustrate the problem:

In this example, x is a global variable that is defined outside of the function my_func(). However, when we try to print the value of x inside the function, we get a UnboundLocalError with the message “local variable ‘x’ referenced before assignment”.

This is because the += operator implicitly creates a local variable within the function’s scope, which shadows the global variable of the same name. Since we’re trying to access the value of x before it’s been assigned a value within the local scope, the interpreter raises an error.

To fix this, you can use the global keyword to explicitly refer to the global variable within the function’s scope:

However, in the above example, the global keyword tells Python that we want to modify the value of the global variable x, rather than creating a new local variable. This allows us to access and modify the global variable within the function’s scope, without causing any errors.

Local variable ‘version’ referenced before assignment ubuntu-drivers

This error occurs with Ubuntu version drivers. To solve this error, you can re-specify the version information and give a split as 2 –

Here, p_name means package name.

With the help of the threading module, you can avoid using global variables in multi-threading. Make sure you lock and release your threads correctly to avoid the race condition.

When a variable that is created locally is called before assigning, it results in Unbound Local Error in Python. The interpreter can’t track the variable.

Therefore, we have examined the local variable referenced before the assignment Exception in Python. The differences between a local and global variable declaration have been explained, and multiple solutions regarding the issue have been provided.

Trending Python Articles

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 ⚡️

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

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.

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 *

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)

• 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
• How to Fix - UnboundLocalError: Local variable Referenced Before Assignment in Python
• Python | Accessing variable value from code scope
• Access environment variable values in Python
• Get Variable Name As String In Python
• How to use Pickle to save and load Variables in Python?
• Undefined Variable Nameerror In Python
• How to Reference Elements in an Array in Python
• Difference between Local Variable and Global variable
• Unused local variable in Python
• Unused variable in for loop in Python
• Assign Function to a Variable in Python
• JavaScript ReferenceError - Can't access lexical declaration`variable' before initialization
• Global and Local Variables in Python
• Pass by reference vs value in Python
• __file__ (A Special variable) in Python
• Variables under the hood in Python
• __name__ (A Special variable) in Python
• PYTHONPATH Environment Variable in Python
• Julia local Keyword | Creating a local variable in Julia

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

The “local variable referenced before assignment” error occurs when you try to use a local variable before it has been assigned a value. This is a general programming concept describing the situation typically arises in situations where you declare a variable within a function but then try to access or modify it before actually assigning a value to it.

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

Here’s an example to illustrate this error:

In this example, you would encounter the above error because you’re trying to print the value of x before it has been assigned a value. To fix this, you should assign a value to x before attempting to access it:

In the corrected version, the local variable x is assigned a value before it’s used, preventing the error.

Keep in mind that Python treats variables inside functions as local unless explicitly stated otherwise using the global keyword (for global variables) or the nonlocal keyword (for variables in nested functions).

If you encounter this error and you’re sure that the variable should have been assigned a value before its use, double-check your code for any logical errors or typos that might be causing the variable to not be assigned properly.

Using the global keyword

If you have a global variable named letter and you try to modify it inside a function without declaring it as global, you will get error.

This is because Python assumes that any variable that is assigned a value inside a function is a local variable, unless you explicitly tell it otherwise.

To fix this error, you can use the global keyword to indicate that you want to use the global variable:

Using nonlocal keyword

The nonlocal keyword is used to work with variables inside nested functions, where the variable should not belong to the inner function. It allows you to modify the value of a non-local variable in the outer scope.

For example, if you have a function outer that defines a variable x , and another function inner inside outer that tries to change the value of x , you need to use the nonlocal keyword to tell Python that you are referring to the x defined in outer , not a new local variable in inner .

Here is an example of how to use the nonlocal keyword:

If you don’t use the nonlocal keyword, Python will create a new local variable x in inner , and the value of x in outer will not be changed:

You might also like

UndboundLocalError: local variable referenced before assignment

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

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

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

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

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

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

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

Thanks again

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

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

Related Topics

Ubuntu 22.04.1 Nvidia Driver (Open Kernel) Nvidia-Driver-515-Open Issue

sudo ubuntu-drivers autoinstall [sudo] password for username: Traceback (most recent call last): File “/usr/bin/ubuntu-drivers”, line 513, in greet() File “/usr/lib/python3/dist-packages/click/core.py”, line 1128, in call return self.main(*args, **kwargs) File “/usr/lib/python3/dist-packages/click/core.py”, line 1053, in main rv = self.invoke(ctx) File “/usr/lib/python3/dist-packages/click/core.py”, line 1659, in invoke return _process_result(sub_ctx.command.invoke(sub_ctx)) File “/usr/lib/python3/dist-packages/click/core.py”, line 1395, in invoke return ctx.invoke(self.callback, **ctx.params) File “/usr/lib/python3/dist-packages/click/core.py”, line 754, in invoke return __callback(*args, **kwargs) File “/usr/lib/python3/dist-packages/click/decorators.py”, line 84, in new_func return ctx.invoke(f, obj, *args, **kwargs) File “/usr/lib/python3/dist-packages/click/core.py”, line 754, in invoke return __callback(*args, **kwargs) File “/usr/bin/ubuntu-drivers”, line 432, in autoinstall command_install(config) File “/usr/bin/ubuntu-drivers”, line 187, in command_install UbuntuDrivers.detect.nvidia_desktop_pre_installation_hook(to_install) File “/usr/lib/python3/dist-packages/UbuntuDrivers/detect.py”, line 839, in nvidia_desktop_pre_installation_hook with_nvidia_kms = version >= 470 UnboundLocalError: local variable ‘version’ referenced before assignment

1.when i run sudo ubuntu-drivers autoinstall this error appear. 2.When i choose using nvidia (open kernel) metapackage from nvidia-driver-515-open(proprietart,tested) my pc will hang after restart and cannot do anything. 3.to solve this error i need to reinstall ubuntu or if have luck i can use recovery and change the driver to default driver. nvidia-bug-report.log.gz (165.9 KB)

hello, i tried to install nvidia-driver today on 22.04 aswell, and incurred into same/similar error. what did is just “sudo apt-get upgrade” and restart, afterwards the error went away and i could install nvidia-driver , but did not use “autoinstall”; instead i went to “additional drivers” and selected from there the newer package “Nvidia-515”. after restart, would check , and if desktop would not load, then i change GUI from Wayland to Xorg, which helped in my case (unity desktop would not load either properly with nvidia-driver)

Hello, I also see the same error on Kubuntu 22.04 when running

sudo ubuntu-drivers install

I ran into the same issue after installing 515 drivers for my RTX A4000. I was able to make my machine boot again by disabling secure boot (not sure if that was necessary) and changing a setting in the “Storage” section of the BIOS. Changed “RAID On” to “AHCI/NVME”. Now it boots again and I can use the GPU for computations and such things. But now my Soundcard and Bluetooth no longer work. So I woudl appreciate a better solution

This is preventing me from upgrading to 520 driver which appears to be required for the 5.17.0 ubuntu kernel. X starts and then hangs when booing /boot/vmlinuz-5.17.0-1020-oem kernel.

Found workaround.

Hi there, we came across this issue during an install of a new Ubuntu.

The error is from the way the installer tries to parse the version (an int) from the driver name. The code assumes the version is the last part of the driver name, which in most cases is fine e.g. nvidia-driver-520 . In my scenario the driver name was nvidia-driver-520-open and the code ends up trying to parse open as an int and fails, throwing the undefined error a few lines later as version failed to be parsed.

Solution: This will only work if you have access to /usr/lib/python3/dist-packages/UbuntuDrivers and have edit permission. offending line: /usr/lib/python3/dist-packages/UbuntuDrivers/detect.py:835 find following line: version = int(package_name.split('-')[-1]) modify to: version = int(package_name.split('-')[2]) Note this is assuming the version is the 3rd word in the driver name split by ‘-’

Understandably when trying to parse values from names only these issues can happen from time to time. A long-term fix would need the insight of the naming conventions of current and for future driver package names.

I’m having the exact same issue. I updated when prompted, only to get a black screen, no boot scenario afterwards. I’ve reinstalled Ubuntu literally about 50 times now and tried to install every available driver, with every possible installer method. All result in black screen after reboot. Nvidia drivers cannot currently be installed on Ubuntu Jellyfish.

I might have to install WINDOWS to WORK 😭

This thread has given me peace however. I will no longer chase a solution to an unfixable problem.

I’ll check back next driver update. 🙏

Same over here. Please don’t tell me everyone is giving up on this issue? I just purchased this laptop for the sole purpose of utilising the GPU to speed up ML model training, and now you’re saying this is not going to happen? I would rather burn this laptop than work on Windows.

Has anyone attempted the driver edit mentioned above? I tried to edit the file “/usr/lib/python3/dist-packages/UbuntuDrivers/detect.py”, but was denied permission, and after a few other attempts at driver installations and workarounds, my whole system is very laggy and my file explorer won’t open at all.

I will attempt a few other things, before possibly having to re-install Ubuntu.

[ Update ] Like @levimulkey , I have tried every combination of driver installations I can find. While I did manage to install the " nvidia-driver-520-open " driver as well as " nvidia-driver-520 ", both via 'Software & Updates", when running “ nvidia-smi ” there is still a “No Driver Found” message?

Via the command line, I managed to install the Nvidia driver/package which I opened and registered on, and even found version 11.8 was installed with nvidia-smi. Alas the CUDA toolkit was still not installed. So after installing, the Nvidia driver was once again not found, or at least not recognised by a python test script.

The purpose of this driver installation is not for gaming and HAS TO function with CUDA/CUDNN, which seem to be clashing with each other.

My entire system has black-screened multiple times as well as file-explorer freezing permanently, and I have also re-installed Ubuntu many times now.

Is it my turn to give up on this also, and go over to my nightmare OS. Has Nvidia always been Windows focussed?

I’m running Ubuntu 22.04 with the standard 5.15.0.52 kernel and both the 515 and the 520 drivers (NOT the 515-open or 520-open) from Ubuntu’s standard repositories work fine on GTX970 and RTX 3080 hardware. I use the Software & Updates/Additional Drivers to select the driver, and reboot. On a fresh Ubuntu install, if you don’t elect to install the Nvidia drivers within the install, the first reboot will take the word “nomodeset” added to the kernel parameters in the grub boot menu, so the nouveau driver does not produce a black screen. Then run the Software & Updates to select a driver so “nomodeset” is no longer needed.

With the Nvidia driver installed and working, I’d suggest using the Nvidia …run script to install CUDA. Uncheck any offer of an Nvidia driver, and use the options to skip any system location for libs. Turn off the icon option too. Then you may take ownership of /usr/local and run the …run script as a normal user, not an admin. After installation, everything will be in /usr/local/cuda-11.8, so follow the recommendations to add those locations to your PATH and LD_LIBRARY_PATH. Restore the permissions on /usr/local, and you should have a working CUDA that is independent of the Nvidia driver in use, and works through kernel updates (which rebuild the driver). Search the askubuntu.com site for more installation answers.

Okay so just a final follow-up with all the steps taken to fully solve the driver to CUDA issue on my system: Hope it leads someone in the right direction.

System Specs:

MSI - 11th Gen Intel® Core™ i7-11800H @ 2.30GHz × 16 NVIDIA Corporation GA106M [GeForce RTX 3060 Mobile 16Gb RAM running Ubuntu 22.04

STEPS TAKEN:

Install anaconda:, install nvidia driver:.

V01 [2021]:

Create and run Conda Virtual with Spyder 5:

USE CONDA-FORGE:

Install CUDA Toolkit:

Install conda-forge:

Install CUDA & CUDNN: With CONDA:

Install missing packages:

• Included cudatoolkit / cudnn / matplotlib / pandas / numpy
• Non Included: torchvision / tensorflow-gpu / sklearn / keras

================================================================

Check Installations:

================================================================ conda activate spyder-env

Driver Check:

CUDA Check:

Tensoflow Check:

Sub-Packages Check:

Final Results:

================================================================ cat /proc/driver/nvidia/version

conda list cudatoolkit conda list cudnn

nvcc --version

In Python, the GPU is now recognised:

There are 2 issues here.

• The ubuntu-drivers script mistakenly parsing “open” as a int ( Solution discussed above)
• The other is even when installed through the Additional Drivers tab or by fixing script, I ended up with a kernel panic with “Out of Memory … press any key to continue”. Where the only option was reboot and get to grub menu and choose a older kernel version. Then reverting to xorg nouveau driver.

For Issue 2 - Downgrading to nvidia driver 470 worked for me.

I solved the problem by moving to a newer distro of Ubuntu.

@levimulkey Could you please share which distro is that you moved to ? Does it mean you are able to use the latest 520 drivers in that distro ?

There are on-going threads in launchpad for these in ubuntu

After, downgrading to 470, Tried again from GUI (Additional Drivers) to move to 515. This worked!

Note : I did follow these steps to handle suspend issue with 470 before trying 515 again.

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.

Ubuntu 22.04.1, got stuck with the option "Continue using a manually installed driver"

I'd be glad if someone could help me out. I'm running 22.04.1 on a notebook with Nvidia GTX1650. I tried changing the Nvidia driver from 510 to 515 in the "Additional Drivers" GUI. The change failed, probably due to temporary internet disconnect, I tried it a couple of more times, it failed again, until I tried to play it safe and switch to the noveau driver. However, at that point I got stuck with the option "Continue using a manually installed driver". I reviewed this previous string in depth: Not able to change the Nvidia driver in ubuntu 20.04 , but couldn't find an answer applicable to my scenario.

E.g. when I try

I get nearly identical output

And when I try removing all Nvidia packages:

But I can't find out which argument is the correct one.

So to sum up, seems like I'm stuck with the integrated UHD Graphics at the moment, unable to use my Nvidia. Settings > About: enter image description here

Any ideas on how I can get a valid Nvidia driver installed again?

I connected to a more stable network, then ran:

This installed the somewhat older 470 driver. Now the Continue using a manually installed driver is gone in the Additional Drivers GUI, and the other options are active, not grayed out.

• It worked for me too.. the only difference is that I had to use driver version 390, which is the latest version available for my old nvidia laptop video card NVS 4200M. sudo apt install nvidia-driver-390 Thank you ! –  Asim Ali Oct 30, 2022 at 3:37
• This worked for me too, but after installing 470 I installed 525. For whichever reason, if I try to change driver in the GUI I just get an "X" pop up. However, now I have the latest driver. –  mic Dec 3, 2022 at 18:22

You must log in to answer this question.

Not the answer you're looking for browse other questions tagged drivers nvidia ., hot network questions.

• Sum of square roots (as an algebraic number)
• Transformer with same size symbol meaning
• Reaction of RuCl3 and alkaline solution
• Why don't professors seem to use learning strategies like spaced repetition and note-taking?
• How was damno derived from damnum?
• What is the point of triggering of a national snap election immediately after losing the EU elections?
• How do I tell which kit lens option is more all-purpose?
• What does "far right tilt" actually mean in the context of the EU in 2024?
• Build the first 6 letters of an Italian codice fiscale (tax identification number)
• Word for a country declaring independence from an empire
• Who would I call to inspect the integrity of a truss after an electrician cut into it?
• Do we know how the SpaceX Starship stack handles engine shutdowns?
• How can I tell whether an HDD uses CMR or SMR?
• Is it theoretically possible for the sun to go dark?
• How to Adjust Comparator Output Voltage for Circuit to Work at 3.3V Instead of 5V?
• Replacement the UPS APC Smart-UPS SRT 10000 battery while it's powered on?
• Have I ruined my AC by running it with the outside cover on?
• ytableau specify boxframe for each \ydiagram
• Is the angular orbital velocity of a satellite in a circular orbit constant?
• Why a truly uninformative prior does not exist?
• How often does systemd journal collect/read logs from sources
• Book recommendation introduction to model theory
• sculpt mode symmetry mirror mode not working
• Asterisk in violin sheet music

Navigation Menu

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 You must be signed in to change notification settings

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 'cublas_path' referenced before assignment on importing torch #91691

sanvyruz commented Jan 4, 2023 • edited by pytorch-bot bot

• 👍 1 reaction

sanvyruz commented Jan 4, 2023 • edited

Sorry, something went wrong.

sanvyruz commented Jan 9, 2023

atalman commented Feb 3, 2023 • edited

ptrblck commented Feb 3, 2023

sanvyruz commented Feb 6, 2023 • edited

ptrblck commented Feb 6, 2023

sanvyruz commented Feb 7, 2023

Successfully merging a pull request may close this issue.

• Stack Overflow Public questions & answers
• Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers
• Talent Build your employer brand
• Advertising Reach developers & technologists worldwide
• Labs The future of collective knowledge sharing
• About the company

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.

"local variable 's' referenced before assignment" error. How do I fix?

I am trying to make a list that takes in numbers and removes all the old numbers in the list and returns s the result, but it keeps on giving me the error "local variable 's' referenced before assignment" How do I fix it?

• 1 This won't work. Do not modify lists while iterating over them. Copy the list and operate on the copy. –  nneonneo Jun 18, 2014 at 15:06
• If you are modifying lst directly, why not just return lst ? You probably either want to initialize s to a copy of lst at the top of the function and then remove from s , or just return lst . Also, I'd bet you actually want s.extend(lst) , not append . –  Silas Ray Jun 18, 2014 at 15:07
• You're trying to add to s before you create it. Have you used assignment before? The tutorial may be useful to you. –  Kevin Jun 18, 2014 at 15:07
• Why do you want to return a list containing a single list? –  jonrsharpe Jun 18, 2014 at 15:07
• exactly what the error says... code reads line by line in this case, and s is defined after you append to s. that's like writing before you get a pen. –  corvid Jun 18, 2014 at 15:11

4 Answers 4

It isn't clear why you want to return a list containing a single list. Why not just return the "purified" list?

This uses a list comprehension . Demo:

For more flexibility, you could also make a version where you pass your own rule:

You need to assign s to empty list first and then append.

As @jonrsharpe suggested, don't remove the item while iterating over list. The good approach is using the list comprehension :

• Or just return [lst] - but note that there is an issue with modifying lst . –  jonrsharpe Jun 18, 2014 at 15:07
• That's a great attempt at spelling my username! –  jonrsharpe Jun 18, 2014 at 15:37
• Sorry about that mate. Fixed :) –  pynovice Jun 18, 2014 at 15:48

You should not remove items while iterating, because the list should not be changed when using iterators (this is what you do with "in").

btw: This can be done very fast with numpy.

• I understand but why did you put the [] around item? –  user2626734 Jun 18, 2014 at 15:20
• s.append(item) would be more typical than s += [item] , which creates two new lists –  jonrsharpe Jun 18, 2014 at 15:36

Simplify your life with filter:

Your Answer

Reminder: Answers generated by artificial intelligence tools are not allowed on Stack Overflow. Learn more

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 python or ask your own question .

• Featured on Meta
• The 2024 Developer Survey Is Live
• The return of Staging Ground to Stack Overflow
• The [tax] tag is being burninated
• Policy: Generative AI (e.g., ChatGPT) is banned

Hot Network Questions

• Have I ruined my AC by running it with the outside cover on?
• Tying shoes on Shabbat if you don’t plan to untie them in a day
• Is boosting all frequencies on an EQ equivalent to a flat EQ
• Need help in creating this chain link
• Am I seeing double? What kind of helicopter is this, and how many blades does it actually have?
• Who would I call to inspect the integrity of a truss after an electrician cut into it?
• What is my new plant Avocado tree leaves dropping?
• Is it allowed to use patents for new inventions?
• How can I adjust a Rheem water heater thermostat?
• What does "far right tilt" actually mean in the context of the EU in 2024?
• It suggests vs It would suggest
• NES Emulator in C
• What percentage of light gets scattered by a mirror?
• Movie I saw in the 80s where a substance oozed off of movie stairs leaving a wet cat behind
• Can I setup new electrical combo panel on new retrofit wood?
• Smallest Harmonic number greater than N
• Group with a translation invariant ultrafilter
• How do I tell which kit lens option is more all-purpose?
• What is the point of triggering of a national snap election immediately after losing the EU elections?
• Best way to halve 12V battery voltage for 6V device, while still being able to measure the battery level?
• Which program is used in this shot of the movie "The Wrong Woman"
• Why was the client spooked when he saw the professor's face?
• Calculating the fibres of a scheme morphism are proper but the morphism is not proper
• What do humans do uniquely, that computers apparently will not be able to?