unboundlocalerror local variable 'issue' referenced before assignment

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.

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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

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UnboundLocalError Local variable Referenced Before Assignment in Python

Handling errors is an integral part of writing robust and reliable Python code. One common stumbling block that developers often encounter is the “UnboundLocalError” raised within a try-except block. This error can be perplexing for those unfamiliar with its nuances but fear not – in this article, we will delve into the intricacies of the UnboundLocalError and provide a comprehensive guide on how to effectively use try-except statements to resolve it.

What is UnboundLocalError Local variable Referenced Before Assignment in Python?

The UnboundLocalError occurs when a local variable is referenced before it has been assigned a value within a function or method. This error typically surfaces when utilizing try-except blocks to handle exceptions, creating a puzzle for developers trying to comprehend its origins and find a solution.

Why does UnboundLocalError: Local variable Referenced Before Assignment Occur?

below, are the reasons of occurring “Unboundlocalerror: Try Except Statements” in Python :

Variable Assignment Inside Try Block

Reassigning a global variable inside except block.

• Accessing a Variable Defined Inside an If Block

In the below code, example_function attempts to execute some_operation within a try-except block. If an exception occurs, it prints an error message. However, if no exception occurs, it prints the value of the variable result outside the try block, leading to an UnboundLocalError since result might not be defined if an exception was caught.

In below code , modify_global function attempts to increment the global variable global_var within a try block, but it raises an UnboundLocalError. This error occurs because the function treats global_var as a local variable due to the assignment operation within the try block.

Solution for UnboundLocalError Local variable Referenced Before Assignment

Below, are the approaches to solve “Unboundlocalerror: Try Except Statements”.

Initialize Variables Outside the Try Block

Avoid reassignment of global variables.

In modification to the example_function is correct. Initializing the variable result before the try block ensures that it exists even if an exception occurs within the try block. This helps prevent UnboundLocalError when trying to access result in the print statement outside the try block.

Below, code calculates a new value ( local_var ) based on the global variable and then prints both the local and global variables separately. It demonstrates that the global variable is accessed directly without being reassigned within the function.

In conclusion , To fix “UnboundLocalError” related to try-except statements, ensure that variables used within the try block are initialized before the try block starts. This can be achieved by declaring the variables with default values or assigning them None outside the try block. Additionally, when modifying global variables within a try block, use the `global` keyword to explicitly declare them.

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Python local variable referenced before assignment Solution

When you start introducing functions into your code, you’re bound to encounter an UnboundLocalError at some point. This error is raised when you try to use a variable before it has been assigned in the local context .

In this guide, we talk about what this error means and why it is raised. We walk through an example of this error in action to help you understand how you can solve it.

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What is unboundlocalerror: local variable referenced before assignment.

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

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

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

Let’s take a look at how to solve this error.

An Example Scenario

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

We start by declaring two variables:

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

Finally, we call our function:

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

Let’s run our code and see what happens:

An error has been raised.

The Solution

Our code returns an error because we reference “letter” before we assign it.

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

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

We solve this problem in two ways. First, we can add an else statement to our code. This ensures we declare “letter” before we try to return it:

Let’s try to run our code again:

Our code successfully prints out the student’s grade.

If you are using an “if” statement where you declare a variable, you should make sure there is an “else” statement in place. This will make sure that even if none of your if statements evaluate to True, you can still set a value for the variable with which you are going to work.

Alternatively, we could use the “global” keyword to make our global keyword available in the local context in our calculate_grade() function. However, this approach is likely to lead to more confusing code and other issues. In general, variables should not be declared using “global” unless absolutely necessary . Your first, and main, port of call should always be to make sure that a variable is correctly defined.

In the example above, for instance, we did not check that the variable “letter” was defined in all use cases.

That’s it! We have fixed the local variable error in our code.

The UnboundLocalError: local variable referenced before assignment error is raised when you try to assign a value to a local variable before it has been declared. You can solve this error by ensuring that a local variable is declared before you assign it a value.

Now you’re ready to solve UnboundLocalError Python errors like a professional developer !

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Python UnboundLocalError: local variable referenced before assignment

by Suf | Programming , Python , Tips

If you try to reference a local variable before assigning a value to it within the body of a function, you will encounter the UnboundLocalError: local variable referenced before assignment.

The preferable way to solve this error is to pass parameters to your function, for example:

Alternatively, you can declare the variable as global to access it while inside a function. For example,

This tutorial will go through the error in detail and how to solve it with code examples .

Table of contents

What is scope in python, unboundlocalerror: local variable referenced before assignment, solution #1: passing parameters to the function, solution #2: use global keyword, solution #1: include else statement, solution #2: use global keyword.

Scope refers to a variable being only available inside the region where it was created. A variable created inside a function belongs to the local scope of that function, and we can only use that variable inside that function.

A variable created in the main body of the Python code is a global variable and belongs to the global scope. Global variables are available within any scope, global and local.

UnboundLocalError occurs when we try to modify a variable defined as local before creating it. If we only need to read a variable within a function, we can do so without using the global keyword. Consider the following example that demonstrates a variable var created with global scope and accessed from test_func :

If we try to assign a value to var within test_func , the Python interpreter will raise the UnboundLocalError:

This error occurs because when we make an assignment to a variable in a scope, that variable becomes local to that scope and overrides any variable with the same name in the global or outer scope.

var +=1 is similar to var = var + 1 , therefore the Python interpreter should first read var , perform the addition and assign the value back to var .

var is a variable local to test_func , so the variable is read or referenced before we have assigned it. As a result, the Python interpreter raises the UnboundLocalError.

Example #1: Accessing a Local Variable

Let’s look at an example where we define a global variable number. We will use the increment_func to increase the numerical value of number by 1.

Let’s run the code to see what happens:

The error occurs because we tried to read a local variable before assigning a value to it.

We can solve this error by passing a parameter to increment_func . This solution is the preferred approach. Typically Python developers avoid declaring global variables unless they are necessary. Let’s look at the revised code:

We have assigned a value to number and passed it to the increment_func , which will resolve the UnboundLocalError. Let’s run the code to see the result:

We successfully printed the value to the console.

We also can solve this error by using the global keyword. The global statement tells the Python interpreter that inside increment_func , the variable number is a global variable even if we assign to it in increment_func . Let’s look at the revised code:

Let’s run the code to see the result:

Example #2: Function with if-elif statements

Let’s look at an example where we collect a score from a player of a game to rank their level of expertise. The variable we will use is called score and the calculate_level function takes in score as a parameter and returns a string containing the player’s level .

In the above code, we have a series of if-elif statements for assigning a string to the level variable. Let’s run the code to see what happens:

The error occurs because we input a score equal to 40 . The conditional statements in the function do not account for a value below 55 , therefore when we call the calculate_level function, Python will attempt to return level without any value assigned to it.

We can solve this error by completing the set of conditions with an else statement. The else statement will provide an assignment to level for all scores lower than 55 . Let’s look at the revised code:

In the above code, all scores below 55 are given the beginner level. Let’s run the code to see what happens:

We can also create a global variable level and then use the global keyword inside calculate_level . Using the global keyword will ensure that the variable is available in the local scope of the calculate_level function. Let’s look at the revised code.

In the above code, we put the global statement inside the function and at the beginning. Note that the “default” value of level is beginner and we do not include the else statement in the function. Let’s run the code to see the result:

Congratulations on reading to the end of this tutorial! The UnboundLocalError: local variable referenced before assignment occurs when you try to reference a local variable before assigning a value to it. Preferably, you can solve this error by passing parameters to your function. Alternatively, you can use the global keyword.

If you have if-elif statements in your code where you assign a value to a local variable and do not account for all outcomes, you may encounter this error. In which case, you must include an else statement to account for the missing outcome.

For further reading on Python code blocks and structure, go to the article: How to Solve Python IndentationError: unindent does not match any outer indentation level .

Go to the  online courses page on Python  to learn more about Python for data science and machine learning.

Have fun and happy researching!

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[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!

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Demystifying Python‘s UnboundLocalError

As a Python developer, chances are you have encountered the confusing UnboundLocalError at some point. This error happens when you try to access a local variable in a function or method, but Python cannot find the variable defined in the local scope.

The UnboundLocalError can be frustrating for beginners, leaving them scratching their heads wondering what went wrong. However, once you understand the causes behind this error, it becomes much easier to prevent and fix.

In this comprehensive guide, we will demystify Python‘s UnboundLocalError by covering:

• What triggers the UnboundLocalError and common causes
• Actionable solutions and best practices to avoid the error
• Real code examples to demonstrate the error and fixes
• Additional tips and tricks for using local and global variables correctly

So let‘s get started unraveling the UnboundLocalError !

What Triggers An UnboundLocalError in Python?

In Python, variables defined within a function are considered local variables, meaning they only exist within the local scope of that function. Trying to use a local variable that hasn‘t been defined causes an UnboundLocalError .

Here is an example that demonstrates this:

Running this code produces:

Python is telling us that local_var doesn‘t exist within the local scope of my_func() . We tried to print local_var before defining it, resulting in the error.

In addition to accessing undefined local variables, there are a couple other common triggers for UnboundLocalError in Python:

Attempting to modify a global variable:

This will produce an UnboundLocalError on global_var within my_func() , even though global_var is defined globally. Python sees we are assigning to global_var inside the function and treats it as a local variable, but global_var isn‘t defined locally which causes issues.

Closing over a variable in nested functions:

This example also causes an UnboundLocalError , but on x inside inner_func() . Python doesn‘t automatically make x from outer_func() available to nested functions. We have to handle closing over variables more carefully as we‘ll discuss next.

There are solutions for both of these scenarios, but first it‘s useful to learn more about Python‘s variable scoping rules that are the root causes of these issues.

Understanding Python‘s Variable Scope Rules

The key to avoiding UnboundLocalError is learning Python‘s LEGB rule for variable scoping along with best practices for using global and nested local variables.

Python‘s LEGB Variable Scope Rule

Python resolves variables in this order:

• L : Local — Variables defined within function or method
• E : Enclosing — Variables defined in any enclosing functions
• G : Global — Variables defined at the uppermost level
• B : Built-in — Built-in functions and exceptions

So when accessing variables:

• Python first checks for a local variable defined within the current function or method
• Then searches any enclosing functions from inner to outer
• If no local variable is found, searches the global variables
• And lastly checks built-ins

Many UnboundLocalError cases happen due to confusion between local vs global variables in Python. Keeping these precedence rules in mind helps avoid issues.

Best Practices for Local and Global Variables

Here are some key best practices for properly using local and global variables in Python:

• Declare globals needed inside functions: If wanting to modify globals within a function, declare them using global to avoid confusion
• Access values, modify locally: Avoid modifying globals from within functions when possible. Access values, make local copy to modify
• Return values from functions rather than relying on side effects
• Use parameters and returns instead of external state: Rely on parameter passing between functions rather than globals

Keeping these variable usage best practices helps avoid confusing Python‘s LEGB scoping rules, preventing many UnboundLocalError cases.

Now let‘s look at solutions for the specific variable scoping scenarios we saw earlier.

Solutions and Best Practices for Avoiding UnboundLocalError

While sound variable usage recommendations help avoid UnboundLocalError in general, you may still run into tricky situations where Python‘s scoping trips you up. Here are some pro tips for resolving two common specific cases.

Modifying Global Variables

Earlier we saw code for modifying a global variable within a function that failed:

The issue is Python treats global_var as a local variable since we assign to it. The solution is using the global keyword:

By using global we tell Python global_var is defined externally, fixing the scoping issue.

A better practice is avoiding modifications entirely though:

Retrieve global values into local variables, modify locally, return new copies. This results in cleaner code.

Closing Over Variables in Nested Functions

We also saw an UnboundLocalError example with nested functions:

The problem here is that Python doesn‘t automatically make outer_func() ‘s x variable available to reference within inner_func() .

We can fix this by manually closing over x :

Here we pass x as a parameter to make the binding available in inner_func() ‘s scope.

An alternative approach is using a nonlocal declaration:

The nonlocal keyword tells Python to resolve x as a variable from an enclosing function scope. Similar to global but for outer function scopes rather than global.

For better encapsulation, instead of relying on external state through closures, have inner functions accept parameters and return updated values explicitly.

Additional Tips For Properly Using Variables in Python

Here are a few closing tips when leveraging local, enclosing, and global variables in Python:

Use descriptive variable names: Well-named local, global, and nonlocal variables help clarify your code and reduce confusion that leads to UnboundLocalError .

Limit variable scope when reasonable: Keep variables local to the narrowest reasonable scope, and lift to a wider scope only when needed. This reduces likelihood of accidental overlap.

Modularize code using functions: Break code into discrete reusable functions rather than relying on shared global state. Functions encourage scoping variables appropriately.

Leverage linters and type checkers : Tools like pylance , mypy , pylint can help analyze code and flag issues early around improper variable usage which could lead to UnboundLocalError .

Putting some discipline around properly scoping variables this way will help you write clean, modular Python code and squashes UnboundLocalError for good!

Putting It All Together: Fixing a Real UnboundLocalError Example

Let‘s bring all our new knowledge together to fix a real UnboundLocalError case.

Here is some flawed code for a simple ecommerce store application:

This results in:

Why the error? Within add_item() Python sees we are modifying cart and treats it as a local variable. But cart isn‘t defined locally in the function causing failure.

We could fix this by using global to declare cart :

However, relying on the shared global cart is poor practice and couples our functions. Instead, we refactor to explicitly pass cart state between functions:

By eliminating the global variable dependency we decouple things, encourage better scoping practices, and resolve the error.

Putting the major points we covered to work by properly scoping variables and passing state explicitly makes debugging UnboundLocalErrors much easier!

Key Takeaways: Effortlessly Avoiding UnboundLocalError

We have covered a lot of ground demystifying Python‘s UnboundLocalError . Let‘s review the key takeaways:

• UnboundLocalError happens when trying to access undefined local variables within a function
• Also occurs attempting to modify globals or closing over external state improperly
• Understanding Python‘s LEGB variable precedence rule helps avoid issues
• Use best practices like avoiding globals, modularizing appropriately, descriptive names, tools
• Declare global and nonlocal where needed for specific situations
• Pass data between functions explicitly rather than relying on shared state

Learning these variable scoping best practices and solutions for common UnboundLocalError cases will help you become a debugging pro in no time. You are now equipped to leverage local, global, and nested function variables with confidence!

With your new skills, the confusing UnboundLocalError will transform from a source of frustration to an easy fix. So get out there, write some Python code, and put your new understanding to work in squashing this error for good!

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Python 3: UnboundLocalError: local variable referenced before assignment

This error occurs when you are trying to access a variable before it has been assigned a value. Here is an example of a code snippet that would raise this error:

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.

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Fixing Python UnboundLocalError: Local Variable ‘x’ Accessed Before Assignment

Understanding unboundlocalerror.

The UnboundLocalError in Python occurs when a function tries to access a local variable before it has been assigned a value. Variables in Python have scope that defines their level of visibility throughout the code: global scope, local scope, and nonlocal (in nested functions) scope. This error typically surfaces when using a variable that has not been initialized in the current function’s scope or when an attempt is made to modify a global variable without proper declaration.

Solutions for the Problem

To fix an UnboundLocalError, you need to identify the scope of the problematic variable and ensure it is correctly used within that scope.

Method 1: Initializing the Variable

Make sure to initialize the variable within the function before using it. This is often the simplest fix.

Method 2: Using Global Variables

If you intend to use a global variable and modify its value within a function, you must declare it as global before you use it.

Method 3: Using Nonlocal Variables

If the variable is defined in an outer function and you want to modify it within a nested function, use the nonlocal keyword.

That’s it. Happy coding!

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UndboundLocalError: local variable referenced before assignment

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

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

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

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

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

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

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

Thanks again

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

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

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Fixing UnboundLocalError: local variable 'name' referenced before assignment in python

Reading a text file with names and birthdays in a function printBook(), returning name, birthday and referencing it later on in another function I get:

UnboundLocalError: local variable 'name' referenced before assignment

• 4 What happens if there are no lines in the file? –  Chris Commented Feb 17, 2020 at 12:51
• What do you want it to return? The last name, month and date? –  Simon Crane Commented Feb 17, 2020 at 12:52
• @chris i've got a specific file i'm working on there so didn't think of preventing this –  stafino Commented Feb 17, 2020 at 13:22
• @SimonCrane yes return name, month, date that could be used later in other functions –  stafino Commented Feb 17, 2020 at 13:22

2 Answers 2

Although there is already an accepted answer, I believe it does not deal with the real issue in the question. Elton's answer will prevent the exception from being thrown, but will not render the desired output, from what I understood of the question.

The Real Issue

The issue here is that when calling the printBook() function inside the givenMonthPrintsElse() function, there is nothing left to read in the file, therefore leading to no line objects, which skips the for loop, returning three unassigned variables ( name , month and date ).

This happens because the with ... open statement is misplaced. Once the file object is read on the printBook() function, the cursor is at EOF (end of file), thus nothing left to be read. Calling again another function that requires the output of printBook() under the same with ... open statement results in empty output. You can confirm this by commenting out the first call to printBook() , and you should get the desired output.

Solution (okay)

The simplest way I can think that solves it is moving the with ... open statement inside the printBook() function, like this:

An important remark here is that, although it solves the issue, I suppose this is not the real intent of the OP, since it only allows to use the last line of the file (even though it reads through all of them). That is because name , month and date are overwritten at each iteration/line of the file, and only the last ones are returned by printBook() .

Solution (better)

So, as following, a simple structure allows to access any line of the file:

This way, in the book object you have a list of the content of all the lines, as strings, that afterwards you can split and process as desired. The printNameDate() function uses a list of strings (as the one returned by fileToList() and a line number (named entry )) to get the desired output.

You're declaring the name variable inside the for loop and accessing outside (in the return statement).

The easiest way to solve this is by declaring the variables before the for loop, so the printBook function would be something like:

PS: As @Chris pointed out, the problem isn't just because you're accessing the variables outside the loop (python accepts it), but because when the loop doesn't run, your variables aren't declared.

• 2 Actually, declaring the variables inside the loop works fine. Loops don't create a new scope. The problem is that if the loop doesn't run , the variables don't get created. –  Chris Commented Feb 17, 2020 at 13:23

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UnboundLocalError: local variable 'eval_manager' referenced before assignment #125

1790387702 commented Aug 24, 2021

tuntun-sd commented Aug 24, 2021

Sorry, something went wrong.

DerrickXuNu commented Aug 24, 2021

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