unboundlocalerror local variable 'valid' referenced before assignment

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!

Suf is a senior advisor in data science with deep expertise in Natural Language Processing, Complex Networks, and Anomaly Detection. Formerly a postdoctoral research fellow, he applied advanced physics techniques to tackle real-world, data-heavy industry challenges. Before that, he was a particle physicist at the ATLAS Experiment of the Large Hadron Collider. Now, he’s focused on bringing more fun and curiosity to the world of science and research online.

• Suf https://researchdatapod.com/author/soofyserial/ How to Solve Python TypeError: ‘str’ object does not support item assignment
• Suf https://researchdatapod.com/author/soofyserial/ How to Solve Python AttributeError: 'list' object has no attribute 'strip'
• Suf https://researchdatapod.com/author/soofyserial/ Python TypeError: list indices must be integers or slices, not str Solution
• Suf https://researchdatapod.com/author/soofyserial/ How to Sort a Dictionary by Value in Python

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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

Borislav Hadzhiev

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Copyright © 2024 Borislav Hadzhiev

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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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[SOLVED] Local Variable Referenced Before Assignment

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

Why Does This Error Occur?

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

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

Local Variable Declarations vs. Global Variable Declarations

Local Variable Referenced Before Assignment Error with Explanation

Try these examples yourself using our Online Compiler.

Let’s look at the following function:

Explanation

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

Using Global Variables

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

Create Functions that Take in Parameters

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

UnboundLocalError: local variable ‘DISTRO_NAME’

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

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

Try and update your Anaconda Navigator with the following command.

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

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

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

DISTRO_NAME = None

Save the file and re-launch Anaconda Navigator.

DJANGO – Local Variable Referenced Before Assignment [Form]

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

Upon running you get the following error:

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

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

Why does the error occur?

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

Local variable Referenced before assignment but it is global

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

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

Here’s an example to help illustrate the problem:

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

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

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

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

Local variable ‘version’ referenced before assignment ubuntu-drivers

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

Here, p_name means package name.

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

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

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

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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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[Bug]: UnboundLocalError: local variable 'h' referenced before assignment #13761

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How to Solve Error - Local Variable Referenced Before Assignment in Python

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• How to Solve Error - Local Variable …

Check the Variable Scope to Fix the local variable referenced before assignment Error in Python

Initialize the variable before use to fix the local variable referenced before assignment error in python, use conditional assignment to fix the local variable referenced before assignment error in python.

This article delves into various strategies to resolve the common local variable referenced before assignment error. By exploring methods such as checking variable scope, initializing variables before use, conditional assignments, and more, we aim to equip both novice and seasoned programmers with practical solutions.

Each method is dissected with examples, demonstrating how subtle changes in code can prevent this frequent error, enhancing the robustness and readability of your Python projects.

The local variable referenced before assignment occurs when some variable is referenced before assignment within a function’s body. The error usually occurs when the code is trying to access the global variable.

The primary purpose of managing variable scope is to ensure that variables are accessible where they are needed while maintaining code modularity and preventing unexpected modifications to global variables.

We can declare the variable as global using the global keyword in Python. Once the variable is declared global, the program can access the variable within a function, and no error will occur.

The below example code demonstrates the code scenario where the program will end up with the local variable referenced before assignment error.

In this example, my_var is a global variable. Inside update_var , we attempt to modify it without declaring its scope, leading to the Local Variable Referenced Before Assignment error.

We need to declare the my_var variable as global using the global keyword to resolve this error. The below example code demonstrates how the error can be resolved using the global keyword in the above code scenario.

In the corrected code, we use the global keyword to inform Python that my_var references the global variable.

When we first print my_var , it displays the original value from the global scope.

After assigning a new value to my_var , it updates the global variable, not a local one. This way, we effectively tell Python the scope of our variable, thus avoiding any conflicts between local and global variables with the same name.

Ensure that the variable is initialized with some value before using it. This can be done by assigning a default value to the variable at the beginning of the function or code block.

The main purpose of initializing variables before use is to ensure that they have a defined state before any operations are performed on them. This practice is not only crucial for avoiding the aforementioned error but also promotes writing clear and predictable code, which is essential in both simple scripts and complex applications.

In this example, the variable total is used in the function calculate_total without prior initialization, leading to the Local Variable Referenced Before Assignment error. The below example code demonstrates how the error can be resolved in the above code scenario.

In our corrected code, we initialize the variable total with 0 before using it in the loop. This ensures that when we start adding item values to total , it already has a defined state (in this case, 0).

This initialization is crucial because it provides a starting point for accumulation within the loop. Without this step, Python does not know the initial state of total , leading to the error.

Conditional assignment allows variables to be assigned values based on certain conditions or logical expressions. This method is particularly useful when a variable’s value depends on certain prerequisites or states, ensuring that a variable is always initialized before it’s used, thereby avoiding the common error.

In this example, message is only assigned within the if and elif blocks. If neither condition is met (as with guest ), the variable message remains uninitialized, leading to the Local Variable Referenced Before Assignment error when trying to print it.

The below example code demonstrates how the error can be resolved in the above code scenario.

In the revised code, we’ve included an else statement as part of our conditional logic. This guarantees that no matter what value user_type holds, the variable message will be assigned some value before it is used in the print function.

This conditional assignment ensures that the message is always initialized, thereby eliminating the possibility of encountering the Local Variable Referenced Before Assignment error.

Throughout this article, we have explored multiple approaches to address the Local Variable Referenced Before Assignment error in Python. From the nuances of variable scope to the effectiveness of initializations and conditional assignments, these strategies are instrumental in developing error-free code.

The key takeaway is the importance of understanding variable scope and initialization in Python. By applying these methods appropriately, programmers can not only resolve this specific error but also enhance the overall quality and maintainability of their code, making their programming journey smoother and more rewarding.

UnboundLocalError: local variable 'values' referenced before assignment in lr_scheduler

This my code:

how to solve this error?

Could you post a code snippet to reproduce this issue, please? This dummy example runs fine:

Need any other snippet code? I’m not sure how much complete code you need

A minimal and executable code snippet would be great. Could you try to remove unnecessary functions and use some random inputs, so that we can reproduce this issue locally?

I’m sorry for my slow response。This code might suit your needs。

You can see that the first call to scheduler. Step should have been fault-free because it printed the following print statement, as well as eval for the model. But an error should have occurred on the second call

I forgot the code that came out and this is the following code

Now it turns out that if you use annotated code, you get an error, while unannotated code works fine

I really don’t know why, is it the data problem that caused the error

Thanks for the code so far. Could you also post the code you are using to initialize the model, optimizer, and scheduler? Also, could you try to run the code on the CPU only and check, if you see the same error? If not, could you rerun the GPU code using CUDA_LAUNCH_BLOCKING=1 python script.py args and post the stack trace again?

This is my code:

It was very embarrassing that I could not debug with CPU because of the machine, but if I used GPU to debug, the error result did not change

Could you call scheudler.get_lr() before the error is thrown and check the return value, please?

I am very sorry for replying to you a few days later, because I am a sophomore student in university and I have a lot of things to do recently, so I didn’t deal with this problem for a few days. As you requested, I added this line of code. The problem is that it returned the value successfully without any problems during the first epoch. But after the second epoch, it reported an error

Are you recreating or manipulating the scheduler or optimizer in each epoch somehow?

29/5000 Thank you for answering my question so patiently. My code should not have this problem。 If I use following code,the error will not appear( it will appear when using the annotating code):

Here is my complete training code(When you put scheduler. Step () into each batch iteration,error will apear)：

I had the same error that I think has been fixed.

In the end it seems like the number of epochs you had mentioned in your scheduler was less than the number of epochs you tried training for. I went into %debug in the notebook and tried calling self.get_lr() as suggested. I got this message: *** ValueError: Tried to step 3752 times. The specified number of total steps is 3750

Then with some basic math and a lot of code search I realised that I had specified 5 epochs in my scheduler but called for 10 epochs in my fit function.

Hope this helps.

There is an error with total_steps.i am also getting same error but i rectified it

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

Hello,I had the same error that I can’t solve it. I want to ask you some question about it.Thank you. What is the ‘The specified number of total steps is 3750’? How to change the number of steps? Thank you.

Hi make sure that your dataloader and the scheduler have the same number of iterations. If I remember correctly I got this error when using the OneCycle LR scheduler which needs you to specify the max number of steps as init parameter. Hope this helps! If this isn’t the error you have, then please provide code and try to see what your scheduler.get_lr() method returns.

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UnboundLocalError: local variable referenced before assignment

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

which isn't clear to me. Any suggestions?

• arcgis-10.1
• unboundlocalerror

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

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

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"UnboundLocalError: local variable 'os' referenced before assignment"

OS: macOS Sonoma 14.0 (on an M2 chip) PsychoPy version: 2023.2.3 Standard Standalone? (y/n) y What are you trying to achieve?: Upgrade an experiment from 2021.1.4 to 2023.2.3

Context: Our lab computers were updated and 2021.1.4 was no longer working (tasks would not start). I think it had something to do with losing support for python 2.X, but I decided to just move forward and upgrade versions. Note: I didn’t go straight to 2023.2.3, but tried some other versions along the way.

What did you try to make it work?: Installed v2023.2.3, updated “use version” to 2023.2.3 Other details: window backend = pyglet, audio library = ptb (but experiment has no sound anyway), keyboard backend = PsychToolbox

What specifically went wrong when you tried that?: I got an: "UnboundLocalError: local variable 'os' referenced before assignment" …referring to line 301 in the generated python script which uses os.chdir(_thisDir) early on in the “run” fuction. According to posts on this forum, an UnboundLocalError is usually due to a problematic conditions file. However, an error with the os package in the run function seems to be a deeper problem. Have I totally corrupted the experiment by changing versions?

(somewhat redacted) error output:

Seems like it could be something like this scoping issue , but I have no idea how the py code got this way.

What happens if use version is blank?

Update: if I set useVersion back to v2022.1.0 (one of the previous versions this experiment passed through), the error does not occur and I am able to run the experiment normally. However, this version is on the list of versions to avoid … so I’d rather avoid it if possible.

Hi, I came into the same problem today. And I’m completely new with Psychopy, so you may not find my solution useful given that you may have already tried it. But for other users as new as me, we should put code component before the other stimuli component (e.g., text, image). Psychopy needs to run the code before it runs the stimuli.

I think this will depend on whether you need the results of that code component to display other components in that routine. There are other situations where a code component would need to come at the end of a routine.

My friend was having the same issue as you. We have been able to solve it. The problem is exactly the scoping issue you were referring to.

The problem is this: In the version 2023.2.3, when you build the project, generated code already imports os in global scope. However, unlike some of the older versions (2021, 2022) all the code related to the routines are under a function run(). At the beginning of this run() function, os module is used with the function os.chdir(). Crucial part is that, this call is being made before any other routine. The problem arises when in one of your routines, you write the statement “import os”. When that happens, the scope issue arises because os is already defined globally, and you are defining it locally, but you already tried to access it before you define it locally. So the exact situation in stackoverflow post happens.

Why wasn’t it happening in older versions? Because in those versions, when you build the project, all the code related to your routines are still under the global scope, unlike newer versions where its under the local scope of run() function.

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UnboundLocalError: local variable <function> referenced before assignment [duplicate]

I have read similar questions on this topic such as from here , here , and here , and others on scope, but I still cannot understand why answers do not explain these test cases.

First see this:

So then this must work:

Returns: UnboundLocalError: local variable 'take_sum' referenced before assignment

Question 1 : How is take_sum referenced before assignment if it exists in the global scope and the if-statement in main evalutes to False?

So let's add an else clause to the if-statement:

Returns: SyntaxError: name 'take_sum' is assigned to before global declaration

Question 2 : But how is this possible if the error from TEST 1 says take_sum was referenced but NOT assigned?

Subsequently, switching the clauses in the if-statement works:

Question 3 : Why does switching clauses (compared to TEST 2) work?

• "How is take_sum referenced before assignment if it exists in the global scope and the if-statement in main evalutes to False?" In the same way as in the first link you found: the existence in global scope is not relevant , because there is assignment code , the presence of which is checked ahead of time . It does not matter that the conditional code doesn't run, just like it didn't matter, for that question, that the assignment happens after the usage. –  Karl Knechtel Commented Feb 6, 2023 at 12:53

2 Answers 2

Whether a name used in a function is a global variable or a local one is determined at compile time, not at run time. Your functions that cause exceptions are trying to have it both ways, to either access a global, or provide their own local variable replacement, but Python's scoping rules don't allow that. It needs to be local or global, and can't be in a nebulous either/or state.

In your Test 1, the function raises an exception because the compiler saw that the code could assign to take_sum as a local variable, and so it makes all the references to take_sum in the code be local. You can no longer look up the global variable take_sum in the normal way once that determination has been made.

A global statement is in effect a compiler directive to change the assumption that an assignment makes a variable local. Subsequent assignments will be made globally, not locally. It's not something that executes at runtime, which is why your other two test cases are so confusing to you.

Test 2 fails because you're trying to tell the compiler that take_sum is a global after it has already seen some of your code make a local assignment to that name. In Test 3, the global statement comes first, so it makes the assignment (in the other branch!) assign to a global variable. It doesn't actually matter that the global statement was in a different branch than the assignment, the compiler interprets the global statement at compile time, not at runtime when the conditional logic of the if s and elif s gets handled.

It might help your understanding of what is going on to disassemble some of the main functions you've written using the dis.dis function in the standard library. You'll see that there are two different sets of bytecodes used for loading and storing of variables, LOAD_GLOBAL / STORE_GLOBAL for global variables (used in all your functions to get names like print and globals ), and LOAD_FAST / STORE_FAST which are used for local variables (like a , b and c in take_sum ). The compiler behavior I talked about above boils down to which bytecode it chooses for each lookup or assignment.

If I rename the main function in Test 1 to test1 , here's what I get when I disassemble it:

Notice that the lookup of take_sum on line 5 is on byte 20 in the bytecode, where it uses LOAD_FAST . This is the bytecode that causes the UnboundLocalError , since there has been no local assigned if the global function exists.

Now, lets look at Test 3:

This time the lookup of take_sum happens on bytecode 40, and it's a LOAD_GLOBAL (which succeeds since there is a global variable of that name).

Take this example

You're pre-defining a variable reference to be global

Output: <function main.<locals>.<lambda> at 0x7f08c2c54820>

Remove, the not , 'take_sum' is not defined , as expected

Now, create a global one

Output this time is the outer one.

Remove the not , then the local one

So, if you try to make global take_sum in the else statement, then as the error says, assigned to before global declaration , and seems like the global gets evaluated without respect for the conditional

• That's where I don't follow: the if-condition in TEST 1 evaluates to False, so there should be no assignment, and when the cursor returns to the print statement, it should print the location of take_sum in memory like in TEST 3... What am I missing –  DSH Commented Aug 12, 2021 at 20:44
• Notice that you get the same error when using if False . I suspect you are trying to do something very different than overriding a global function with a local one? –  OneCricketeer Commented Aug 12, 2021 at 20:48

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