variable referenced before assignment error in python

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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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How to Fix Local Variable Referenced Before Assignment Error in Python

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.

Fixing local variable referenced before assignment error

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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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.

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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.

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Local Variable Referenced Before Assignment in Python

• Python How-To's
• Local Variable Referenced Before …

This tutorial will explain why the error local variable referenced before assignment occurs and how it can be resolved.

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. As the global variables have global scope and can be accessed from anywhere within the program, the user usually tries to use the global variable within a function.

In Python, we do not have to declare or initialized the variable before using it; a variable is always considered local by default. Therefore when the program tries to access the global variable within a function without specifying it as global, the code will return the local variable referenced before assignment error, since the variable being referenced is considered a local variable.

the Solution of local variable referenced before assignment Error in Python

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.

We need to declare the count 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.

Local variable referenced before assignment in Python

Last updated: Feb 17, 2023 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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Web Developer

Copyright © 2023 Borislav Hadzhiev

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How can I fix a ‘local variable referenced before assignment’ error in Python?

• Last edited: March 17, 2023
• Programming Advice , Python

Table of Contents

Are you struggling with the ‘local variable referenced before assignment’ error in Python? This blog post will provide a few strategies to help you fix this issue. We’ll discuss how to make sure that you’ve assigned a value to the variable, use the ‘global’ keyword if necessary, add input parameters for functions, and initialize variables before loops or conditionals. By following these steps, you can resolve this error quickly and easily.

Programming Guide

The ‘local variable referenced before assignment’ error in Python occurs when you’re trying to use a local variable in a function before it has been assigned a value. To fix this error, there are a few strategies that you can follow:

1. Make sure you’ve assigned a value to the variable before referencing it.

2. If the variable is meant to be a global variable, use the ‘global’ keyword to indicate that you’re referring to the global variable and not creating a local variable.

3. If the variable is meant to be an input parameter for the function, ensure that you’ve added it as a parameter in the function definition, and pass the value when calling the function.

4. If you’re encountering this error in a loop or conditional, make sure the variable is initialized before the loop or conditional.

By ensuring that you’ve properly assigned a value to the variable in question and correctly identified its intended use (local, global, or input parameter), you can resolve the error.

The strategies outlined in this blog post can help you resolve the ‘local variable referenced before assignment’ error in Python. To fix this issue, make sure that you’ve assigned a value to the variable before referencing it, use the ‘global’ keyword if necessary, add input parameters for functions when needed and initialize variables prior to loops or conditionals.

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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 UnboundLocalError: local variable 'x' referenced before assignment in Python

by Nathan Sebhastian

Posted on May 26, 2023

One error you might encounter when running Python code is:

This error commonly occurs when you reference a variable inside a function without first assigning it a value.

You could also see this error when you forget to pass the variable as an argument to your function.

Let me show you an example that causes this error and how I fix it in practice.

How to reproduce this error

Suppose you have a variable called name declared in your Python code as follows:

Next, you created a function that uses the name variable as shown below:

When you execute the code above, you’ll get this error:

This error occurs because you both assign and reference a variable called name inside the function.

Python thinks you’re trying to assign the local variable name to name , which is not the case here because the original name variable we declared is a global variable.

How to fix this error

To resolve this error, you can change the variable’s name inside the function to something else. For example, name_with_title should work:

As an alternative, you can specify a name parameter in the greet() function to indicate that you require a variable to be passed to the function.

When calling the function, you need to pass a variable as follows:

This code allows Python to know that you intend to use the name variable which is passed as an argument to the function as part of the newly declared name variable.

Still, I would say that you need to use a different name when declaring a variable inside the function. Using the same name might confuse you in the future.

Here’s the best solution to the error:

Now it’s clear that we’re using the name variable given to the function as part of the value assigned to name_with_title . Way to go!

The UnboundLocalError: local variable 'x' referenced before assignment occurs when you reference a variable inside a function before declaring that variable.

To resolve this error, you need to use a different variable name when referencing the existing variable, or you can also specify a parameter for the function.

I hope this tutorial is useful. See you in other tutorials.

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Python local variable referenced before assignment Solution

Posted in PROGRAMMING LANGUAGE /   PYTHON

Vinay Khatri Last updated on September 21, 2023

Table of Content

The most common error you may encounter while working with Python and user-defined functions is UnboundLocalError: local variable 'name' referenced before assignment . The reason for this error occurrence is we try to access a variable before it has been assigned in the local scope or context of the function.

In this Python guide, we will walk through this Python error and discuss why this error occurs and how to solve it. We will also look at some examples so that you can get a better idea about this Python error.

The error: UnboundLocalError: local variable referenced before assignment?

The error statement UnboundLocalError: local variable referenced before assignment is divided into two statements

• UnboundLocalError: It is a Python error type that occurs when we mishandle the Python local variables.
• the local variable referenced before assignment : This is the error message that tells that we are trying to access or assign a new value to a Python local variable before its initialization.

Error Reasons

There are two main reasons why your Python program is showing this error.

• You are trying to create a new local variable with the same name as the global variable and using the value of the global variable.
• Or, have created a local variable inside a function using the if..else statement, and it never gets assigned, and you are accessing it.

This is the major scenario where the Python learner commits a mistake. When they try to create a new local variable with the same name as the global variable after accessing the global variable into the function.

Once you have accessed the global variables into a Python function, you can not create a local variable with the same name. If you do, you will receive the UnboundLocalError: local variable referenced before assignment error.

Break the code

In the above example, we are getting the error because we are trying to create a new local variable name and accessing the global variable name value using the statement name = name + lname at line 4.

When Python executes that statement, it gets confused between the local and global variables and treats both variables as local variables. By that time, Python does not find the value of the right hand name so it throws the error.

The solution for this example is very simple. Although we can access the value of a global variable inside a function, we can not alter it. In case you want to access the global variable and change its value, you can use the Python global keyword.

Onces we have accessed the gloable variable inside a function wihtout using global keyword, we can not create a new local variable by the same name.

Another common reason why we receive this error is when we create a local variable inside a Python if..else conditional statement, and it never gets initialized because the condition was False.

The age value is 12 , which means the statement inside the if age>18: condition block did not execute. That leads to no assignment value to the adult variable, but at the backend, when Python executes its program line by line and initializes the variable adult but did not assign any value to it. So when we accessed the variable at line 7, it threw the error.

When we access a variable inside a local scope, we need to make sure that we initialize and assign a value to it before accessing it. If we are creating a new variable inside the if statement, we also need to make sure that there must be an else statement that also assigns the value to the variable if the condition is False.

In our above example, the value of adult only get assigned when the condition is true, so all we need to do is create an else statement that also assigns a value to the variable adult when the condition is False.

In this Python tutorial, we discussed one of the most common Python functions error local variable referenced before assignment . The error raises when we try to access a local variable before its assignment inside a function. We often encounter this error when we try to access a global variable with the same name as the local variable or create a local variable inside a function that never gets assigned.

If you are still getting this error in your Python program, please comment down your code and query in the comment section, and we will try to debug it for you.

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Vinay Khatri I am a Full Stack Developer with a Bachelor's Degree in Computer Science, who also loves to write technical articles that can help fellow developers.

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UnboundLocalError: local variable 'pred' referenced before assignment

I'm try to classify AD diseases. When I trying to calculate the probability of a random image from test data it's showing this error. Kindly help.

This is a function that fetches a random image and displays a pie chart showing the probability distribution of which target value the image belongs to, represented as percentages. In this way, it will be seen which class the model gives the highest probability to

• deep-learning
• conv-neural-network

• It looks like the line: pred = model.predict(... may not execute since it is inside a for loop. If that happens (ie the for loop has no items to iterate over), then the next line will result in the error reported. –  quamrana Sep 14 at 10:39
• The loop is never run, test_data.skip(5).take(1) is empty. –  Nikolaj Š. Sep 14 at 10:39
• Thank you! It worked after removing it. –  Since 1995 9 hours ago

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Local variable referenced before assignment in Python

In Python, while working with functions, you can encounter various types of errors. A common error when working with the functions is “ Local variable referenced before assignment ”. The stated error occurs when a local variable is referenced before being assigned any value.

This write-up will provide the possible reasons and the appropriate solutions to the error “Local variable referenced before assignment” with practical examples. The following aspects are discussed in this write-up in detail:

Reason: Reference a Local Variable

Solution 1: mark the variable globally, solution 2: using function parameter value, solution 3: using nonlocal keyword.

The main reason for the “ local variable referenced before assignment ” error in Python is using a variable that does not have local scope. This also means referencing a local variable without assigning it a value in a function.

The variable initialized inside the function will only be accessed inside the function, and these variables are known as local variables. To use variables in the entire program, variables must be initialized globally. The below example illustrates how the “ UnboundLocalError ” occurs in Python.

In the above snippet, the “ Student ” variable is not marked as global, so when it is accessed inside the function, the Python interpreter returns an error.

Note: We can access the outer variable inside the function, but when the new value is assigned to a variable, the “UnboundLocalError” appears on the screen.

To solve this error, we must mark the “ Student ” variable as a global variable using the keyword “ global ” inside the function.

Within the function, we can assign a new value to the student variable without any error. Let’s have a look at the below snippet for a detailed understanding:

In the above code, the local variable is marked as a “ global ” variable inside the function. We can easily reference the variable before assigning the variable in the program.

The above snippet proves that the global keyword resolves the “unboundLocalError”.

Passing a value as an argument to the function will also resolve the stated error. The function accepts the variable as an argument and uses the argument value inside the function. Let’s have a look at the given below code block for a better understanding:

In the above code, the variable is referenced before assigning the value inside the user-defined function. The program executes successfully without any errors because the variable is passed as a parameter value of the function.

The above output shows the value of the function when the function is accessed in the program without any “ Local Variable referenced ” error.

The “ nonlocal ” keyword is utilized in the program to assign a new value to a local variable of function in the nested function. Here is an example of code:

In the above code, the keyword “ nonlocal ” is used to mark the local variable of the outer function as nonlocal. After making the variable nonlocal, we can reference it before assigning a value without any error.

The above output shows the value of the inner function without any “ local variable referenced ” error in a program.

The “ Local variable referenced before assignment ” appears in Python due to assigning a value to a variable that does not have a local scope. To fix this error, the global keyword, return statement, and nonlocal nested function is used in Python script. The global keywords are used with variables to make it able to access inside and outside the function. The return statement is also used to return the variable’s new value back to function and display the result on the screen. This Python guide presented a detailed overview of the reason and solutions for the error “Local variable referenced before assignment” in Python.

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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. To clarify, a variable is assigned in a function, that variable is local. Because it is assumed that when you define a variable inside a function, you only need to access it inside that function.

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

An example of Local variable referenced before assignment

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

Firstly, we start by declaring two variables:

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

Finally, we call our function:

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

Let’s run our code of Local variable referenced before assignment and see what happens:

Here is an error!

The Solution of Local variable referenced before assignment

The code returns an error: Unboundlocalerror local variable referenced before assignment because we reference “letter” before we assign it.

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

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

Therefore, this problem of variable referenced before assignment could be solved in two ways. Firstly, we can add an  else  statement to our code. This ensures we declare “letter” before we try to return it:

Let’s try to run our code again:

Our code successfully prints out the student’s grade. This approach is good because it lets us keep “letter” in the local context. To clarify, we could even remove the “letter = “F”” statement from the top of our code because we do not use it in the global context.

Alternatively, we could use the “global” keyword to make our global keyword available in the local context in our  calculate_grade()  function:

We use the “global” keyword at the start of our function.

This keyword changes the scope of our variable to a global variable. This means the “return” statement will no longer treat “letter” like a local variable. Let’s run our code. Our code returns: F.

The code works successfully! Let’s try it using a different grade number by setting the value of “numerical” to a new number:

Our code returns: B.

Finally, we have fixed the local variable referenced before assignment error in the code.

To sum up, as you can see, the UnboundLocalError: local variable referenced before assignment error is raised when you try to assign a value to a local variable before it has been declared. Then, you can solve this error by ensuring that a local variable is declared before you assign it a value. Moreover, if a variable is declared globally that you want to access in a function, you can use the “global” keyword to change its value. In case you have any inquiry, let’s CONTACT US . With a lot of experience in Mobile app development services , we will surely solve it for you instantly.

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• fix python error , Local variable referenced before assignment , python , python dictionary , python error , python learning , UnboundLocalError , UnboundLocalError in Python

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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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UnboundLocalError: local variable referenced before assignment

Hi everyone,

This is a question I had when I was doing Jupyter Notebook practice. First, I defined the function get_y() and tested it:

Then I wrote another function:

When I run the testing codes:

The error shows:

UnboundLocalError: local variable ‘get_y’ referenced before assignment

I tried to fix it by replacing “get_y” variable to “y”, and it worked. I googled this error but I am still a little confused: I didn’t find my “get_y” variable referenced before. Is it because there is a conflict between variable name and the function name?

Where is it referenced? Where is it assigned? Use the trace back to help discover one of the endpoints, then trace back to the other.

An UnboundLocalError is raised when a local variable is referenced before it has been assigned. In most cases this will occur when trying to modify a local variable before it is actually assigned within the local scope. Python doesn’t have variable declarations, so it has to figure out the scope of variables itself. It does so by a simple rule: If there is an assignment to a variable inside a function, that variable is considered local.

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

Get variable name causing UnboundLocalError

When I invoke a def, I get the UnboundLocalError message. Is there a way using an except statement to ge the name of the variable causing the error? I have the following code, but it does not give me the name of the variable causing the error.

The .args of this UnboundLocalError will be a tuple that contains the complete error message string. For example:

However, the exact contents of the string are allowed to vary with the Python version.

More importantly, you should not be writing code in the first place where this is a concern. If you are trying to debug the code, then don’t catch exceptions that indicate a problem in the code ; exception handling is for problems that result from bad input to the program. If you are for example doing some strange thing with dynamically created variables, so that you can’t really be sure what variables you will have just by reading and debugging the code, then don’t do that ; instead, use dictionaries and lists to hold structured data.

This is way too complicated for the kind of task that you describe, and it gives the impression that you are just trying to look up code and use whatever seems to work, without understanding it. This is also a bad habit. (In particular, it makes no real sense to use stuff like sys.exc_info inside an ordinary exception handler; e already means the same object as exception_object in this code, and you can more easily get the type as type(e) and the traceback as e.__traceback__ . Normal efforts to log errors won’t care about the traceback, either; if you did care then you should normally just let the program crash and read it. As for the type, you should normally be specifying a particular exception type that you’re interested in with except anyway, like in my example - so you should already know the type.)

Many thanks for you response. The exception was there to catch an error from bad date and write to an error log.

I removed the exception code and got the name of the local variable that caused the issue. I did not intentionally use a variable before assignment.

I corrected my error in the code and the problem is solved.

Thanks again. YouR response is very much appreciated. Wallis

Introduction to Programming

Variables in python.

• The purpose of a variable is to store information within a program while it is running.
• A variable is a named storage location in computer memory. Use the name to access the value.
• To store a value in a variable, use the = operator (called the assignment operator).
• An = sign in Python is nothing like an equal sign in mathematics. Think of it more like an arrow going from right to left. The expression on the right is evaluated and then stored in the variable named on the left.
• For example, the line of code hourly_wage = 16.75 stores the value 16.75 in the variable called hourly_wage
• You can change the value of a variable with another assignment statement, such as hourly_wage = 17.25
• Every value has a type ( int for integers, float for decimals, str for text). In Python, when you store a value in a variable (with = ), that variable then automatically has a type. For example, after the above assignment, hourly_wage is of type float

Rules and conventions for naming variables in Python

• The first character must be a letter or an underscore. For now, stick to letters for the first character.
• The remaining characters must be letters, numbers or underscores.
• No spaces are allowed in variable names.
• Legal examples: _pressure , pull , x_y , r2d2
• Invalid examples, these are NOT legal variable names: 4th_dimension , %profit , x*y , four(4) , repo man
• In Python, it's a conventiion to use snake case to name variables. This means that we use all lower-case letters and we separate words in the variable name with underscores. Examples include age , x_coordinate , hourly_wage , user_password
• If the value stored in a variable is a true constant (in other words, its value will never change throughout the program), then we use all capital letters: COURSE_ENROLLMENT_LIMIT , MAX_PASSWORD_ATTEMPTS .
• For high quality code, it is crucial that you choose names for variables that are descriptive. The variable names must help the reader of your program to understand your intention.

Typical way we visualize variables

We usually draw variables by putting the value in a box, and labelling the box with the name of the variable:

Types of variables

Each variable has a name, a value, and a type. Types are necessary because different kinds of data are stored differently within the computer's memory. For now, we will learn three different types, for storing signed (positive or negative) whole numbers, signed decimals, and text.

Creating a variable with an assignment operator

A variable is created or declared when we assign a value to it using the assignment operator = . In Python, the code looks like this: variable_name = <value> .

Notice that the left hand side of an assignment must be a variable name. Non-example:

After creating a variable, you can change the value stored in a variable with another assignment operator at any time. This is called reassignment .

Finding out the type of a variable or value

The type() function in Python will return the type of either a variable or a value. Here's an example that shows how to use it:

The output of the above code will be:

Casting (changing the type) of a variable or value

You can change the type of a value (called “casting”) using the int() , float() and str() functions. For example:

• int(23.7) (truncates the float value 23.7 to the int value 23. This is different from rounding - the decimal part is discarded, regardless of whether it is larger or smaller than 0.5.
• float(23) (outputting the result will give 23.0 rather than 23)
• str(23) (converts the integer 23 to the text "23")
• int("23") (converts the string "23" into a numerical integer value 23)
• float("23") (converts the string "23" into a numerical decimal value 23.0)
• int("23.5") results in an error
• float("hello") results in an error

Doing arithmetic in Python

Here are the basic arithmetic operators in Python. In these examples, assume

An example of a use of the modulus operator is to determine if an integer is even or odd. Note that if x is an integer, then x%2 takes the value 0 or 1 . So x % 2 == 0 is True when x is even and false when x is odd.

Another example of integer division and modulus: When we divide 3 by 4, we get a quotient of 0 and a remainder of 3. So 3//4 results in 0 and 3%4 results in 3.

Warning note: In Python, ^ is not an exponent!

Order of operations

The order of operations in Python is similar to the order you are familiar with in math: parentheses, then exponentiation, then multiplication/division/modulus in order from left to right, then addition/subtraction in order from left to right.