function unboundlocalerror local variable referenced before assignment

[SOLVED] Local Variable Referenced Before Assignment

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

Why Does This Error Occur?

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

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

Local Variable Declarations vs. Global Variable Declarations

Local Variable Referenced Before Assignment Error with Explanation

Try these examples yourself using our Online Compiler.

Let’s look at the following function:

Explanation

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

Using Global Variables

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

Create Functions that Take in Parameters

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

UnboundLocalError: local variable ‘DISTRO_NAME’

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

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

Try and update your Anaconda Navigator with the following command.

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

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

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

DISTRO_NAME = None

Save the file and re-launch Anaconda Navigator.

DJANGO – Local Variable Referenced Before Assignment [Form]

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

Upon running you get the following error:

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

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

Why does the error occur?

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

Local variable Referenced before assignment but it is global

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

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

Here’s an example to help illustrate the problem:

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

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

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

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

Local variable ‘version’ referenced before assignment ubuntu-drivers

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

Here, p_name means package name.

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

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

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

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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

Web Developer

Copyright © 2024 Borislav Hadzhiev

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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How to fix UnboundLocalError: local variable 'x' referenced before assignment in Python

by Nathan Sebhastian

Posted on May 26, 2023

Reading time: 2 minutes

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

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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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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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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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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UnboundLocalError: local variable 'fig' referenced before assignment

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

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

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

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

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

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

@adamschroeder I updated it as a new topic

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

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

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

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

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

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

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

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

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

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

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• Python »
• 3.14.0a0 Documentation »
• 파이썬 언어 레퍼런스 »
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4.1. 프로그램의 구조 ¶

파이썬 프로그램은 코드 블록으로 만들어집니다. 블록 (block) 은 한 단위로 실행되는 한 조각의 파이썬 프로그램 텍스트입니다. 다음과 같은 것들이 블록입니다: 모듈, 함수 바디, 클래스 정의. 대화형으로 입력되는 각 명령은 블록입니다. 스크립트 파일(표준 입력을 통해 인터프리터로 제공되는 파일이나 인터프리터에 명령행 인자로 지정된 파일)은 코드 블록입니다. 스크립트 명령( -c 옵션으로 인터프리터 명령행에 지정된 명령)은 코드 블록입니다. -m 인자를 사용하여 명령 줄에서 최상위 수준 스크립트로 (모듈 __main__ 으로) 실행되는 모듈도 코드 블록입니다. 내장함수 eval() 과 exec() 로 전달되는 문자열 인자도 코드 블록입니다.

코드 블록은 실행 프레임 (execution frame) 에서 실행됩니다. 프레임은 몇몇 관리를 위한 정보(디버깅에 사용됩니다)를 포함하고, 코드 블록의 실행이 끝난 후에 어디서 어떻게 실행을 계속할 것인지를 결정합니다.

4.2. 이름과 연결(binding) ¶

4.2.1. 이름의 연결 ¶.

이름 (Names) 은 객체를 가리킵니다. 이름은 이름 연결 연산 때문에 만들어집니다.

The following constructs bind names:

formal parameters to functions,

class definitions,

function definitions,

assignment expressions,

targets that are identifiers if occurring in an assignment:

for loop header,

after as in a with statement, except clause, except* clause, or in the as-pattern in structural pattern matching,

in a capture pattern in structural pattern matching

import statements.

type statements.

type parameter lists .

The import statement of the form from ... import * binds all names defined in the imported module, except those beginning with an underscore. This form may only be used at the module level.

del 문에 나오는 대상 역시 이 목적에서 연결된 것으로 간주합니다(실제 의미가 이름을 연결 해제하는 것이기는 해도).

각 대입이나 임포트 문은 클래스나 함수 정의 때문에 정의되는 블록 내에 등장할 수 있고, 모듈 수준(최상위 코드 블록)에서 등장할 수도 있습니다.

만약 이름이 블록 내에서 연결되면, nonlocal 이나 global 로 선언되지 않는 이상, 그 블록의 지역 변수입니다. 만약 이름이 모듈 수준에서 연결되면, 전역 변수입니다. (모듈 코드 블록의 변수들 지역이면서 전역입니다.) 만약 변수가 코드 블록에서 사용되지만, 거기에서 정의되지 않았으면 자유 변수 (free variable) 입니다.

프로그램 텍스트에 등장하는 각각의 이름들은 다음에 나오는 이름 검색(name resolution) 규칙에 따라 확정되는 이름의 연결 (binding) 을 가리킵니다.

4.2.2. 이름의 검색(resolution) ¶

스코프 (scope) 는 블록 내에서 이름의 가시성(visibility)을 정의합니다. 지역 변수가 블록에서 정의되면, 그것의 스코프는 그 블록을 포함합니다. 만약 정의가 함수 블록에서 이루어지면, 포함된 블록이 그 이름에 대해 다른 결합을 만들지 않는 이상, 스코프는 정의하고 있는 것 안에 포함된 모든 블록으로 확대됩니다.

이름이 코드 블록 내에서 사용될 때, 가장 가깝게 둘러싸고 있는 스코프에 있는 것으로 검색됩니다. 코드 블록이 볼 수 있는 모든 스코프의 집합을 블록의 환경 (environment) 이라고 부릅니다.

이름이 어디에서도 발견되지 않으면 NameError 예외가 발생합니다. 만약 현재 스코프가 함수 스코프이고, 그 이름이 사용되는 시점에 아직 연결되지 않은 지역 변수면 UnboundLocalError 예외가 발생합니다. UnboundLocalError 는 NameError 의 서브 클래스입니다.

If a name binding operation occurs anywhere within a code block, all uses of the name within the block are treated as references to the current block. This can lead to errors when a name is used within a block before it is bound. This rule is subtle. Python lacks declarations and allows name binding operations to occur anywhere within a code block. The local variables of a code block can be determined by scanning the entire text of the block for name binding operations. See the FAQ entry on UnboundLocalError for examples.

If the global statement occurs within a block, all uses of the names specified in the statement refer to the bindings of those names in the top-level namespace. Names are resolved in the top-level namespace by searching the global namespace, i.e. the namespace of the module containing the code block, and the builtins namespace, the namespace of the module builtins . The global namespace is searched first. If the names are not found there, the builtins namespace is searched next. If the names are also not found in the builtins namespace, new variables are created in the global namespace. The global statement must precede all uses of the listed names.

global 문은 같은 블록의 이름 연결 연산과 같은 스코프를 갖습니다. 자유 변수의 경우 가장 가까이서 둘러싸는 스코프가 global 문을 포함한다면, 그 자유 변수는 전역으로 취급됩니다.

The nonlocal statement causes corresponding names to refer to previously bound variables in the nearest enclosing function scope. SyntaxError is raised at compile time if the given name does not exist in any enclosing function scope. Type parameters cannot be rebound with the nonlocal statement.

모듈의 이름 공간은 모듈이 처음 임포트될 때 자동으로 만들어집니다. 스크립트의 메인 모듈은 항상 __main__ 이라고 불립니다.

Class definition blocks and arguments to exec() and eval() are special in the context of name resolution. A class definition is an executable statement that may use and define names. These references follow the normal rules for name resolution with an exception that unbound local variables are looked up in the global namespace. The namespace of the class definition becomes the attribute dictionary of the class. The scope of names defined in a class block is limited to the class block; it does not extend to the code blocks of methods. This includes comprehensions and generator expressions, but it does not include annotation scopes , which have access to their enclosing class scopes. This means that the following will fail:

However, the following will succeed:

4.2.3. Annotation scopes ¶

Type parameter lists and type statements introduce annotation scopes , which behave mostly like function scopes, but with some exceptions discussed below. Annotations currently do not use annotation scopes, but they are expected to use annotation scopes in Python 3.13 when PEP 649 is implemented.

Annotation scopes are used in the following contexts:

Type parameter lists for generic type aliases .

Type parameter lists for generic functions . A generic function’s annotations are executed within the annotation scope, but its defaults and decorators are not.

Type parameter lists for generic classes . A generic class’s base classes and keyword arguments are executed within the annotation scope, but its decorators are not.

The bounds, constraints, and default values for type parameters ( lazily evaluated ).

The value of type aliases ( lazily evaluated ).

Annotation scopes differ from function scopes in the following ways:

Annotation scopes have access to their enclosing class namespace. If an annotation scope is immediately within a class scope, or within another annotation scope that is immediately within a class scope, the code in the annotation scope can use names defined in the class scope as if it were executed directly within the class body. This contrasts with regular functions defined within classes, which cannot access names defined in the class scope.

Expressions in annotation scopes cannot contain yield , yield from , await , or := expressions. (These expressions are allowed in other scopes contained within the annotation scope.)

Names defined in annotation scopes cannot be rebound with nonlocal statements in inner scopes. This includes only type parameters, as no other syntactic elements that can appear within annotation scopes can introduce new names.

While annotation scopes have an internal name, that name is not reflected in the __qualname__ of objects defined within the scope. Instead, the __qualname__ of such objects is as if the object were defined in the enclosing scope.

Added in version 3.12: Annotation scopes were introduced in Python 3.12 as part of PEP 695 .

버전 3.13에서 변경: Annotation scopes are also used for type parameter defaults, as introduced by PEP 696 .

4.2.4. Lazy evaluation ¶

The values of type aliases created through the type statement are lazily evaluated . The same applies to the bounds, constraints, and default values of type variables created through the type parameter syntax . This means that they are not evaluated when the type alias or type variable is created. Instead, they are only evaluated when doing so is necessary to resolve an attribute access.

Here the exception is raised only when the __value__ attribute of the type alias or the __bound__ attribute of the type variable is accessed.

This behavior is primarily useful for references to types that have not yet been defined when the type alias or type variable is created. For example, lazy evaluation enables creation of mutually recursive type aliases:

Lazily evaluated values are evaluated in annotation scope , which means that names that appear inside the lazily evaluated value are looked up as if they were used in the immediately enclosing scope.

Added in version 3.12.

4.2.5. builtins 와 제한된 실행 ¶

CPython 구현 상세: 사용자는 __builtins__ 를 건드리지 말아야 합니다; 이것은 구현 세부사항입니다. 내장 이름 공간의 값을 변경하고 싶은 사용자는 builtins 모듈을 import 하고 그것의 어트리뷰트를 적절하게 수정해야 합니다.

코드 블록의 실행과 연관된 내장 이름 공간은, 사실 전역 이름 공간의 이름 __builtins__ 를 조회함으로써 발견됩니다. 이것은 딕셔너리나 모듈이어야 합니다(후자의 경우 모듈의 딕셔너리가 사용됩니다). 기본적으로, __main__ 모듈에 있을 때는 __builtins__ 가 내장 모듈 builtins 이고, 다른 모듈에 있을 때는 __builtins__ 는 builtins 모듈의 딕셔너리에 대한 별칭입니다.

4.2.6. 동적 기능과의 상호작용 ¶

자유 변수에 대해 이름 검색은 컴파일 시점이 아니라 실행 시점에 이루어집니다. 이것은 다음과 같은 코드가 42를 출력한다는 것을 뜻합니다:

eval() 과 exec() 함수는 이름 검색을 위한 완전한 환경에 대한 접근권이 없습니다. 이름은 호출자의 지역과 전역 이름 공간에서 검색될 수 있습니다. 자유 변수는 가장 가까이 둘러싼 이름 공간이 아니라 전역 이름 공간에서 검색됩니다. [ 1 ] exec() 과 eval() 함수에는 전역과 지역 이름 공간을 재정의할 수 있는 생략 가능한 인자가 있습니다. 만약 단지 한 이름 공간만 주어지면, 그것이 두 가지 모두로 사용됩니다.

예외는 에러나 예외적인 조건을 처리하기 위해 코드 블록의 일반적인 제어 흐름을 깨는 수단입니다. 에러가 감지된 지점에서 예외를 일으킵니다(raised) ; 둘러싼 코드 블록이나 직접적 혹은 간접적으로 에러가 발생한 코드 블록을 호출한 어떤 코드 블록에서건 예외는 처리될 수 있습니다.

파이썬 인터프리터는 실행 시간 에러(0으로 나누는 것 같은)를 감지할 때 예외를 일으킵니다. 파이썬 프로그램은 raise 문을 사용해서 명시적으로 예외를 일으킬 수 있습니다. 예외 처리기는 try … except 문으로 지정됩니다. 그런 문장에서 finally 구는 정리(cleanup) 코드를 지정하는 데 사용되는데, 예외를 처리하는 것이 아니라 앞선 코드에서 예외가 발생하건 그렇지 않건 실행됩니다.

파이썬은 에러 처리에 “종결 (termination)” 모델을 사용합니다; 예외 처리기가 뭐가 발생했는지 발견할 수 있고, 바깥 단계에서 실행을 계속할 수는 있지만, 에러의 원인을 제거한 후에 실패한 연산을 재시도할 수는 없습니다(문제의 코드 조각을 처음부터 다시 시작시키는 것은 예외입니다).

예외가 어디서도 처리되지 않을 때, 인터프리터는 프로그램의 실행을 종료하거나, 대화형 메인 루프로 돌아갑니다. 두 경우 모두, 예외가 SystemExit 인 경우를 제외하고, 스택 트레이스백을 인쇄합니다.

Exceptions are identified by class instances. The except clause is selected depending on the class of the instance: it must reference the class of the instance or a non-virtual base class thereof. The instance can be received by the handler and can carry additional information about the exceptional condition.

예외 메시지는 파이썬 API 일부가 아닙니다. 그 내용은 파이썬의 버전이 바뀔 때 경고 없이 변경될 수 있고, 코드는 여러 버전의 인터프리터에서 실행될 수 있는 코드는 이것에 의존하지 말아야 합니다.

섹션 try 문 에서 try 문, raise 문 에서 raise 문에 대한 설명이 제공됩니다.

• 4.1. 프로그램의 구조
• 4.2.1. 이름의 연결
• 4.2.2. 이름의 검색(resolution)
• 4.2.3. Annotation scopes
• 4.2.4. Lazy evaluation
• 4.2.5. builtins 와 제한된 실행
• 4.2.6. 동적 기능과의 상호작용