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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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[SOLVED] Local Variable Referenced Before Assignment

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

[Fixed] typeerror can’t compare datetime.datetime to datetime.date

Local Variable Referenced Before Assignment Error with Explanation

Try these examples yourself using our Online Compiler.

Let’s look at the following function:

Local Variable Referenced Before Assignment Error

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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[Fixed] nameerror: name Unicode is not defined

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

Understanding unboundlocalerror.

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

Solutions for the Problem

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

Method 1: Initializing the Variable

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

Method 2: Using Global Variables

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

Method 3: Using Nonlocal Variables

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

That’s it. Happy coding!

Next Article: Fixing Python TypeError: Descriptor 'lower' for 'str' Objects Doesn't Apply to 'dict' Object

Previous Article: Python TypeError: write() argument must be str, not bytes

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Local variable referenced before assignment in Python

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Last updated: Feb 17, 2023 Reading time · 4 min

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# 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 .

unboundlocalerror local variable name referenced before assignment

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.

mark variable as global

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 global variables in functions

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.

variables declared in function cannot be accessed in 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.

return value from the function

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.

pass global variable as argument to 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.

assign value to local variable from outer scope

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.

not using nonlocal prints 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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How to fix UnboundLocalError: local variable 'x' referenced before assignment in Python

by Nathan Sebhastian

Posted on May 26, 2023

Reading time: 2 minutes

exception unboundlocalerror local variable referenced before assignment

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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How to Fix Local Variable Referenced Before Assignment Error in Python

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!

exception unboundlocalerror local variable referenced before assignment

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Python local variable referenced before assignment Solution

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

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

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What is unboundlocalerror: local variable referenced before assignment.

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

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

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

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

An Example Scenario

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

We start by declaring two variables:

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

Finally, we call our function:

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

Let’s run our code and see what happens:

An error has been raised.

The Solution

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

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

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

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

Let’s try to run our code again:

Our code successfully prints out the student’s grade.

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

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

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

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

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

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

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Python UnboundLocalError: local variable referenced before assignment

by Suf | Programming , Python , Tips

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

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

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

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

Table of contents

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

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

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

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

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

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

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

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

Example #1: Accessing a Local Variable

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

Let’s run the code to see what happens:

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

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

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

We successfully printed the value to the console.

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

Let’s run the code to see the result:

Example #2: Function with if-elif statements

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

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

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

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

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

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

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

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

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

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

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

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Feb 7, 2018 6:00:01 PM | Python Exception Handling - UnboundLocalError

A deep dive into the UnboundLocalError in Python, with code samples illustrating the difference between NameErrors and UnboundLocalErrors.

Making our way through our in-depth Python Exception Handling series, today we'll be getting into the UnboundLocalError . An UnboundLocalError is raised when a local variable is referenced before it has been assigned. This error is a subclass of the Python NameError we explored in another recent article .

Throughout the remainder of this post we'll examine the UnboundLocalError in more detail, starting with where it sits in the larger Python Exception Class Hierarchy . We'll also look at some functional sample code showing the slight difference between NameErrors and UnboundLocalErrors , and how you can avoid UnboundLocalErrors with special statements found in Python like global . Let the games begin!

The Technical Rundown

All Python exceptions inherit from the BaseException class, or extend from an inherited class therein. The full exception hierarchy of this error is:

  • UnboundLocalError

Full Code Sample

Below is the full code sample we'll be using in this article. It can be copied and pasted if you'd like to play with the code yourself and see how everything works.

from gw_utility.book import Book from gw_utility.logging import Logging

def main(): try: # Increment local count. increment_local_count()

# Set local book title. set_local_book_title("The Silmarillion")

# Set global book title. set_global_book_title("The Silmarillion")

# Disassemble functions. Logging.line_separator("DISASSEMBLY OF increment_count.", 60) disassemble_object(increment_local_count)

Logging.line_separator("DISASSEMBLY OF set_local_book_title.", 60) disassemble_object(set_local_book_title)

Logging.line_separator("DISASSEMBLY OF set_global_book_title.", 60) disassemble_object(set_global_book_title) except NameError as error: # Output expected NameErrors. Logging.log_exception(error) except Exception as exception: # Output unexpected Exceptions. Logging.log_exception(exception, False)

def increment_local_count(): """Increment count by one and output new value.

:return: None """ try: Logging.line_separator("Incrementing LOCAL count.", 60) count += 1 Logging.log("Count incremented to: {}".format(count)) except UnboundLocalError as error: # Output expected UnboundLocalErrors. Logging.log_exception(error) except Exception as exception: # Output unexpected Exceptions. Logging.log_exception(exception, False)

def set_local_book_title(title): """Set title property of local book to passed value and output.

:param title: Title to be set. :return: None """ try: Logging.line_separator("Setting LOCAL book title to '{}'.".format(title), 60) book.title = title Logging.log(book) except UnboundLocalError as error: # Output expected UnboundLocalErrors. Logging.log_exception(error) except Exception as exception: # Output unexpected Exceptions. Logging.log_exception(exception, False)

global_book = Book("The Hobbit", "J.R.R. Tolkien", 365, datetime.date(1977, 9, 15))

def set_global_book_title(title): """Set title property of global_book to passed value and output.

:param title: Title to be set. :return: None """ try: Logging.line_separator("Setting GLOBAL book title to '{}'.".format(title), 60) global global_book global_book.title = title Logging.log(global_book) except UnboundLocalError as error: # Output expected UnboundLocalErrors. Logging.log_exception(error) except Exception as exception: # Output unexpected Exceptions. Logging.log_exception(exception, False)

def disassemble_object(value): """Outputs disassembly of passed object.

:param value: Object to be disassembled. :return: None """ dis.dis(value)

if __name__ == "__main__": main()

class Book: author: str page_count: int publication_date: datetime.date title: str

def __eq__(self, other): """Determines if passed object is equivalent to current object.""" return self.__dict__ == other.__dict__

def __init__(self, title: str = None, author: str = None, page_count: int = None, publication_date: datetime.date = None): """Initializes Book instance.

:param title: Title of Book. :param author: Author of Book. :param page_count: Page Count of Book. :param publication_date: Publication Date of Book. """ self.author = author self.page_count = page_count self.publication_date = publication_date self.title = title

def __getattr__(self, name: str): """Returns the attribute matching passed name.""" # Get internal dict value matching name. value = self.__dict__.get(name) if not value: # Raise AttributeError if attribute value not found. raise AttributeError(f'{self.__class__.__name__}.{name} is invalid.') # Return attribute value. return value

def __len__(self): """Returns the length of title.""" return len(self.title)

def __str__(self): """Returns a formatted string representation of Book.""" date = '' if self.publication_date is None else f', published on {self.publication_date.__format__("%B %d, %Y")}' pages = '' if self.page_count is None else f' at {self.page_count} pages' return f'\'{self.title}\' by {self.author}{pages}{date}.'

class Logging: separator_character_default = '-' separator_length_default = 40

@classmethod def __output(cls, *args, sep: str = ' ', end: str = '\n', file=None): """Prints the passed value(s) to the console.

:param args: Values to output. :param sep: String inserted between values, default a space. :param end: String appended after the last value, default a newline. :param file: A file-like object (stream); defaults to the current sys.stdout. :return: None """ print(*args, sep=sep, end=end, file=file)

@classmethod def line_separator(cls, value: str = None, length: int = separator_length_default, char: str = separator_character_default): """Print a line separator with inserted text centered in the middle.

:param value: Inserted text to be centered. :param length: Total separator length. :param char: Separator character. """ output = value

# If no value passed, output separator of length. if value == None or len(value) == 0: output = f'{char * length}' elif len(value) < length: # Update length based on insert length, less a space for margin. length -= len(value) + 2 # Halve the length and floor left side. left = math.floor(length / 2) right = left # If odd number, add dropped remainder to right side. if length % 2 != 0: right += 1

# Surround insert with separators. output = f'{char * left} {value} {char * right}'

cls.__output(output)

@classmethod def log(cls, *args, sep: str = ' ', end: str = '\n', file=None): """Prints the passed value(s) to the console.

:param args: Values to output. :param sep: String inserted between values, default a space. :param end: String appended after the last value, default a newline. :param file: A file-like object (stream); defaults to the current sys.stdout. """ cls.__output(*args, sep=sep, end=end, file=file)

@classmethod def log_exception(cls, exception: BaseException, expected: bool = True): """Prints the passed BaseException to the console, including traceback.

:param exception: The BaseException to output. :param expected: Determines if BaseException was expected. """ output = "[{}] {}: {}".format('EXPECTED' if expected else 'UNEXPECTED', type(exception).__name__, exception) cls.__output(output) exc_type, exc_value, exc_traceback = sys.exc_info() traceback.print_tb(exc_traceback)

When Should You Use It?

As the name suggests, UnboundLocalErrors are only raised when improperly referencing an unassigned local variable. In most cases this will occur when trying to modify a local variable before it is actually assigned within the local scope. To illustrate we'll get right into our sample code and the increment_local_count() function:

As you can see the increment_local_count() function does just what the name implies: trying to increment the local count variable by one and then outputting the result. However, there's a distinct lack of assignment for the count variable in the local scope of our function block, so executing this code raises an UnboundLocalError :

That sort of makes sense. Since no count variable could be located by the parser no resolution can occur for the increment statement. However, this may look very similar to the NameError we looked at previously, which is raised when "global or local names are not found." So, what makes UnboundLocalError different from NameError ? We can illustrate this difference in our second test function, set_locaL_book_title(title) :

Just as with the increment test we're trying to use a local variable ( book ) that has not been assigned within the local function scope. However, executing this function raises a NameError , rather than an UnboundLocalError :

As we can see by the produced error message the difference here is that, since we're referencing the title property of book , the compiler assumes that book is actually a global name, so the CPython interpreter evaluates the instruction before it even reaches the secondary instruction that references the title property we're ultimately attempting to reference. We can use the dis module to disassemble our functions and see the full bytecode that the CPython interpreter actually processes during execution. We won't go into full detail of these instructions and how the interpreter parses them, but check out our NameError article from last week for more details.

Here is the bytecode instruction set for the book.title = title source code line in set_local_book_title(title) :

As we can see, the first instruction is LOAD_GLOBAL using the book argument, indicating that the interpreter thinks book is a global. This is why a NameError is produced, even though book is actually an undefined local in this case.

One way to resolve this is by using the special global statement to reference a global variable that is assigned outside the local function scope. Here we see global_book is assigned to a value outside of the set_global_book_title(title) function scope:

However, if we run this test function we're able to successfully update the global_book.title property:

Just as within set_local_book_title(title) , the bytecode of set_global_book_title(title) shows the global_book.title = title source code statement contains the LOAD_GLOBAL instruction for the global_book object, but our use of the global statement informs the interpreter to actually seek out the globally-scoped name for reference:

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Written By: Frances Banks

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></center></p><h2>Local variable referenced before assignment: The UnboundLocalError in Python</h2><p>When you start introducing functions into your code, you’re bound to encounter an UnboundLocalError at some point. Because you try to use a local variable referenced before assignment. So, in this guide, we talk about what this error means and why it is raised. We walk through an example in action to help you understand how you can solve it.</p><p>Source: careerkarma</p><p><center><img style=

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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exception unboundlocalerror local variable referenced before assignment

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UnboundLocalError Local Variable 'index' Referenced Before Assignment

exception unboundlocalerror local variable referenced before assignment

.css-13lojzj{position:absolute;padding-right:0.25rem;margin-left:-1.25rem;left:0;height:100%;display:-webkit-inline-box;display:-webkit-inline-flex;display:-ms-inline-flexbox;display:inline-flex;-webkit-align-items:center;-webkit-box-align:center;-ms-flex-align:center;align-items:center;display:none;} .css-b94zdx{width:1rem;height:1rem;} The Problem

The exception UnboundLocalError: local variable 'index' referenced before assignment happens in Python when you use a global variable in a function that also defines a local version of the same variable.

Because index is defined globally as well as inside the foo() function, Python throws an exception if you try to use the variable inside foo before it’s declared.

The Solution

In this case, we want to rename the variable to avoid the conflict.

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Local variable referenced before assignment in Python

The “local variable referenced before assignment” error occurs in Python when you try to use a local variable before it has been assigned a value.

This error typically arises in situations where you declare a variable within a function but then try to access or modify it before actually assigning a value to it.

Here’s an example to illustrate this error:

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

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

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

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

Using the global keyword

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

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

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

Using nonlocal keyword

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

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

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

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

exception unboundlocalerror local variable referenced before assignment

Understanding UnboundLocalError in Python

If you're closely following the Python tag on StackOverflow , you'll notice that the same question comes up at least once a week. The question goes on like this:

Why, when run, this results in the following error:

There are a few variations on this question, with the same core hiding underneath. Here's one:

Running the lst.append(5) statement successfully appends 5 to the list. However, substitute it for lst += [5] , and it raises UnboundLocalError , although at first sight it should accomplish the same.

Although this exact question is answered in Python's official FAQ ( right here ), I decided to write this article with the intent of giving a deeper explanation. It will start with a basic FAQ-level answer, which should satisfy one only wanting to know how to "solve the damn problem and move on". Then, I will dive deeper, looking at the formal definition of Python to understand what's going on. Finally, I'll take a look what happens behind the scenes in the implementation of CPython to cause this behavior.

The simple answer

As mentioned above, this problem is covered in the Python FAQ. For completeness, I want to explain it here as well, quoting the FAQ when necessary.

Let's take the first code snippet again:

So where does the exception come from? Quoting the FAQ:

This is because when you make an assignment to a variable in a scope, that variable becomes local to that scope and shadows any similarly named variable in the outer scope.

But x += 1 is similar to x = x + 1 , so it should first read x , perform the addition and then assign back to x . As mentioned in the quote above, Python considers x a variable local to foo , so we have a problem - a variable is read (referenced) before it's been assigned. Python raises the UnboundLocalError exception in this case [1] .

So what do we do about this? The solution is very simple - Python has the global statement just for this purpose:

This prints 11 , without any errors. The global statement tells Python that inside foo , x refers to the global variable x , even if it's assigned in foo .

Actually, there is another variation on the question, for which the answer is a bit different. Consider this code:

This kind of code may come up if you're into closures and other techniques that use Python's lexical scoping rules. The error this generates is the familiar UnboundLocalError . However, applying the "global fix":

Doesn't help - another error is generated: NameError: global name 'x' is not defined . Python is right here - after all, there's no global variable named x , there's only an x in external . It may be not local to internal , but it's not global. So what can you do in this situation? If you're using Python 3, you have the nonlocal keyword. Replacing global by nonlocal in the last snippet makes everything work as expected. nonlocal is a new statement in Python 3, and there is no equivalent in Python 2 [2] .

The formal answer

Assignments in Python are used to bind names to values and to modify attributes or items of mutable objects. I could find two places in the Python (2.x) documentation where it's defined how an assignment to a local variable works.

One is section 6.2 "Assignment statements" in the Simple Statements chapter of the language reference:

Assignment of an object to a single target is recursively defined as follows. If the target is an identifier (name): If the name does not occur in a global statement in the current code block: the name is bound to the object in the current local namespace. Otherwise: the name is bound to the object in the current global namespace.

Another is section 4.1 "Naming and binding" of the Execution model chapter:

If a name is bound in a block, it is a local variable of that block. [...] When a name is used in a code block, it is resolved using the nearest enclosing scope. [...] If the name refers to a local variable that has not been bound, a UnboundLocalError exception is raised.

This is all clear, but still, another small doubt remains. All these rules apply to assignments of the form var = value which clearly bind var to value . But the code snippets we're having a problem with here have the += assignment. Shouldn't that just modify the bound value, without re-binding it?

Well, no. += and its cousins ( -= , *= , etc.) are what Python calls " augmented assignment statements " [ emphasis mine ]:

An augmented assignment evaluates the target (which, unlike normal assignment statements, cannot be an unpacking) and the expression list, performs the binary operation specific to the type of assignment on the two operands, and assigns the result to the original target . The target is only evaluated once. An augmented assignment expression like x += 1 can be rewritten as x = x + 1 to achieve a similar, but not exactly equal effect. In the augmented version, x is only evaluated once. Also, when possible, the actual operation is performed in-place, meaning that rather than creating a new object and assigning that to the target, the old object is modified instead. With the exception of assigning to tuples and multiple targets in a single statement, the assignment done by augmented assignment statements is handled the same way as normal assignments . Similarly, with the exception of the possible in-place behavior, the binary operation performed by augmented assignment is the same as the normal binary operations.

So when earlier I said that x += 1 is similar to x = x + 1 , I wasn't telling all the truth, but it was accurate with respect to binding. Apart for possible optimization, += counts exactly as = when binding is considered. If you think carefully about it, it's unavoidable, because some types Python works with are immutable. Consider strings, for example:

The first line binds x to the value "abc". The second line doesn't modify the value "abc" to be "abcdef". Strings are immutable in Python . Rather, it creates the new value "abcdef" somewhere in memory, and re-binds x to it. This can be seen clearly when examining the object ID for x before and after the += :

Note that some types in Python are mutable. For example, lists can actually be modified in-place:

id(y) didn't change after += , because the object y referenced was just modified. Still, Python re-bound y to the same object [3] .

The "too much information" answer

This section is of interest only to those curious about the implementation internals of Python itself.

One of the stages in the compilation of Python into bytecode is building the symbol table [4] . An important goal of building the symbol table is for Python to be able to mark the scope of variables it encounters - which variables are local to functions, which are global, which are free (lexically bound) and so on.

When the symbol table code sees a variable is assigned in a function, it marks it as local. Note that it doesn't matter if the assignment was done before usage, after usage, or maybe not actually executed due to a condition in code like this:

We can use the symtable module to examine the symbol table information gathered on some Python code during compilation:

This prints:

So we see that x was marked as local in foo . Marking variables as local turns out to be important for optimization in the bytecode, since the compiler can generate a special instruction for it that's very fast to execute. There's an excellent article here explaining this topic in depth; I'll just focus on the outcome.

The compiler_nameop function in Python/compile.c handles variable name references. To generate the correct opcode, it queries the symbol table function PyST_GetScope . For our x , this returns a bitfield with LOCAL in it. Having seen LOCAL , compiler_nameop generates a LOAD_FAST . We can see this in the disassembly of foo :

The first block of instructions shows what x += 1 was compiled to. You will note that already here (before it's actually assigned), LOAD_FAST is used to retrieve the value of x .

This LOAD_FAST is the instruction that will cause the UnboundLocalError exception to be raised at runtime, because it is actually executed before any STORE_FAST is done for x . The gory details are in the bytecode interpreter code in Python/ceval.c :

Ignoring the macro-fu for the moment, what this basically says is that once LOAD_FAST is seen, the value of x is obtained from an indexed array of objects [5] . If no STORE_FAST was done before, this value is still NULL , the if branch is not taken [6] and the exception is raised.

You may wonder why Python waits until runtime to raise this exception, instead of detecting it in the compiler. The reason is this code:

Suppose something_true is a function that returns True , possibly due to some user input. In this case, x = 1 binds x locally, so the reference to it in x += 1 is no longer unbound. This code will then run without exceptions. Of course if something_true actually turns out to return False , the exception will be raised. Python has no way to resolve this at compile time, so the error detection is postponed to runtime.

exception unboundlocalerror local variable referenced before assignment

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UnboundLocalError: local variable 'boxprops' referenced before assignment #3647

@catskillsresearch

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【Python】成功解决UnboundLocalError: local variable ‘a‘ referenced before assignment(几种场景下的解决方案)

exception unboundlocalerror local variable referenced before assignment

【Python】成功解决UnboundLocalError: local variable ‘a’ referenced before assignment(几种场景下的解决方案)

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🌵文章目录🌵

🐛一、什么是unboundlocalerror?, 🛠️二、如何解决unboundlocalerror?, 🌐三、实际场景中的解决方案, 📖四、深入理解作用域与变量生命周期, 🔍五、举一反三:其他常见错误与陷阱, 💡六、总结与最佳实践, 🎉结语.

  在Python编程中, UnboundLocalError: local variable 'a' referenced before assignment 这个错误常常让初学者感到困惑。这个错误表明 你尝试在一个函数内部引用了一个局部变量,但是在引用之前并没有对它进行赋值 。换句话说, Python解释器在函数的作用域内找到了一个变量的引用,但是这个变量并没有在引用它之前被定义或赋值 。

下面是一个简单的例子,演示了如何触发这个错误:

在这个例子中,我们尝试在 a 被赋值之前就打印它的值,这会导致 UnboundLocalError 。

  要解决 UnboundLocalError ,你需要 确保在引用局部变量之前,该变量已经被正确地赋值 。这可以通过几种不同的方式实现。

确保在引用局部变量之前,该变量已经被正确赋值。

如果你打算在函数内部引用的是全局变量,那么需要使用 global 关键字来明确指定。

如果你希望变量有一个默认值,你可以使用函数的参数来提供这个默认值。

在某些情况下,你可能需要在使用变量之前检查它是否已经被定义。这可以通过使用 try-except 块来实现。

  在实际编程中, UnboundLocalError 可能会出现在更复杂的场景中。下面是一些实际案例及其解决方案。

场景1:在循环中引用和修改变量

在正确示例中,我们在循环中累加 i 到 total ,并在循环结束后打印 total 。注意,我们在累加之前已经对 total 进行了初始化,避免了 UnboundLocalError 。

场景2:在条件语句中引用变量

在正确示例中,我们在 if 语句中根据 x 的值计算 y ,然后在 if 语句外部打印 y 的值。我们使用了 if-else 语句确保了 y 在引用之前一定会被定义。

  在解决 UnboundLocalError 时,理解Python中的作用域和变量生命周期至关重要。作用域决定了变量的可见性,即变量在哪里可以被访问。而变量的生命周期则关系到变量的创建和销毁的时机。局部变量只在函数内部可见,并且当函数执行完毕后,它们的生命周期就结束了。全局变量在整个程序中都是可见的,它们的生命周期则与程序的生命周期一致。

  除了 UnboundLocalError 之外,Python编程中还有其他一些与变量作用域和生命周期相关的常见错误和陷阱。例如,不小心修改了全局变量而没有意识到,或者在循环中意外地创建了一个新的变量而不是更新现有的变量。避免这些错误的关键在于保持对变量作用域和生命周期的清晰理解,并谨慎地使用 global 关键字。

  解决 UnboundLocalError 的关键在于确保在引用局部变量之前已经对其进行了赋值。这可以通过在引用前赋值、使用全局变量、使用默认值或检查变量是否已定义等方式实现。同时,深入理解作用域和变量生命周期对于避免此类错误至关重要。最佳实践包括:

  • 在函数内部使用局部变量时,确保在引用之前已经对其进行了赋值。
  • 如果需要在函数内部修改全局变量,请使用 global 关键字明确声明。
  • 尽量避免在函数内部意外地创建新的全局变量。
  • 对于复杂的逻辑,使用明确的变量命名和注释来提高代码的可读性和可维护性。

通过遵循这些最佳实践,你可以减少 UnboundLocalError 的发生,并编写出更加健壮和可靠的Python代码。

  通过本文的学习,相信你已经对 UnboundLocalError 有了更深入的理解,并掌握了解决这一错误的几种方法。在实际编程中,遇到问题时不要害怕,要勇于探索和实践。通过不断学习和积累经验,你会逐渐成为一名优秀的Python程序员。加油!🚀

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  1. "Fixing UnboundLocalError: Local Variable Referenced Before Assignment

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  2. UnboundLocalError: Local Variable Referenced Before Assignment

    exception unboundlocalerror local variable referenced before assignment

  3. UnboundLocalError: local variable referenced before assignment

    exception unboundlocalerror local variable referenced before assignment

  4. Local Variable Referenced Before Assignment Solved Error In Python

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  5. Exception has occurred: UnboundLocalError local variable 'depth

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COMMENTS

  1. Python 3: UnboundLocalError: local variable referenced before assignment

    The following code gives the error UnboundLocalError: local variable 'Var1' referenced before assignment: Var1 = 1 Var2 = 0 def function(): if Var2 == 0 and Var1 > 0: print("Result 1") elif Var2 == 1 and Var1 > 0: print("Result 2") elif Var1 < 1: print("Result 3") Var1 -= 1 function() How can I fix this? python python-3.x scope Share

  2. 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. Syntax:

  3. UnboundLocalError: local variable referenced before assignment in

    UnboundLocalError: local variable referenced before assignment in Python [duplicate] Ask Question Asked 3 years, 9 months ago Modified 3 years, 9 months ago Viewed 4k times 0 This question already has answers here : UnboundLocalError trying to use a variable (supposed to be global) that is (re)assigned (even after first use) (14 answers)

  4. "UnboundLocalError: local variable referenced before assignment" after

    UnboundLocalError: local variable 'T' referenced before assignment The first print works correctly but the second causes an exception. I tried making T a global variable but then both answers are the same. What is going wrong, and how can I fix it? python Share Improve this question Follow edited Feb 6, 2023 at 11:52 Karl Knechtel 62.1k 11 113 159

  5. [SOLVED] Local Variable Referenced Before Assignment

    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.

  6. Fixing Python UnboundLocalError: Local Variable 'x' Accessed Before

    The UnboundLocalError in Python occurs when a function tries to access a local variable before it has been assigned a value. Variables in Python have scope that defines their level of visibility throughout the code: global scope, local scope, and nonlocal (in nested functions) scope.

  7. 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. main.py

  8. How to fix UnboundLocalError: local variable 'x' referenced before

    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.

  9. How to Fix Local Variable Referenced Before Assignment Error in Python

    UnboundLocalError: local variable 'value' referenced before assignment The issue is that in this line: PYTHON value = value + 1 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.

  10. Python local variable referenced before assignment Solution

    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: UnboundLocalError: local variable referenced before assignment Python has a simple rule to determine the scope of a variable.

  11. Python UnboundLocalError: local variable referenced before assignment

    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: test_var = 0 def test_func(test_var): test_var += 1 return test_var

  12. Python 3: UnboundLocalError: local variable referenced before assignment

    UnboundLocalError: local variable 'x' referenced before assignment In this example, the variable x is being accessed before it is assigned a value, which is causing the error. To fix this, you can either move the assignment of the variable x before the print statement, or give it an initial value before the print statement. def example ():

  13. Python Exception Handling

    An UnboundLocalError is raised when a local variable is referenced before it has been assigned. This error is a subclass of the Python NameError we explored in another recent article. Throughout the remainder of this post we'll examine the UnboundLocalError in more detail, starting with where it sits in the larger Python Exception Class Hierarchy.

  14. UnboundLocalError: local variable 'x' referenced before assignment

    UnboundLocalError: local variable referenced before assignment. 1. ... Error: local variable referenced before assignment in ArcPy. 1. python - psycopg2.errors.RaiseException find_srid() - could not find the corresponding SRID. 3. Using ogr2ogr in Python scripts. Hot Network Questions Show how close a user is to being unsung

  15. Local variable referenced before assignment: The UnboundLocalError

    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: 1 UnboundLocalError: local variable referenced before assignment Python has a simple rule to determine the scope of a variable.

  16. UnboundLocalError Local Variable 'index' Referenced Before Assignment

    The exception UnboundLocalError: local variable 'index' referenced before assignment happens in Python when you use a global variable in a function that also defines a local version of the same variable. if index == 0:

  17. Local variable referenced before assignment in Python

    The "local variable referenced before assignment" error occurs in Python when you try to use a local variable before it has been assigned a value. This error typically arises in situations where you declare a variable within a function but then try to access or modify it before actually assigning a value to it.

  18. Understanding UnboundLocalError in Python

    I could find two places in the Python (2.x) documentation where it's defined how an assignment to a local variable works. One is section 6.2 "Assignment statements" in the Simple Statements chapter of the language reference: Assignment of an object to a single target is recursively defined as follows. If the target is an identifier (name):

  19. UnboundLocalError: local variable 'eval_manager' referenced before

    During handling of the above exception, another exception occurred: Traceback (most recent call last): ... UnboundLocalError: local variable 'eval_manager' referenced before assignment WARNING: sensor object went out of the scope but the sensor is still alive in the simulation: Actor 88 (sensor.other.gnss) ...

  20. UnboundLocalError: local variable <var> referenced before assignment

    UnboundLocalError: local variable <var> referenced before assignment [duplicate] Ask Question Asked 2 years, 9 months ago Modified 2 years, 9 months ago Viewed 1k times 2 This question already has answers here : How can a name be "unbound" in Python? What code can cause an `UnboundLocalError`? (3 answers) Closed 17 days ago.

  21. UnboundLocalError: local variable 'boxprops' referenced before ...

    There is a for loop in the routine in categorical.py. The variable boxprops is set for various cases inside the for loop. It is undefined before the loop and used afterwards. It just doesn't imagine the case where there is no data and the for loop has 0 iterations. Setting boxprops = None before the start of the loop cures the problem.

  22. 【Python】成功解决UnboundLocalError: local variable 'a' referenced before

    一、什么是UnboundLocalError? 在Python编程中,UnboundLocalError: local variable 'a' referenced before assignment这个错误常常让初学者感到困惑。这个错误表明你尝试在一个函数内部引用了一个局部变量,但是在引用之前并没有对它进行赋值。换句话说,Python解释器在函数的作用域内找到了一个变量的引用,但是这个 ...

  23. UnboundLocalError: local variable 'image' referenced before assignment

    UnboundLocalError: local variable 'image' referenced before assignment Ask Question Asked 2 years, 8 months ago Modified 2 years, 8 months ago Viewed 3k times 1 I'm running this code, def extract_features(filename, model): try: image = Image.open(filename) except: print("ERROR: Couldn't open image!