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.
Take your skills to the next level ⚡️
I'm sending out an occasional email with the latest tutorials on programming, web development, and statistics. Drop your email in the box below and I'll send new stuff straight into your inbox!
Hello! This website is dedicated to help you learn tech and data science skills with its step-by-step, beginner-friendly tutorials. Learn statistics, JavaScript and other programming languages using clear examples written for people.
Learn more about this website
Connect with me on Twitter
Or LinkedIn
Type the keyword below and hit enter
Click to see all tutorials tagged with:
Fix "local variable referenced before assignment" in Python
Introduction
If you're a Python developer, you've probably come across a variety of errors, like the "local variable referenced before assignment" error. This error can be a bit puzzling, especially for beginners and when it involves local/global variables.
Today, we'll explain this error, understand why it occurs, and see how you can fix it.
The "local variable referenced before assignment" Error
The "local variable referenced before assignment" error in Python is a common error that occurs when a local variable is referenced before it has been assigned a value. This error is a type of UnboundLocalError , which is raised when a local variable is referenced before it has been assigned in the local scope.
Here's a simple example:
Running this code will throw the "local variable 'x' referenced before assignment" error. This is because the variable x is referenced in the print(x) statement before it is assigned a value in the local scope of the foo function.
Even more confusing is when it involves global variables. For example, the following code also produces the error:
But wait, why does this also produce the error? Isn't x assigned before it's used in the say_hello function? The problem here is that x is a global variable when assigned "Hello ". However, in the say_hello function, it's a different local variable, which has not yet been assigned.
We'll see later in this Byte how you can fix these cases as well.
Fixing the Error: Initialization
One way to fix this error is to initialize the variable before using it. This ensures that the variable exists in the local scope before it is referenced.
Let's correct the error from our first example:
In this revised code, we initialize x with a value of 1 before printing it. Now, when you run the function, it will print 1 without any errors.
Fixing the Error: Global Keyword
Another way to fix this error, depending on your specific scenario, is by using the global keyword. This is especially useful when you want to use a global variable inside a function.
No spam ever. Unsubscribe anytime. Read our Privacy Policy.
Here's how:
In this snippet, we declare x as a global variable inside the function foo . This tells Python to look for x in the global scope, not the local one . Now, when you run the function, it will increment the global x by 1 and print 1 .
Similar Error: NameError
An error that's similar to the "local variable referenced before assignment" error is the NameError . This is raised when you try to use a variable or a function name that has not been defined yet.
Running this code will result in a NameError :
In this case, we're trying to print the value of y , but y has not been defined anywhere in the code. Hence, Python raises a NameError . This is similar in that we are trying to use an uninitialized/undefined variable, but the main difference is that we didn't try to initialize y anywhere else in our code.
Variable Scope in Python
Understanding the concept of variable scope can help avoid many common errors in Python, including the main error of interest in this Byte. But what exactly is variable scope?
In Python, variables have two types of scope - global and local. A variable declared inside a function is known as a local variable, while a variable declared outside a function is a global variable.
Consider this example:
In this code, x is a global variable, and y is a local variable. x can be accessed anywhere in the code, but y can only be accessed within my_function . Confusion surrounding this is one of the most common causes for the "variable referenced before assignment" error.
In this Byte, we've taken a look at the "local variable referenced before assignment" error and another similar error, NameError . We also delved into the concept of variable scope in Python, which is an important concept to understand to avoid these errors. If you're seeing one of these errors, check the scope of your variables and make sure they're being assigned before they're being used.
Building Your First Convolutional Neural Network With Keras
Most resources start with pristine datasets, start at importing and finish at validation. There's much more to know. Why was a class predicted? Where was...
© 2013- 2024 Stack Abuse. All rights reserved.
Local variable referenced before assignment in Python
Last updated: Feb 17, 2023 Reading time · 4 min
# Local variable referenced before assignment in Python
The Python "UnboundLocalError: Local variable referenced before assignment" occurs when we reference a local variable before assigning a value to it in a function.
To solve the error, mark the variable as global in the function definition, e.g. global my_var .
Here is an example of how the error occurs.
We assign a value to the name variable in the function.
# Mark the variable as global to solve the error
To solve the error, mark the variable as global in your function definition.
If a variable is assigned a value in a function's body, it is a local variable unless explicitly declared as global .
# Local variables shadow global ones with the same name
You could reference the global name variable from inside the function but if you assign a value to the variable in the function's body, the local variable shadows the global one.
Accessing the name variable in the function is perfectly fine.
On the other hand, variables declared in a function cannot be accessed from the global scope.
The name variable is declared in the function, so trying to access it from outside causes an error.
Make sure you don't try to access the variable before using the global keyword, otherwise, you'd get the SyntaxError: name 'X' is used prior to global declaration error.
# Returning a value from the function instead
An alternative solution to using the global keyword is to return a value from the function and use the value to reassign the global variable.
We simply return the value that we eventually use to assign to the name global variable.
# Passing the global variable as an argument to the function
You should also consider passing the global variable as an argument to the function.
We passed the name global variable as an argument to the function.
If we assign a value to a variable in a function, the variable is assumed to be local unless explicitly declared as global .
# Assigning a value to a local variable from an outer scope
If you have a nested function and are trying to assign a value to the local variables from the outer function, use the nonlocal keyword.
The nonlocal keyword allows us to work with the local variables of enclosing functions.
Had we not used the nonlocal statement, the call to the print() function would have returned an empty string.
Printing the message variable on the last line of the function shows an empty string because the inner() function has its own scope.
Changing the value of the variable in the inner scope is not possible unless we use the nonlocal keyword.
Instead, the message variable in the inner function simply shadows the variable with the same name from the outer scope.
# Discussion
As shown in this section of the documentation, when you assign a value to a variable inside a function, the variable:
- Becomes local to the scope.
- Shadows any variables from the outer scope that have the same name.
The last line in the example function assigns a value to the name variable, marking it as a local variable and shadowing the name variable from the outer scope.
At the time the print(name) line runs, the name variable is not yet initialized, which causes the error.
The most intuitive way to solve the error is to use the global keyword.
The global keyword is used to indicate to Python that we are actually modifying the value of the name variable from the outer scope.
- If a variable is only referenced inside a function, it is implicitly global.
- If a variable is assigned a value inside a function's body, it is assumed to be local, unless explicitly marked as global .
If you want to read more about why this error occurs, check out [this section] ( this section ) of the docs.
# Additional Resources
You can learn more about the related topics by checking out the following tutorials:
- SyntaxError: name 'X' is used prior to global declaration
Borislav Hadzhiev
Web Developer
Copyright © 2024 Borislav Hadzhiev
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!
- Privacy Policy
- Terms of Service
Explore your training options in 10 minutes Get Started
- Graduate Stories
- Partner Spotlights
- Bootcamp Prep
- Bootcamp Admissions
- University Bootcamps
- Coding Tools
- Software Engineering
- Web Development
- Data Science
- Tech Guides
- Tech Resources
- Career Advice
- Online Learning
- Internships
- Apprenticeships
- Tech Salaries
- Associate Degree
- Bachelor's Degree
- Master's Degree
- University Admissions
- Best Schools
- Certifications
- Bootcamp Financing
- Higher Ed Financing
- Scholarships
- Financial Aid
- Best Coding Bootcamps
- Best Online Bootcamps
- Best Web Design Bootcamps
- Best Data Science Bootcamps
- Best Technology Sales Bootcamps
- Best Data Analytics Bootcamps
- Best Cybersecurity Bootcamps
- Best Digital Marketing Bootcamps
- Los Angeles
- San Francisco
- Browse All Locations
- Digital Marketing
- Machine Learning
- See All Subjects
- Bootcamps 101
- Full-Stack Development
- Career Changes
- View all Career Discussions
- Mobile App Development
- Cybersecurity
- Product Management
- UX/UI Design
- What is a Coding Bootcamp?
- Are Coding Bootcamps Worth It?
- How to Choose a Coding Bootcamp
- Best Online Coding Bootcamps and Courses
- Best Free Bootcamps and Coding Training
- Coding Bootcamp vs. Community College
- Coding Bootcamp vs. Self-Learning
- Bootcamps vs. Certifications: Compared
- What Is a Coding Bootcamp Job Guarantee?
- How to Pay for Coding Bootcamp
- Ultimate Guide to Coding Bootcamp Loans
- Best Coding Bootcamp Scholarships and Grants
- Education Stipends for Coding Bootcamps
- Get Your Coding Bootcamp Sponsored by Your Employer
- GI Bill and Coding Bootcamps
- Tech Intevriews
- Our Enterprise Solution
- Connect With Us
- Publication
- Reskill America
- Partner With Us
- Resource Center
- Bachelor’s Degree
- Master’s Degree
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.
Find your bootcamp match
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 !
About us: Career Karma is a platform designed to help job seekers find, research, and connect with job training programs to advance their careers. Learn about the CK publication .
What's Next?
Get matched with top bootcamps
Ask a question to our community, take our careers quiz.
Leave a Reply Cancel reply
Your email address will not be published. Required fields are marked *
[SOLVED] Local Variable Referenced Before Assignment
Python treats variables referenced only inside a function as global variables. Any variable assigned to a function’s body is assumed to be a local variable unless explicitly declared as global.
Why Does This Error Occur?
Unboundlocalerror: local variable referenced before assignment occurs when a variable is used before its created. Python does not have the concept of variable declarations. Hence it searches for the variable whenever used. When not found, it throws the error.
Before we hop into the solutions, let’s have a look at what is the global and local variables.
Local Variable Declarations vs. Global Variable Declarations
Local Variable Referenced Before Assignment Error with Explanation
Try these examples yourself using our Online Compiler.
Let’s look at the following function:
Explanation
The variable myVar has been assigned a value twice. Once before the declaration of myFunction and within myFunction itself.
Using Global Variables
Passing the variable as global allows the function to recognize the variable outside the function.
Create Functions that Take in Parameters
Instead of initializing myVar as a global or local variable, it can be passed to the function as a parameter. This removes the need to create a variable in memory.
UnboundLocalError: local variable ‘DISTRO_NAME’
This error may occur when trying to launch the Anaconda Navigator in Linux Systems.
Upon launching Anaconda Navigator, the opening screen freezes and doesn’t proceed to load.
Try and update your Anaconda Navigator with the following command.
If solution one doesn’t work, you have to edit a file located at
After finding and opening the Python file, make the following changes:
In the function on line 159, simply add the line:
DISTRO_NAME = None
Save the file and re-launch Anaconda Navigator.
DJANGO – Local Variable Referenced Before Assignment [Form]
The program takes information from a form filled out by a user. Accordingly, an email is sent using the information.
Upon running you get the following error:
We have created a class myForm that creates instances of Django forms. It extracts the user’s name, email, and message to be sent.
A function GetContact is created to use the information from the Django form and produce an email. It takes one request parameter. Prior to sending the email, the function verifies the validity of the form. Upon True , .get() function is passed to fetch the name, email, and message. Finally, the email sent via the send_mail function
Why does the error occur?
We are initializing form under the if request.method == “POST” condition statement. Using the GET request, our variable form doesn’t get defined.
Local variable Referenced before assignment but it is global
This is a common error that happens when we don’t provide a value to a variable and reference it. This can happen with local variables. Global variables can’t be assigned.
This error message is raised when a variable is referenced before it has been assigned a value within the local scope of a function, even though it is a global variable.
Here’s an example to help illustrate the problem:
In this example, x is a global variable that is defined outside of the function my_func(). However, when we try to print the value of x inside the function, we get a UnboundLocalError with the message “local variable ‘x’ referenced before assignment”.
This is because the += operator implicitly creates a local variable within the function’s scope, which shadows the global variable of the same name. Since we’re trying to access the value of x before it’s been assigned a value within the local scope, the interpreter raises an error.
To fix this, you can use the global keyword to explicitly refer to the global variable within the function’s scope:
However, in the above example, the global keyword tells Python that we want to modify the value of the global variable x, rather than creating a new local variable. This allows us to access and modify the global variable within the function’s scope, without causing any errors.
Local variable ‘version’ referenced before assignment ubuntu-drivers
This error occurs with Ubuntu version drivers. To solve this error, you can re-specify the version information and give a split as 2 –
Here, p_name means package name.
With the help of the threading module, you can avoid using global variables in multi-threading. Make sure you lock and release your threads correctly to avoid the race condition.
When a variable that is created locally is called before assigning, it results in Unbound Local Error in Python. The interpreter can’t track the variable.
Therefore, we have examined the local variable referenced before the assignment Exception in Python. The differences between a local and global variable declaration have been explained, and multiple solutions regarding the issue have been provided.
Trending Python Articles
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!
Share this:
- Click to share on Facebook (Opens in new window)
- Click to share on LinkedIn (Opens in new window)
- Click to share on Reddit (Opens in new window)
- Click to share on Pinterest (Opens in new window)
- Click to share on Telegram (Opens in new window)
- Click to share on WhatsApp (Opens in new window)
- Click to share on Twitter (Opens in new window)
- Click to share on Tumblr (Opens in new window)
- Free Python 3 Tutorial
- Control Flow
- Exception Handling
- Python Programs
- Python Projects
- Python Interview Questions
- Python Database
- Data Science With Python
- Machine Learning with Python
- How to Fix: ValueError: Unknown Label Type: 'Continuous' in Python
- How To Check If Variable Is Empty In Python?
- How To Fix - Python RuntimeWarning: overflow encountered in scalar
- How To Resolve Issue " Threading Ignores Keyboardinterrupt Exception" In Python?
- AttributeError: can’t set attribute in Python
- How to Fix ImportError: Cannot Import name X in Python
- AttributeError: __enter__ Exception in Python
- Modulenotfounderror: No Module Named 'httpx' in Python
- How to Fix - SyntaxError: (Unicode Error) 'Unicodeescape' Codec Can't Decode Bytes
- SyntaxError: ‘return’ outside function in Python
- How to fix "SyntaxError: invalid character" in Python
- No Module Named 'Sqlalchemy-Jsonfield' in Python
- Python Code Error: No Module Named 'Aiohttp'
- Fix Python Attributeerror: __Enter__
- Python Runtimeerror: Super() No Arguments
- Filenotfounderror: Errno 2 No Such File Or Directory in Python
- ModuleNotFoundError: No module named 'dotenv' in Python
- Nameerror: Name Plot_Cases_Simple Is Not Defined in Python
- How To Avoid Notimplementederror In Python?
How to Fix – UnboundLocalError: Local variable Referenced Before Assignment in Python
Developers often encounter the UnboundLocalError Local Variable Referenced Before Assignment error in Python. In this article, we will see what is local variable referenced before assignment error in Python and how to fix it by using different approaches.
What is UnboundLocalError: Local variable Referenced Before Assignment?
This error 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.
Below, are the reasons by which UnboundLocalError: Local variable Referenced Before Assignment error occurs in Python :
Nested Function Variable Access
Global variable modification.
In this code, the outer_function defines a variable ‘x’ and a nested inner_function attempts to access it, but encounters an UnboundLocalError due to a local ‘x’ being defined later in the inner_function.
In this code, the function example_function tries to increment the global variable ‘x’, but encounters an UnboundLocalError since it’s treated as a local variable due to the assignment operation within the function.
Solution for Local variable Referenced Before Assignment in Python
Below, are the approaches to solve “Local variable Referenced Before Assignment”.
In this code, example_function successfully modifies the global variable ‘x’ by declaring it as global within the function, incrementing its value by 1, and then printing the updated value.
In this code, the outer_function defines a local variable ‘x’, and the inner_function accesses and modifies it as a nonlocal variable, allowing changes to the outer function’s scope from within the inner function.
Please Login to comment...
- Python Errors
- Python How-to-fix
- 10 Best HuggingChat Alternatives and Competitors
- Best Free Android Apps for Podcast Listening
- Google AI Model: Predicts Floods 7 Days in Advance
- Who is Devika AI? India's 'AI coder', an alternative to Devin AI
- 30 OOPs Interview Questions and Answers (2024)
Improve your Coding Skills with Practice
What kind of Experience do you want to share?
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
IMAGES
VIDEO
COMMENTS
File "weird.py", line 5, in main. print f(3) UnboundLocalError: local variable 'f' referenced before assignment. Python sees the f is used as a local variable in [f for f in [1, 2, 3]], and decides that it is also a local variable in f(3). You could add a global f statement: def f(x): return x. def main():
DataFrame is a 2-dimensional labeled data structure with columns of potentially different types. You can think of it like a spreadsheet or SQL table, or a dict of Series objects. It is generally the most commonly used pandas object. Like Series, DataFrame accepts many different kinds of input: Dict of 1D ndarrays, lists, dicts, or Series.
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.
Building Your First Convolutional Neural Network With Keras # python # artificial intelligence # machine learning # tensorflow Most resources start with pristine datasets, start at importing and finish at validation.
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.
value = value + 1 print (value) increment() If you run this code, you'll get. BASH. 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 ...
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. If a variable is assigned in a function, that variable is local. This is because it is assumed that when you define a ...
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. ... Therefore, we have examined the local variable referenced before the assignment Exception in Python. The differences between a local and global variable ...
UnboundLocalError: local variable referenced before assignment. Example #1: Accessing a Local Variable. Solution #1: Passing Parameters to the Function. Solution #2: Use Global Keyword. Example #2: Function with if-elif statements. Solution #1: Include else statement. Solution #2: Use global keyword. Summary.
Output. Hangup (SIGHUP) Traceback (most recent call last): File "Solution.py", line 7, in <module> example_function() File "Solution.py", line 4, in example_function x += 1 # Trying to modify global variable 'x' without declaring it as global UnboundLocalError: local variable 'x' referenced before assignment Solution for Local variable Referenced Before Assignment in Python
"UnboundLocalError: local variable referenced before assignment" when calling a function 0 "local variable referenced before assignment" but assigning the variable is the first thing i do
In order for you to modify test1 while inside a function you will need to do define test1 as a global variable, for example: test1 = 0. def test_func(): global test1. test1 += 1. test_func() However, if you only need to read the global variable you can print it without using the keyword global, like so: test1 = 0.
I'm having a strange scope problem. The following code is a cell in a Jupyter Notebook. I have assigned p1_total_value and p2_total_value within correlation_check(). Then, later in correlation_chec...
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.
I'm trying to create an empty global pandas dataframe and then append to it within data processing functions. I'm having trouble figuring out what's going wrong. Here's a minimal example: from dat...
Use global statement to handle that: def three_upper(s): #check for 3 upper letter. global count. for i in s: From docs: All variable assignments in a function store the value in the local symbol table; whereas variable references first look in the local symbol table, then in the global symbol table, and then in the table of built-in names.
local variable feed referenced before the assignment at fo.write(column1[feed])#,column2[feed],urls[feed],'200','image created','/n') Any idea why? python; ... = val does not raise a "local variable referenced before assignment" error? The reason is that this is not a bare name assignment. Instead, it causes Python to make the function call dct ...
Applying a Cythonized function (like the string methods) to whole columns is generally far faster than looping over rows when the DataFrame is large. - unutbu Sep 13, 2017 at 19:49
The problem is that your assignment creates a local variable v, that shadows your global variable v. This happens at parse/load time. This happens at parse/load time. That's why python knows that you tried to use the local variable before giving it any value.
2. Because you are assigning x in the first case, python will assume x is a local variable (which is the default). In the second case you aren't assigning it so it will check the global scope. If you want this example to work, you have to pass the outer x into the inner function like so: def printHam(x=x): x = x+1. print x.
local variable 'encrypted' might be referenced before assignment . is a warning generated by the linter. This is because the linter sees that encrypted is assigned values inside two if conditions. if question.lower() == 'yes' or question.lower() == 'y':