time.sleep(1) unboundlocalerror local variable 'time' referenced before assignment

[SOLVED] Local Variable Referenced Before Assignment

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

Why Does This Error Occur?

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

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

Local Variable Declarations vs. Global Variable Declarations

Local Variable Referenced Before Assignment Error with Explanation

Try these examples yourself using our Online Compiler.

Let’s look at the following function:

Explanation

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

Using Global Variables

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

Create Functions that Take in Parameters

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

UnboundLocalError: local variable ‘DISTRO_NAME’

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

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

Try and update your Anaconda Navigator with the following command.

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

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

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

DISTRO_NAME = None

Save the file and re-launch Anaconda Navigator.

DJANGO – Local Variable Referenced Before Assignment [Form]

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

Upon running you get the following error:

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

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

Why does the error occur?

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

Local variable Referenced before assignment but it is global

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

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

Here’s an example to help illustrate the problem:

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

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

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

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

Local variable ‘version’ referenced before assignment ubuntu-drivers

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

Here, p_name means package name.

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

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

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

Trending Python Articles

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

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

How to reproduce this error

How to fix this error.

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

Take your skills to the next level ⚡️

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)

Local variable referenced before assignment in Python

Last updated: Apr 8, 2024 Reading time · 4 min

# Local variable referenced before assignment in Python

The Python "UnboundLocalError: Local variable referenced before assignment" occurs when we reference a local variable before assigning a value to it in a function.

To solve the error, mark the variable as global in the function definition, e.g. global my_var .

Here is an example of how the error occurs.

We assign a value to the name variable in the function.

# Mark the variable as global to solve the error

To solve the error, mark the variable as global in your function definition.

If a variable is assigned a value in a function's body, it is a local variable unless explicitly declared as global .

# Local variables shadow global ones with the same name

You could reference the global name variable from inside the function but if you assign a value to the variable in the function's body, the local variable shadows the global one.

Accessing the name variable in the function is perfectly fine.

On the other hand, variables declared in a function cannot be accessed from the global scope.

The name variable is declared in the function, so trying to access it from outside causes an error.

Make sure you don't try to access the variable before using the global keyword, otherwise, you'd get the SyntaxError: name 'X' is used prior to global declaration error.

# Returning a value from the function instead

An alternative solution to using the global keyword is to return a value from the function and use the value to reassign the global variable.

We simply return the value that we eventually use to assign to the name global variable.

# Passing the global variable as an argument to the function

You should also consider passing the global variable as an argument to the function.

We passed the name global variable as an argument to the function.

If we assign a value to a variable in a function, the variable is assumed to be local unless explicitly declared as global .

# Assigning a value to a local variable from an outer scope

If you have a nested function and are trying to assign a value to the local variables from the outer function, use the nonlocal keyword.

The nonlocal keyword allows us to work with the local variables of enclosing functions.

Had we not used the nonlocal statement, the call to the print() function would have returned an empty string.

Printing the message variable on the last line of the function shows an empty string because the inner() function has its own scope.

Changing the value of the variable in the inner scope is not possible unless we use the nonlocal keyword.

Instead, the message variable in the inner function simply shadows the variable with the same name from the outer scope.

# Discussion

As shown in this section of the documentation, when you assign a value to a variable inside a function, the variable:

• Becomes local to the scope.
• Shadows any variables from the outer scope that have the same name.

The last line in the example function assigns a value to the name variable, marking it as a local variable and shadowing the name variable from the outer scope.

At the time the print(name) line runs, the name variable is not yet initialized, which causes the error.

The most intuitive way to solve the error is to use the global keyword.

The global keyword is used to indicate to Python that we are actually modifying the value of the name variable from the outer scope.

• If a variable is only referenced inside a function, it is implicitly global.
• If a variable is assigned a value inside a function's body, it is assumed to be local, unless explicitly marked as global .

If you want to read more about why this error occurs, check out [this section] ( this section ) of the docs.

# Additional Resources

You can learn more about the related topics by checking out the following tutorials:

• SyntaxError: name 'X' is used prior to global declaration

Borislav Hadzhiev

Web Developer

Copyright © 2024 Borislav Hadzhiev

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

Series: Common Errors in Python and How to Fix Them

Related Articles

• Python Warning: Secure coding is not enabled for restorable state
• 4 ways to install Python modules on Windows without admin rights
• Python TypeError: object of type ‘NoneType’ has no len()
• Python: How to access command-line arguments (3 approaches)
• Understanding ‘Never’ type in Python 3.11+ (5 examples)
• Python: 3 Ways to Retrieve City/Country from IP Address
• Using Type Aliases in Python: A Practical Guide (with Examples)
• Python: Defining distinct types using NewType class
• Using Optional Type in Python (explained with examples)
• Python: How to Override Methods in Classes
• Python: Define Generic Types for Lists of Nested Dictionaries
• Python: Defining type for a list that can contain both numbers and strings

Search tutorials, examples, and resources

• PHP programming
• Symfony & Doctrine
• Laravel & Eloquent
• Tailwind CSS
• Sequelize.js
• Mongoose.js

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 *

• Python Course
• Python Basics
• Interview Questions
• Python Quiz
• Popular Packages
• Python Projects
• Practice Python
• AI With Python
• Learn Python3
• Python Automation
• Python Web Dev
• DSA with Python
• Python OOPs
• Dictionaries

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.

Please Login to comment...

Similar reads.

• Python Programs
• Python Errors
• Best External Hard Drives for Mac in 2024: Top Picks for MacBook Pro, MacBook Air & More
• How to Watch NFL Games Live Streams Free
• OpenAI o1 AI Model Launched: Explore o1-Preview, o1-Mini, Pricing & Comparison
• How to Merge Cells in Google Sheets: Step by Step Guide
• #geekstreak2024 – 21 Days POTD Challenge Powered By Deutsche Bank

Improve your Coding Skills with Practice

What kind of Experience do you want to share?

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

• Password Generator
• HTML Editor
• HTML Encoder
• JSON Beautifier
• CSS Beautifier
• Markdown Convertor
• Find the Closest Tailwind CSS Color
• Phrase encrypt / decrypt
• Browser Feature Detection
• Number convertor
• CSS Maker text shadow
• CSS Maker Text Rotation
• CSS Maker Out Line
• CSS Maker RGB Shadow
• CSS Maker Transform
• CSS Maker Font Face
• Color Picker
• Colors CMYK
• Color mixer
• Color Converter
• Color Contrast Analyzer
• Color Gradient
• String Length Calculator
• MD5 Hash Generator
• Sha256 Hash Generator
• String Reverse
• URL Encoder
• URL Decoder
• Base 64 Encoder
• Base 64 Decoder
• Extra Spaces Remover
• String to Lowercase
• String to Uppercase
• Word Count Calculator
• Empty Lines Remover
• HTML Tags Remover
• Binary to Hex
• Hex to Binary
• Rot13 Transform on a String
• String to Binary
• Duplicate Lines Remover

Python 3: UnboundLocalError: local variable referenced before assignment

This error occurs when you are trying to access a variable before it has been assigned a value. Here is an example of a code snippet that would raise this error:

Watch a video course Python - The Practical Guide

The error message will be:

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.

Both will work without any error.

Related Resources

• Using global variables in a function
• "Least Astonishment" and the Mutable Default Argument
• Why is "1000000000000000 in range(1000000000000001)" so fast in Python 3?
• HTML Basics
• Javascript Basics
• TypeScript Basics
• React Basics
• Angular Basics
• Sass Basics
• Vue.js Basics
• Python Basics
• Java Basics
• NodeJS Basics

UndboundLocalError: local variable referenced before assignment

Hello all, I’m using PsychoPy 2023.2.3 Win 10 x64bits

What I’m trying to do? The experiment will show in the middle of the screen an abstracted stimuli (B1 or B2), and after valid click on it, the stimulus will remain on the middle of the screen and three more stimuli will appear in the cornor of the screen.

I’m having this erro (attached above), a simple error, but I can not see where the error is. Also the experiment isn’t working proberly and is the old version (I don’t know but someone are having troubles with this version of PscyhoPy)? ba_training_block.xlsx (13.8 KB) SMTS.psyexp (91.6 KB) stimuli, instructions and parameters.xlsx (12.8 KB)

You have a routine called sample but you also use that name for your image file in sample_box .

I changed the name of the routine for ‘stimulus_sample’ and manteined the image file in sample_box as ‘sample’. But, the error still remain. But it do not happen all the time, this is very interesting…

Can u give it a look again? (I made some minor changes here)

Here the exp file ba_training_block.xlsx (13.7 KB) SMTS.psyexp (89.7 KB) stimuli, instructions and parameters.xlsx (12.8 KB)

Thanks again

Please could you confirm/show the new error message? Is it definitely still related to sample?

I think you have blank rows in your spreadsheet. The loop claims that there are 19 conditions but I think you only want 12. Without a value for sample_category sample doesn’t get set. With random presentation this will happen at a random point.

Related Topics

• All Communities Products ArcGIS Pro ArcGIS Survey123 ArcGIS Online ArcGIS Enterprise Data Management Geoprocessing ArcGIS Web AppBuilder ArcGIS Experience Builder ArcGIS Dashboards ArcGIS Spatial Analyst ArcGIS CityEngine All Products Communities Industries Education Water Resources State & Local Government Transportation Gas and Pipeline Water Utilities Roads and Highways Telecommunications Natural Resources Electric Public Safety All Industries Communities Developers Python JavaScript Maps SDK Native Maps SDKs ArcGIS API for Python ArcObjects SDK ArcGIS Pro SDK Developers - General ArcGIS REST APIs and Services ArcGIS Online Developers File Geodatabase API Game Engine Maps SDKs All Developers Communities Global Comunidad Esri Colombia - Ecuador - Panamá ArcGIS 開発者コミュニティ Czech GIS ArcNesia Esri India GeoDev Germany ArcGIS Content - Esri Nederland Esri Italia Community Comunidad GEOTEC Esri Ireland Používatelia ArcGIS All Global Communities All Communities Developers User Groups Industries Services Community Resources Global Events Learning Networks ArcGIS Topics GIS Life View All Communities
• ArcGIS Ideas
• Community Resources Community Help Documents Community Blog Community Feedback Member Introductions Community Ideas All Community Resources
• All Communities
• local variable 'row' referenced before assignment ...
• GeoNet, The Esri Community - STAGE
• ArcGIS AppStudio
• ArcGIS AppStudio Questions
• Subscribe to RSS Feed
• Mark Topic as New
• Mark Topic as Read
• Float this Topic for Current User
• Printer Friendly Page

local variable 'row' referenced before assignment

• Mark as New
• Report Inappropriate Content

Solved! Go to Solution.

• Previous Topic

View solution in original post

• Terms of Use
• Community Guidelines
• Community Resources
• Contact Community Team
• Trust Center
• Contact Esri

Navigation Menu

Search code, repositories, users, issues, pull requests..., provide feedback.

We read every piece of feedback, and take your input very seriously.

Saved searches

Use saved searches to filter your results more quickly.

To see all available qualifiers, see our documentation .

• Notifications You must be signed in to change notification settings

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement . We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

[Re-appeared] DPO training error UnboundLocalError: local variable 'num_patches' referenced before assignment #2012

Lopa07 commented Sep 11, 2024

Jintao-Huang commented Sep 12, 2024

• 👍 1 reaction

Sorry, something went wrong.

No branches or pull requests

• Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers
• Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand
• OverflowAI GenAI features for Teams
• OverflowAPI Train & fine-tune LLMs
• Labs The future of collective knowledge sharing
• About the company Visit the blog

Collectives™ on Stack Overflow

Find centralized, trusted content and collaborate around the technologies you use most.

Q&A for work

Connect and share knowledge within a single location that is structured and easy to search.

Get early access and see previews of new features.

Unboundlocalerror: local variable ' pulse_start' is referenced before assignment

It stated that unboundlocalerror: local variable 'pulse_start' referenced before assignment . I run it previously and so sudden error happen.

• What do you expect to happen if GPIO.input(ECHO) never equals 0 or 1? –  DavidG Commented Apr 23, 2018 at 8:17

2 Answers 2

You need to declare your variables before the while loop, like so:

you need to declare pulse_start and pulse_end first just in case you don't go in the loop

• 2 pulse_start will be accessible after the end of the loop. –  Colin Ricardo Commented Apr 23, 2018 at 8:26
• 1 ...But won't be assigned if you never enter the loop (i.e. if GPIO.input(ECHO) does not equal 0), as stated in the error message... –  DavidG Commented Apr 23, 2018 at 8:27

Your Answer

Reminder: Answers generated by artificial intelligence tools are not allowed on Stack Overflow. Learn more

Sign up or log in

Post as a guest.

Required, but never shown

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy .

Not the answer you're looking for? Browse other questions tagged python python-2.7 sensors or ask your own question .

• The Overflow Blog
• Looking under the hood at the tech stack that powers multimodal AI
• Featured on Meta
• Join Stack Overflow’s CEO and me for the first Stack IRL Community Event in...
• User activation: Learnings and opportunities
• What does a new user need in a homepage experience on Stack Overflow?
• Announcing the new Staging Ground Reviewer Stats Widget

Hot Network Questions

• How am I supposed to solder this tiny component with pads UNDER it?
• Why a relay frequently clicks when a battery is low?
• PCB design references and roadmap
• Is it possible to monitor the current drawn by a computer from an outlet on the computer?
• Is it possible to make sand from bones ? Would it have the same properties as regular sand?
• How can I prove that this expression defines the area of the quadrilateral?
• Horror short film about a guy trying to test a VR game with spiders in a house. He wakes up and realizes the game hasn't started
• Writing in first person for fiction novel, how to portray her inner dialogue and drag it out to make a chapter long enough?
• Removing undermount sink
• My one-liner 'delete old files' command finds the right files but will not delete them
• What early 60s puppet show similar to fireball XL5 used the phrase "Meson Power?"
• Why is Germany looking to import workers from Kenya, specifically?
• A function to convert numbers from scientific notation to plain decimal
• Why did mire/bog skis fall out of use?
• If a mount provokes opportunity attacks, can its rider be targeted?
• "First et al.", many authors with same surname, and IEEE citations
• How to assign a definition locally?
• Model looks dented but geometry is correct
• How can "chemical-free" surface cleaners work?
• Change of variable, u = y(x)
• Is it ethical to request partial reimbursement to present something from my previous position?
• What's "jam" mean in "The room reeled and he jammed his head down" (as well as the sentence itself)?
• How to identify and uninstall packages installed as dependencies during custom backport creation?
• Returning to the US for 2 weeks after a short stay around 6 months prior with an ESTA but a poor entry interview - worried about visiting again