How to fix UnboundLocalError: local variable 'x' referenced before assignment in Python
by Nathan Sebhastian
Posted on May 26, 2023
Reading time: 2 minutes
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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.
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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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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!
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Local variable referenced before assignment in Python
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Last updated: Apr 8, 2024 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 'bins_x_grouped' referenced before assignment unboundlocalerror local variable name referenced before assignment](https://bobbyhadz.com/images/blog/python-unboundlocalerror-local-variable-name-referenced-before-assignment/unboundlocalerror-local-variable-name-referenced-before-assignment.webp)
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.
![unboundlocalerror local variable 'bins_x_grouped' referenced before assignment mark variable as global](https://bobbyhadz.com/images/blog/python-unboundlocalerror-local-variable-name-referenced-before-assignment/mark-variable-as-global.webp)
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.
![unboundlocalerror local variable 'bins_x_grouped' referenced before assignment accessing global variables in functions](https://bobbyhadz.com/images/blog/python-unboundlocalerror-local-variable-name-referenced-before-assignment/accessing-global-variables-in-functions.webp)
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.
![unboundlocalerror local variable 'bins_x_grouped' referenced before assignment variables declared in function cannot be accessed in global scope](https://bobbyhadz.com/images/blog/python-unboundlocalerror-local-variable-name-referenced-before-assignment/variables-declared-in-function-cannot-be-accessed-in-global-scope.webp)
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.
![unboundlocalerror local variable 'bins_x_grouped' referenced before assignment return value from the function](https://bobbyhadz.com/images/blog/python-unboundlocalerror-local-variable-name-referenced-before-assignment/return-value-from-the-function.webp)
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.
![unboundlocalerror local variable 'bins_x_grouped' referenced before assignment pass global variable as argument to function](https://bobbyhadz.com/images/blog/python-unboundlocalerror-local-variable-name-referenced-before-assignment/pass-global-variable-as-argument-to-function.webp)
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.
![unboundlocalerror local variable 'bins_x_grouped' referenced before assignment assign value to local variable from outer scope](https://bobbyhadz.com/images/blog/python-unboundlocalerror-local-variable-name-referenced-before-assignment/assign-value-to-local-variable-from-outer-scope.webp)
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.
![unboundlocalerror local variable 'bins_x_grouped' referenced before assignment not using nonlocal prints empty string](https://bobbyhadz.com/images/blog/python-unboundlocalerror-local-variable-name-referenced-before-assignment/not-using-nonlocal-prints-empty-string.webp)
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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Python UnboundLocalError: local variable referenced before assignment
by Suf | Programming , Python , Tips
If you try to reference a local variable before assigning a value to it within the body of a function, you will encounter the UnboundLocalError: local variable referenced before assignment.
The preferable way to solve this error is to pass parameters to your function, for example:
Alternatively, you can declare the variable as global to access it while inside a function. For example,
This tutorial will go through the error in detail and how to solve it with code examples .
Table of contents
What is scope in python, unboundlocalerror: local variable referenced before assignment, solution #1: passing parameters to the function, solution #2: use global keyword, solution #1: include else statement, solution #2: use global keyword.
Scope refers to a variable being only available inside the region where it was created. A variable created inside a function belongs to the local scope of that function, and we can only use that variable inside that function.
A variable created in the main body of the Python code is a global variable and belongs to the global scope. Global variables are available within any scope, global and local.
UnboundLocalError occurs when we try to modify a variable defined as local before creating it. If we only need to read a variable within a function, we can do so without using the global keyword. Consider the following example that demonstrates a variable var created with global scope and accessed from test_func :
If we try to assign a value to var within test_func , the Python interpreter will raise the UnboundLocalError:
This error occurs because when we make an assignment to a variable in a scope, that variable becomes local to that scope and overrides any variable with the same name in the global or outer scope.
var +=1 is similar to var = var + 1 , therefore the Python interpreter should first read var , perform the addition and assign the value back to var .
var is a variable local to test_func , so the variable is read or referenced before we have assigned it. As a result, the Python interpreter raises the UnboundLocalError.
Example #1: Accessing a Local Variable
Let’s look at an example where we define a global variable number. We will use the increment_func to increase the numerical value of number by 1.
Let’s run the code to see what happens:
The error occurs because we tried to read a local variable before assigning a value to it.
We can solve this error by passing a parameter to increment_func . This solution is the preferred approach. Typically Python developers avoid declaring global variables unless they are necessary. Let’s look at the revised code:
We have assigned a value to number and passed it to the increment_func , which will resolve the UnboundLocalError. Let’s run the code to see the result:
We successfully printed the value to the console.
We also can solve this error by using the global keyword. The global statement tells the Python interpreter that inside increment_func , the variable number is a global variable even if we assign to it in increment_func . Let’s look at the revised code:
Let’s run the code to see the result:
Example #2: Function with if-elif statements
Let’s look at an example where we collect a score from a player of a game to rank their level of expertise. The variable we will use is called score and the calculate_level function takes in score as a parameter and returns a string containing the player’s level .
In the above code, we have a series of if-elif statements for assigning a string to the level variable. Let’s run the code to see what happens:
The error occurs because we input a score equal to 40 . The conditional statements in the function do not account for a value below 55 , therefore when we call the calculate_level function, Python will attempt to return level without any value assigned to it.
We can solve this error by completing the set of conditions with an else statement. The else statement will provide an assignment to level for all scores lower than 55 . Let’s look at the revised code:
In the above code, all scores below 55 are given the beginner level. Let’s run the code to see what happens:
We can also create a global variable level and then use the global keyword inside calculate_level . Using the global keyword will ensure that the variable is available in the local scope of the calculate_level function. Let’s look at the revised code.
In the above code, we put the global statement inside the function and at the beginning. Note that the “default” value of level is beginner and we do not include the else statement in the function. Let’s run the code to see the result:
Congratulations on reading to the end of this tutorial! The UnboundLocalError: local variable referenced before assignment occurs when you try to reference a local variable before assigning a value to it. Preferably, you can solve this error by passing parameters to your function. Alternatively, you can use the global keyword.
If you have if-elif statements in your code where you assign a value to a local variable and do not account for all outcomes, you may encounter this error. In which case, you must include an else statement to account for the missing outcome.
For further reading on Python code blocks and structure, go to the article: How to Solve Python IndentationError: unindent does not match any outer indentation level .
Go to the online courses page on Python to learn more about Python for data science and machine learning.
Have fun and happy researching!
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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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How to Fix Local Variable Referenced Before Assignment Error in Python
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Table of Contents
Fixing local variable referenced before assignment error.
In Python , when you try to reference a variable that hasn't yet been given a value (assigned), it will throw an error.
That error will look like this:
In this post, we'll see examples of what causes this and how to fix it.
Let's begin by looking at an example of this error:
If you run this code, you'll get
The issue is that in this line:
We are defining a local variable called value and then trying to use it before it has been assigned a value, instead of using the variable that we defined in the first line.
If we want to refer the variable that was defined in the first line, we can make use of the global keyword.
The global keyword is used to refer to a variable that is defined outside of a function.
Let's look at how using global can fix our issue here:
Global variables have global scope, so you can referenced them anywhere in your code, thus avoiding the error.
If you run this code, you'll get this output:
In this post, we learned at how to avoid the local variable referenced before assignment error in Python.
The error stems from trying to refer to a variable without an assigned value, so either make use of a global variable using the global keyword, or assign the variable a value before using it.
Thanks for reading!
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UndboundLocalError: local variable referenced before assignment
Hello all, I’m using PsychoPy 2023.2.3 Win 10 x64bits
![unboundlocalerror local variable 'bins_x_grouped' referenced before assignment image](https://global.discourse-cdn.com/business7/uploads/psychopy/original/3X/0/1/0151248d436b66d1e32c264550a4c46ec49ddc35.png)
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)
![unboundlocalerror local variable 'bins_x_grouped' referenced before assignment image](https://global.discourse-cdn.com/business7/uploads/psychopy/original/3X/d/3/d392f5974b0c159115d0a91155ef5b845945e2f3.png)
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?
![unboundlocalerror local variable 'bins_x_grouped' referenced before assignment image](https://global.discourse-cdn.com/business7/uploads/psychopy/optimized/3X/7/5/75f4b1745de0a83978131450c9e2c37a0221d2b1_2_690x387.png)
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.
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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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UnboundLocalError: local variable 'group' referenced before assignment #7
tiramisutes commented Apr 29, 2021
SunPengChuan commented Apr 29, 2021
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tiramisutes commented May 7, 2021
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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():
X implies passing dataframe without y included. UnboundLocalError: local variable 'bins_X_grouped' refere... I can work around this problem by passing the entire dataframe in... along w/ the y, but I'm not sure if this is expected behavior. ... UnboundLocalError: local variable 'bins_X_grouped' referenced before ... local variable 'bins_X ...
UnboundLocalError: local variable 'x' referenced before assignment [duplicate] Ask Question ... line 8, in findPoints results['north'] = (x,y) UnboundLocalError: local variable 'x' referenced before assignment I have tried global and nonlocal, but it does not work. Since I am not gettin any input from outside the function, so I anyways would ...
1. In python and almost all other programming languages, you cannot change the value in a variable unless you declare it. in your code: if exactShareNameFound: x+= line. you did not declare x, but you are referring to it in the above line. If you wanted to append the value of line to x, then declare a variable x before using it, like so:
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
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.
2 Solutions for the Problem. 2.1 Method 1: Initializing the Variable. 2.2 Method 2: Using Global Variables. 2.3 Method 3: Using Nonlocal Variables.
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.
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.
Finally I figured it out. It happens because of the shape of your target variable needs to be changed. The program is expecting a shape like this (len(data),) whereas the target shape using the code above generated a shape (len(data,1)).
There isn't a standard way to handle this situation. Common approaches are: 1. make sure that the variable is initialized in every code path (in your case: including the else case) 2. initialize the variable to some reasonable default value at the beginning. 3. return from the function in the code paths which cannot provide a value for the ...
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.
To prevent UnboundLocalError, the secret is in scope declaration. Declare a variable as global within a function if you're modifying a global variable. Alternatively, use nonlocal for variables in nested functions. python Global variable fix def func(): global var Hello, Global! var = 1 Changed it, see? Nested function fix def outer(): var = 0 def inner(): nonlocal var Outer! Lemme borrow this ...
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 ...
% e) gpu-jupyter_1 | UnboundLocalError: local variable '_' referenced before assignment Should this line work? The text was updated successfully, but these errors were encountered:
UndboundLocalError: local variable referenced before assignment. Coding. MarcelloSilvestre February 29, 2024, 12:17pm 1. Hello all, I'm using PsychoPy 2023.2.3 Win 10 x64bits. I am having a few issues in my experiment, some of the errors I never saw in older versions of Psychopy ... "UnboundLocalError: local variable 'os' referenced before ...
Avoid Reassignment of Global Variables. 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.
UnboundLocalError: local variable 'group' referenced before assignment #7. Closed tiramisutes opened this issue Apr 29, 2021 · 3 comments Closed UnboundLocalError: local variable 'group' referenced before assignment #7. ... local variable 'group' referenced before assignment ...
I have following simple function to get percent values for different cover types from a raster. It gives me following error: UnboundLocalError: local variable 'a' referenced before assignment whic...
UnboundLocalError: local variable 'player1_head' referenced before assignment from turtle import * from random import randint from utils import square, vector player1_xy = vector(-100, 0) player1_aim = vector(4, 0) player1_body = [] player1_head = "It looks like I'm assigning here."
Any variable which is changed or created inside of a function is local, if it hasn't been declared as a global variable. To tell Python, that we want to use the global variable, we have to explicitly state this by using the keyword "global".
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