unboundlocalerror local variable 'http' referenced before assignment

[SOLVED] Local Variable Referenced Before Assignment

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

Why Does This Error Occur?

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

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

Local Variable Declarations vs. Global Variable Declarations

Local Variable Referenced Before Assignment Error with Explanation

Try these examples yourself using our Online Compiler.

Let’s look at the following function:

Explanation

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

Using Global Variables

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

Create Functions that Take in Parameters

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

UnboundLocalError: local variable ‘DISTRO_NAME’

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

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

Try and update your Anaconda Navigator with the following command.

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

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

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

DISTRO_NAME = None

Save the file and re-launch Anaconda Navigator.

DJANGO – Local Variable Referenced Before Assignment [Form]

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

Upon running you get the following error:

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

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

Why does the error occur?

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

Local variable Referenced before assignment but it is global

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

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

Here’s an example to help illustrate the problem:

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

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

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

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

Local variable ‘version’ referenced before assignment ubuntu-drivers

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

Here, p_name means package name.

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

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

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

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Local variable referenced before assignment in Python

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

# Local variable referenced before assignment in Python

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

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

Here is an example of how the error occurs.

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

# Mark the variable as global to solve the error

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

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

# Local variables shadow global ones with the same name

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

Accessing the name variable in the function is perfectly fine.

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

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

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

# Returning a value from the function instead

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

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

# Passing the global variable as an argument to the function

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

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

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

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

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

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

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

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

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

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

# Discussion

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

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

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

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

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

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

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

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

# Additional Resources

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

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

Borislav Hadzhiev

Web Developer

Copyright © 2024 Borislav Hadzhiev

Python UnboundLocalError: local variable referenced before assignment

by Suf | Programming , Python , Tips

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

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

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

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

Table of contents

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

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

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

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

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

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

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

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

Example #1: Accessing a Local Variable

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

Let’s run the code to see what happens:

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

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

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

We successfully printed the value to the console.

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

Let’s run the code to see the result:

Example #2: Function with if-elif statements

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

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

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

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

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

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

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

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

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

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

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

Have fun and happy researching!

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

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.

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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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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.

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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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Understanding UnboundLocalError in Python

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

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

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

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

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

The simple answer

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

Let's take the first code snippet again:

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

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

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

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

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

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

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

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

The formal answer

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

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

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

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

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

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

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

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

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

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

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

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

The "too much information" answer

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

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

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

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

This prints:

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

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

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

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

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

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

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

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UnboundLocalError: local variable 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

which isn't clear to me. Any suggestions?

• arcgis-10.1
• unboundlocalerror

• 1 Because if row.getValue("Value") == 1 might be false and so a never gets assigned. –  Nathan W Commented May 20, 2013 at 2:39
• It has value and do gets assigned. I checked it in arcmap interactive python window but can't get it to work in a stand alone script. –  Ibe Commented May 20, 2013 at 2:44
• 1 your loop will also only give you the values of the last loop iteration as you are returning out of the loop and not doing anything with each value. –  Nathan W Commented May 20, 2013 at 2:49
• You could use 3 x elif and an else to see if any values other than 1-4 are encountered. –  PolyGeo ♦ Commented May 20, 2013 at 3:44
• I tried that way as well but still hung up with error. –  Ibe Commented May 20, 2013 at 4:15

This error is pretty much explained here and it helped me to get assignments and return values for all variables.

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having a problem with error code: UnboundLocalError: local variable 'player' referenced before assignment

hi hoping someone here can actually help me. I asked this question on stack overflow and they closed it and redirected to similar previously asked questions. none of which actually helped me to understand how to correct the error code.

I am following a video tutorial series for learning to code with python/pygame. and i have checked my coding aginst the source material and they are the same, because i typed it as was told to in the video.

here's the full tracebackback including the error:

File "/home/dev/PycharmProjects/game7_meteor_game/game7_meteor_game.py", line 160, in <module>

File "/home/dev/PycharmProjects/game7_meteor_game/game7_meteor_game.py", line 152, in main

GameInterface(num_player=1, screen=screen)

File "/home/dev/PycharmProjects/game7_meteor_game/game7_meteor_game.py", line 82, in GameInterface

if player.cooling_time > 0:

UnboundLocalError: local variable 'player' referenced before assignment.

here's the actual code:

if player.cooling_time > 0: player.cooling_time -= 1

UnboundLocalError in line 384 & frozen participant screen

OS: Win 10 Psychopy: v1.5 Error Message: File “D:\separate_components_lastrun.py”, line 384, in run text_L_C = visual.TextStim(win=win, name=‘text_L_C’, UnboundLocalError: local variable ‘visual’ referenced before assignment ################# Experiment ended with exit code 1 [pid:9836] ################# Previous attempts: established the visual.textStim earlier in the code (got a syntax error). Tried to just use builder to make the text boxes and ended up with code talking about the visual.textstim.

I don’t know what to do to get this block of the code to work reliably. I was able to see the participant screen once but was not able to use the keyboard arrows to respond to the prompt. separate_components_lastrun.py (40.9 KB) separate_components.psyexp (104.9 KB)

You have coder code which should be removed for use in Builder. For example, all Builder experiments automatically import a range of libraries.

You shouldn’t therefore import visual, especially not 33 times (end routine within a loop of 3 repetitions of a spreadsheet with 11 conditions).

Also, PsychoPy will already be running in a window called win so when you execute win = visual.Window() you will break the current window. Are you trying to use two windows?

Here is a thread about how to do that

Thank you so much for sending the thread about it. I’ll look into it asap!

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Python UnboundLocalError: local variable referenced before assignment

I'm confused. What is different about player1_head compared to the other variables I am printing in the code below? As far as I can tell it should behave the same as the others - it's declared in the global scope, no? I don't think it's a typo.

UnboundLocalError: local variable 'player1_head' referenced before assignment

• global-variables

• Is this the complete code? –  Iain Shelvington Commented Aug 3, 2019 at 21:04
• Does this answer your question? UnboundLocalError on local variable when reassigned after first use –  Tomerikoo Commented Feb 10, 2021 at 20:30

2 Answers 2

Because you failed to declare player1_head as a global , in the draw() function it appears to that function that you're printing out local variable player1_head before it has a value:

Instead do:

• 1 No "appears" about it; the presence of an assignment makes player1_head a local variable throughout the scope where the assignment takes place. Locality is not determined on a per-statement basis. –  chepner Commented Aug 3, 2019 at 21:24
• @chepner, I wasn't expressing doubt , I meant it appears to the draw() function that player1_head is local, not that it appears to me. I'll clarify the text. –  cdlane Commented Aug 3, 2019 at 21:50
• But why only that variable needs declaring as global inside the function, not the others? –  Robin Andrews Commented Aug 3, 2019 at 22:40
• @Robin, only that variable needs to be declared global because it is being reset to a new value, the other variable are only being examined or having their methods called. Read up on the Python global keyword -- it is often misunderstood. –  cdlane Commented Aug 4, 2019 at 3:10

The assignment player1_head = player1_xy.copy() in the draw() function is saying to Python that the variable player1_head is a local variable to the function draw() , and since print("head: ", player1_head) is referencing a local variable before its assignment, the error is shown. You can fix this by using player1_head as a global variable (since you're modifying it, same goes for the variable player1_body , since you're doing player1_body.append(player1_head) ), like so:

Note however that you should avoid using global variables when possible, this is one the problems that arises from using them (they can be modified by any function, which can sometimes lead to errors and confusions).

• So even though the print statement comes before the assignment of player1_head in the function, it is still treated as local even though it exists in the global scope? This seems like non-obvious behaviour to me. Something about the order in which the interpreter processes statements within a function? –  Robin Andrews Commented Aug 3, 2019 at 22:44

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UnboundLocalError: local variable 'Normalizer' referenced before assignment #173

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