unboundlocalerror local variable 'val1' referenced before assignment

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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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

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UnboundLocalError Local variable Referenced Before Assignment in Python

Handling errors is an integral part of writing robust and reliable Python code. One common stumbling block that developers often encounter is the “UnboundLocalError” raised within a try-except block. This error can be perplexing for those unfamiliar with its nuances but fear not – in this article, we will delve into the intricacies of the UnboundLocalError and provide a comprehensive guide on how to effectively use try-except statements to resolve it.

What is UnboundLocalError Local variable Referenced Before Assignment in Python?

The UnboundLocalError occurs when a local variable is referenced before it has been assigned a value within a function or method. This error typically surfaces when utilizing try-except blocks to handle exceptions, creating a puzzle for developers trying to comprehend its origins and find a solution.

Why does UnboundLocalError: Local variable Referenced Before Assignment Occur?

below, are the reasons of occurring “Unboundlocalerror: Try Except Statements” in Python :

Variable Assignment Inside Try Block

Reassigning a global variable inside except block.

• Accessing a Variable Defined Inside an If Block

In the below code, example_function attempts to execute some_operation within a try-except block. If an exception occurs, it prints an error message. However, if no exception occurs, it prints the value of the variable result outside the try block, leading to an UnboundLocalError since result might not be defined if an exception was caught.

In below code , modify_global function attempts to increment the global variable global_var within a try block, but it raises an UnboundLocalError. This error occurs because the function treats global_var as a local variable due to the assignment operation within the try block.

Solution for UnboundLocalError Local variable Referenced Before Assignment

Below, are the approaches to solve “Unboundlocalerror: Try Except Statements”.

Initialize Variables Outside the Try Block

Avoid reassignment of global variables.

In modification to the example_function is correct. Initializing the variable result before the try block ensures that it exists even if an exception occurs within the try block. This helps prevent UnboundLocalError when trying to access result in the print statement outside the try block.

Below, code calculates a new value ( local_var ) based on the global variable and then prints both the local and global variables separately. It demonstrates that the global variable is accessed directly without being reassigned within the function.

In conclusion , To fix “UnboundLocalError” related to try-except statements, ensure that variables used within the try block are initialized before the try block starts. This can be achieved by declaring the variables with default values or assigning them None outside the try block. Additionally, when modifying global variables within a try block, use the `global` keyword to explicitly declare them.

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[SOLVED] Local Variable Referenced Before Assignment

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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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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Python local variable referenced before assignment Solution

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

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

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What is unboundlocalerror: local variable referenced before assignment.

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

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

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

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

An Example Scenario

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

We start by declaring two variables:

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

Finally, we call our function:

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

Let’s run our code and see what happens:

An error has been raised.

The Solution

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

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

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

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

Let’s try to run our code again:

Our code successfully prints out the student’s grade.

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

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

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

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

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

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

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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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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!

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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:

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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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[Python] UnboundLocalErrorとは？発生原因や対処法・回避方法を解説

この記事では、Pythonにおける UnboundLocalError の定義や発生原因、対処法、回避方法、応用例について詳しく解説します。

エラーの理解を深めることで、より効率的なプログラミングが可能になります。

• UnboundLocalErrorの基本的な理解
• エラーが発生する具体的なシナリオ
• グローバル変数やローカル変数の正しい使い方
• nonlocalキーワードの活用方法
• 大規模プロジェクトやデータサイエンスにおける変数管理の重要性

UnboundLocalErrorとは?

Pythonにおける UnboundLocalError は、ローカルスコープ内で変数が参照される際に、その変数が初期化されていない場合に発生するエラーです。

このエラーは、特に関数内で変数を使用する際に注意が必要です。

ローカル変数が定義されていない状態でアクセスしようとすると、Pythonはこのエラーを発生させます。

UnboundLocalErrorの定義

UnboundLocalError は、Pythonの組み込み例外の一つで、ローカルスコープ内で変数が参照されるが、その変数が初期化されていない場合に発生します。

具体的には、関数内で変数を使用する際に、その変数がローカルとして扱われるが、実際には値が設定されていない場合にこのエラーが発生します。

UnboundLocalErrorの基本的な例

以下は、 UnboundLocalError が発生する基本的な例です。

このコードを実行すると、 UnboundLocalError: local variable 'value' referenced before assignment というエラーメッセージが表示されます。

これは、 value が関数内で初期化される前に参照されているためです。

UnboundLocalError は、他のエラーといくつかの点で異なります。

以下の表に、主な違いを示します。

このように、 UnboundLocalError は特にローカルスコープに関連するエラーであり、他のエラーとは異なる条件で発生します。

UnboundLocalErrorの発生原因

UnboundLocalError は、主に変数のスコープや初期化に関連する問題から発生します。

以下に、具体的な発生原因を詳しく解説します。

関数内で変数を参照する際、その変数がローカルスコープに存在しない場合、 UnboundLocalError が発生します。

Pythonは、関数内で変数が初期化される前にその変数を参照しようとすると、このエラーを投げます。

このコードでは、 number が初期化される前に参照されているため、エラーが発生します。

グローバル変数とローカル変数の混同

グローバル変数とローカル変数を混同することも、 UnboundLocalError の原因となります。

関数内で同名のローカル変数を定義すると、Pythonはその変数をローカルとして扱います。

これにより、グローバル変数が参照できなくなり、エラーが発生することがあります。

この場合、 value はローカル変数として扱われ、初期化される前に参照されるため、エラーが発生します。

関数内での変数の再定義

関数内で変数を再定義する場合も、 UnboundLocalError が発生することがあります。

特に、同じ関数内で変数を初期化する前にその変数を参照すると、エラーが発生します。

このコードでは、 counter が初期化される前に参照されているため、エラーが発生します。

関数のスコープと変数のライフタイム

Pythonでは、変数のスコープとライフタイムが重要です。

関数内で定義された変数は、その関数のスコープ内でのみ有効です。

関数が終了すると、ローカル変数は消失します。

このため、関数内で変数を参照する際には、必ず初期化されていることを確認する必要があります。

この場合、 temp は条件文内でのみ初期化されているため、条件が満たされない場合に参照されるとエラーが発生します。

スコープの理解は、 UnboundLocalError を回避するために非常に重要です。

UnboundLocalErrorの対処法

UnboundLocalError を回避するためには、変数のスコープや初期化に注意を払う必要があります。

以下に、具体的な対処法を解説します。

グローバル変数の正しい使用方法

グローバル変数を関数内で使用する場合、 global キーワードを使って明示的にその変数がグローバルであることを示すことが重要です。

これにより、ローカル変数として扱われることを防ぎます。

このコードでは、 global キーワードを使用することで、 value がグローバル変数として正しく参照され、エラーが発生しません。

関数内でローカル変数を使用する場合は、必ず初期化してから参照するようにしましょう。

これにより、 UnboundLocalError を回避できます。

このコードでは、 number が初期化されているため、エラーは発生しません。

関数の引数として変数を渡す

関数に変数を引数として渡すことで、ローカル変数の初期化を避けることができます。

これにより、関数内での変数のスコープを明確にし、エラーを防ぐことができます。

このように、引数を使用することで、 UnboundLocalError を回避できます。

nonlocalキーワードの使用

ネストされた関数内で外側の関数の変数を参照したい場合は、 nonlocal キーワードを使用します。

これにより、外側のスコープの変数をローカルとして扱うことができます。

このコードでは、 nonlocal を使用することで、 inner_function 内で outer_function の counter を正しく参照し、エラーを回避しています。

UnboundLocalErrorの回避方法

UnboundLocalError を回避するためには、コーディングスタイルや変数の管理方法に注意を払うことが重要です。

以下に、具体的な回避方法を解説します。

コーディングスタイルの改善

明確で一貫したコーディングスタイルを採用することで、変数のスコープや初期化の問題を減少させることができます。

以下のポイントに注意しましょう。

• 変数の初期化を必ず行う
• 変数のスコープを明確にする
• コメントを使って変数の役割を説明する

これにより、コードの可読性が向上し、エラーを未然に防ぐことができます。

変数名を一貫して使用することで、混乱を避けることができます。

特に、グローバル変数とローカル変数で同じ名前を使用しないようにしましょう。

以下のポイントを考慮してください。

• グローバル変数にはプレフィックスを付ける(例： global_value )
• ローカル変数には関数名や用途に基づいた名前を付ける(例： local_counter )

このようにすることで、変数のスコープを明確にし、 UnboundLocalError を回避できます。

関数の設計とスコープの管理

関数を設計する際には、変数のスコープを意識して管理することが重要です。

以下の点に注意しましょう。

• 変数は必要なスコープ内でのみ定義する
• 関数の引数を利用して外部のデータを渡す
• ネストされた関数を使用する場合は、 nonlocal や global を適切に使用する

これにより、変数のライフタイムを適切に管理し、エラーを防ぐことができます。

テストとデバッグの重要性

コードを書く際には、テストとデバッグを行うことが不可欠です。

以下の方法でエラーを早期に発見し、修正することができます。

• ユニットテストを作成して、関数の動作を確認する
• デバッグツールを使用して、変数のスコープや値を確認する
• エラーメッセージを注意深く読み、問題の原因を特定する

これにより、 UnboundLocalError を含むさまざまなエラーを未然に防ぐことができます。

テストとデバッグは、コードの品質を向上させるための重要なプロセスです。

UnboundLocalError を理解し、回避することは、さまざまなプロジェクトや分野で非常に重要です。

以下に、具体的な応用例を示します。

大規模プロジェクトでの変数管理

大規模なプロジェクトでは、複数の開発者が同じコードベースで作業するため、変数の管理が特に重要です。

以下のポイントに注意することで、 UnboundLocalError を回避できます。

• 明確な命名規則 : 変数名にプロジェクトのコンテキストを反映させることで、混乱を避ける。
• ドキュメンテーション : 変数の役割やスコープを文書化し、チーム全体で共有する。
• コードレビュー : 他の開発者によるレビューを通じて、潜在的なエラーを早期に発見する。

データサイエンスにおける変数のスコープ管理

データサイエンスのプロジェクトでは、データの前処理や分析を行う際に、変数のスコープを適切に管理することが重要です。

以下の方法でエラーを回避できます。

• 関数を利用したデータ処理 : データの前処理や分析を関数に分け、各関数内で変数を初期化する。
• 引数の活用 : データフレームや配列を関数の引数として渡し、スコープを明確にする。
• テストの実施 : 各関数の動作をテストし、エラーを早期に発見する。

Webアプリケーション開発でのエラー回避

Webアプリケーション開発では、ユーザーからの入力や外部APIとの連携が多いため、エラー処理が重要です。

• エラーハンドリング : 例外処理を適切に行い、 UnboundLocalError を含むエラーをキャッチする。
• スコープの明確化 : グローバル変数とローカル変数を明確に区別し、混同を避ける。
• ロギング : エラーが発生した際に、詳細なログを記録し、問題の特定を容易にする。

自動化スクリプトでのエラー防止

自動化スクリプトでは、エラーが発生すると処理が中断されるため、事前にエラーを防ぐことが重要です。

以下の方法で UnboundLocalError を回避できます。

• 変数の初期化 : スクリプトの冒頭で使用する変数を初期化し、参照エラーを防ぐ。
• 関数の利用 : 処理を関数に分け、各関数内で変数を管理することで、スコープを明確にする。
• テストの実施 : スクリプトを実行する前に、ユニットテストを行い、エラーを未然に防ぐ。

これらの応用例を通じて、 UnboundLocalError を理解し、適切に対処することが、さまざまなプロジェクトでの成功に繋がります。

UnboundLocalErrorが発生する具体的なシーンは？

UnboundLocalError は、関数内でローカル変数が初期化される前にその変数を参照しようとした場合に発生します。

ローカル変数は必ず初期化してから参照しましょう。

グローバル変数を使うべきか？

グローバル変数は便利ですが、使用する際には注意が必要です。

グローバル変数を使うべき場合は、以下のような状況です。

• 複数の関数で同じデータを共有する必要がある場合
• 状態を保持する必要がある場合

ただし、グローバル変数を多用すると、コードの可読性が低下し、バグの原因になることがあるため、必要最小限に留めることが推奨されます。

nonlocalキーワードの使い方は？

nonlocal キーワードは、ネストされた関数内で外側の関数の変数を参照する際に使用します。

以下のように使います。

例： nonlocal counter # 関数の外側で宣言されているcounter変数を参照

このように、 nonlocal を使うことで、外側のスコープの変数を正しく参照できます。

UnboundLocalError は、Pythonプログラミングにおいてよく見られるエラーであり、変数のスコープや初期化に関連しています。

この記事では、エラーの発生原因や対処法、回避方法、応用例について詳しく解説しました。

これらの知識を活用することで、エラーを未然に防ぎ、より良いコードを書くことができるでしょう。

ぜひ、実際のプロジェクトでこれらのポイントを意識してみてください。

当サイトはリンクフリーです。出典元を明記していただければ、ご自由に引用していただいて構いません。

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UnboundLocalError: local variable 'values1' referenced before assignment #818

caonetto commented Oct 19, 2022

hyphaltip commented Oct 19, 2022

Sorry, something went wrong.

nextgenusfs commented Oct 19, 2022

caonetto commented Oct 20, 2022

nextgenusfs commented Oct 20, 2022

caonetto commented Oct 21, 2022 • edited Loading

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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.

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Fixing UnboundLocalError: local variable 'name' referenced before assignment in python

Reading a text file with names and birthdays in a function printBook(), returning name, birthday and referencing it later on in another function I get:

UnboundLocalError: local variable 'name' referenced before assignment

• 4 What happens if there are no lines in the file? –  Chris Commented Feb 17, 2020 at 12:51
• What do you want it to return? The last name, month and date? –  Simon Crane Commented Feb 17, 2020 at 12:52
• @chris i've got a specific file i'm working on there so didn't think of preventing this –  stafino Commented Feb 17, 2020 at 13:22
• @SimonCrane yes return name, month, date that could be used later in other functions –  stafino Commented Feb 17, 2020 at 13:22

2 Answers 2

Although there is already an accepted answer, I believe it does not deal with the real issue in the question. Elton's answer will prevent the exception from being thrown, but will not render the desired output, from what I understood of the question.

The Real Issue

The issue here is that when calling the printBook() function inside the givenMonthPrintsElse() function, there is nothing left to read in the file, therefore leading to no line objects, which skips the for loop, returning three unassigned variables ( name , month and date ).

This happens because the with ... open statement is misplaced. Once the file object is read on the printBook() function, the cursor is at EOF (end of file), thus nothing left to be read. Calling again another function that requires the output of printBook() under the same with ... open statement results in empty output. You can confirm this by commenting out the first call to printBook() , and you should get the desired output.

Solution (okay)

The simplest way I can think that solves it is moving the with ... open statement inside the printBook() function, like this:

An important remark here is that, although it solves the issue, I suppose this is not the real intent of the OP, since it only allows to use the last line of the file (even though it reads through all of them). That is because name , month and date are overwritten at each iteration/line of the file, and only the last ones are returned by printBook() .

Solution (better)

So, as following, a simple structure allows to access any line of the file:

This way, in the book object you have a list of the content of all the lines, as strings, that afterwards you can split and process as desired. The printNameDate() function uses a list of strings (as the one returned by fileToList() and a line number (named entry )) to get the desired output.

You're declaring the name variable inside the for loop and accessing outside (in the return statement).

The easiest way to solve this is by declaring the variables before the for loop, so the printBook function would be something like:

PS: As @Chris pointed out, the problem isn't just because you're accessing the variables outside the loop (python accepts it), but because when the loop doesn't run, your variables aren't declared.

• 2 Actually, declaring the variables inside the loop works fine. Loops don't create a new scope. The problem is that if the loop doesn't run , the variables don't get created. –  Chris Commented Feb 17, 2020 at 13:23

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