unboundlocalerror local variable 'n' referenced before assignment

Local variable referenced before assignment in Python

Last updated: Feb 17, 2023 Reading time · 4 min

# Local variable referenced before assignment in Python

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

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

Here is an example of how the error occurs.

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

# Mark the variable as global to solve the error

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

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

# Local variables shadow global ones with the same name

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

Accessing the name variable in the function is perfectly fine.

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

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

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

# Returning a value from the function instead

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

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

# Passing the global variable as an argument to the function

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

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

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

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

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

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

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

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

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

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

# Discussion

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

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

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

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

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

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

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

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

# Additional Resources

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

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

Borislav Hadzhiev

Web Developer

Copyright © 2024 Borislav Hadzhiev

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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!

Next Article: Fixing Python TypeError: Descriptor 'lower' for 'str' Objects Doesn't Apply to 'dict' Object

Previous Article: Python TypeError: write() argument must be str, not bytes

Series: Common Errors in Python and How to Fix Them

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Local variable referenced before assignment in Python

The “local variable referenced before assignment” error occurs in Python when you try to use a local variable before it has been assigned a value.

This error typically arises in situations where you declare a variable within a function but then try to access or modify it before actually assigning a value to it.

Here’s an example to illustrate this error:

In this example, you would encounter the “local variable ‘x’ referenced before assignment” error because you’re trying to print the value of x before it has been assigned a value. To fix this, you should assign a value to x before attempting to access it:

In the corrected version, the local variable x is assigned a value before it’s used, preventing the error.

Keep in mind that Python treats variables inside functions as local unless explicitly stated otherwise using the global keyword (for global variables) or the nonlocal keyword (for variables in nested functions).

If you encounter this error and you’re sure that the variable should have been assigned a value before its use, double-check your code for any logical errors or typos that might be causing the variable to not be assigned properly.

Using the global keyword

If you have a global variable named letter and you try to modify it inside a function without declaring it as global, you will get error.

This is because Python assumes that any variable that is assigned a value inside a function is a local variable, unless you explicitly tell it otherwise.

To fix this error, you can use the global keyword to indicate that you want to use the global variable:

Using nonlocal keyword

The nonlocal keyword is used to work with variables inside nested functions, where the variable should not belong to the inner function. It allows you to modify the value of a non-local variable in the outer scope.

For example, if you have a function outer that defines a variable x , and another function inner inside outer that tries to change the value of x , you need to use the nonlocal keyword to tell Python that you are referring to the x defined in outer , not a new local variable in inner .

Here is an example of how to use the nonlocal keyword:

If you don’t use the nonlocal keyword, Python will create a new local variable x in inner , and the value of x in outer will not be changed:

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How to Fix – UnboundLocalError: Local variable Referenced Before Assignment in Python

Developers often encounter the  UnboundLocalError Local Variable Referenced Before Assignment error in Python. In this article, we will see what is local variable referenced before assignment error in Python and how to fix it by using different approaches.

What is UnboundLocalError: Local variable Referenced Before Assignment?

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

Below, are the reasons by which UnboundLocalError: Local variable Referenced Before Assignment error occurs in  Python :

Nested Function Variable Access

Global variable modification.

In this code, the outer_function defines a variable ‘x’ and a nested inner_function attempts to access it, but encounters an UnboundLocalError due to a local ‘x’ being defined later in the inner_function.

In this code, the function example_function tries to increment the global variable ‘x’, but encounters an UnboundLocalError since it’s treated as a local variable due to the assignment operation within the function.

Solution for Local variable Referenced Before Assignment in Python

Below, are the approaches to solve “Local variable Referenced Before Assignment”.

In this code, example_function successfully modifies the global variable ‘x’ by declaring it as global within the function, incrementing its value by 1, and then printing the updated value.

In this code, the outer_function defines a local variable ‘x’, and the inner_function accesses and modifies it as a nonlocal variable, allowing changes to the outer function’s scope from within the inner function.

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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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UnboundLocalError: local variable 'trialList' referenced before assignment

OS (e.g. Win10): mac 10.15.4 PsychoPy version (e.g. 1.84.x): 2020.1.2 Standard Standalone? (y/n) If not then what?: y What are you trying to achieve?:

I have this old exp that I can still run [github link removed after problem solved ]. I just refactored it: the stimuli are slightly different, but I meant to use the same paradigm [github link removed after problem solved ]. I copied the .psyexp file over, and when I run it, psychopy crushes after instruction presentation and gives me

I am stumped, waiting for help.

This happens for a reason: probably I did something different, but the error message doesn’t make any sense.

So I pip installed psychopy2020.1.2 into my conda python3 env and ran the script from the terminal. Crash, then correct error messages are shown: I fixed it and the script is running now (from both terminal and runner). The runner is not showing correct error messages, a psychopy bug?

I’m currently receiving the same error message about ‘trialList’ being referenced before assignment. What did you do to fix it? Any help would be much appreciated.

I had the same error and found out that it was caused by an oversight in the csv file, which was generated by another script:

There were leading spaces on the column names. After I removed them, things worked.

Hope that helps

CSV file has no spaces or other marking. I haven’t used any other script so far to have interference with this. I would appreciate any help. Thank you in advance!

Hi @ar2 I found a partial solution to this. Copying the cell data into a new csv file and overwriting the existing file solved this error for me.

Though it’s still not clear why this problem is occurring.

Hi, I got the same error. I checked the .cvs: two columns had the same name. I changed them, made sure that all column names differed and the problem got solved. Hope this could help!

Hello, I was getting the same error when I was trying to add my .csv conditions file to a loop. Referring to some of the above answers, I ensured that there were no spaces in my column names. It turns out that Psychopy didn’t like that one of my column names had a slash (I was trying to define a variable as old/new). But I generally use underscore a lot in my column names to make it easier during data cleaning and analysis and that hasn’t given me any errors in Psychopy so far. Hope this helps others who are getting a similar error!

This worked!! thank you!

Hello, I need help with the same issue too!

I tried every suggestion mentioned here but I am still having the same error. I also tried it with .csv format, with numbers under the index column, recreating the excel table, writing choice without the ‘’

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UnboundLocalError: local variable 'cam_image' referenced before assignment #1

Markson-Young commented Dec 18, 2023

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码龄 粉丝数 原力等级 --

训练时遇到UnboundLocalError: local variable 'output_proposals' referenced before assignment

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我在训练时，使用单阶段模型训练，如下

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9 条回答 默认 最新

🕒下午好，题主！目前：🟢[ 在线 ] 📢参考通义千问、由郭老师整理提供！ 📣今日诗号：桃李春风一杯酒，江 湖夜雨十年灯。

根据您提供的代码片段和错误信息 "UnboundLocalError: local variable 'output_proposals' referenced before assignment"，问题在于当 self.two_stage 为 False 时， output_proposals 变量没有被赋值，但在最后返回时却被引用。

要解决这个问题，你需要确保在两种情况下（即 self.two_stage 为 True 和 False ）都对 output_proposals 进行初始化或赋值。由于这段代码似乎是 Deformable DETR 模型的一部分，而在单阶段模式下可能不需要 output_proposals ，因此可以在非两阶段模式下为其赋予一个合理的默认值或者直接移除不必要的返回项。

示例解决方案：

请注意，我在这里假设了一个默认值 torch.zeros(...) ，你需要根据实际情况替换为合适的默认输出。如果单阶段模式下确实不需要 output_proposals ，那么在返回时不包含它即可。

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PythonでUnboundLocalErrorに遭遇したら

データアナリティクス事業本部 機械学習チームの中村( nokomoro3 )です。

本記事ではPythonの変数スコープと、 UnboundLocalError について整理したいと思います。

Pythonでは、関数外の変数であっても、関数内で以下のように参照することができます。

上記のようにインデントの最も浅い部分に定義された変数は、グローバル変数となります。

以下のコードでそれをチェックしてみましょう。

var1はグローバル変数であり、オブジェクトIDも同じものとなっていることが分かります。

グローバル変数を書き換えようとすると

ただし、例えば以下のように参照しつつ再代入しようとするとエラーとなります。

これは関数内で var1 がローカル変数と見なされるようになるためです。

こういった UnboundLocalError に遭遇する機会は意外と経験があることかと思います。

ローカル変数を定義しなおすと

こちらは関数内で変数を定義しなおすとエラーがなくなります。

しかしこれは元のグローバル変数を変更するわけではありません。

グローバル変数は書き変わっておらず、オブジェクトIDも関数内と関数外で異なることが分かります。

つまり、グローバル変数と同じ変数名のものを、ローカル変数として関数内で定義すると、ローカル変数と見なされていることが分かります。

ここまでの話を整理すると

先ほどのエラーとなるコードでは var1 = var1 + "_結合したい文字列" という一行により、ローカル変数としての定義と参照を同時に行っています。

また定義前に参照が実行されるため、 UnboundLocalError が発生しているということとなります。

つまり、グローバル変数と同じ名前の変数を関数内でローカル変数として使用すると、ローカル変数と見なされ、ローカル変数と見なされたものを、定義前に参照しようとすることでエラーとなっています。

よって以下のようなコードも、もちろんエラーとなります。

グローバル変数を書き換えるには

先ほどのコードではグローバル変数は書き換えができなかったですが、グローバル変数を書き換えたい場合は（良いかどうかは置いておいて）、以下のように global 変数名 と書くことで実現できます。

ちなみに、最初とオブジェクトIDが変わるのは再代入により新しい変数となっているだけで、str型がimmutableなためであるため、今回の変数スコープの話とは無関係となります。

またmutableなデータ型の場合、 global 変数名 などと記述しなくとも参照さえできれば書き換えが可能となります。

（mutable、immutableについては機会があれば別途記事にしようと思います）

関数内の関数の場合もエラーとなりうる

これらの挙動はグローバル変数だけではなく、関数内の関数で、その親関数の変数を参照する際にも発生します。

wrapper内で定義された変数が、その中の関数でも変数として定義されているため、未定義前に参照しに行くこととなり、先ほどと同様に UnboundLocalError となることがわかります。

親関数の変数を書き換えたい場合はnonlocal

このような場合に、親関数の変数を書き換えたい時は、 global 変数 の代わりに nonlocal 変数 を記載する必要があります。

関数内の関数内の関数でnonlocalを使うと？

では関数内の関数内の関数でnonlocalを使うとどうなるか？気になるところですが、これはひとつ外側の変数を取ってくるという挙動のようです。

いかがでしたでしょうか。本ブログがPythonの変数スコープや UnboundLocalError に遭遇した方の参考になれば幸いです。

EVENT 【4/30（火）リモート】クラスメソッドの会社説明会を開催します

Event 【5/9（木）】組織的なクラウド統制 はじめの一歩, event 【4/24（水）リモート】フリーランストーク#10 -フリーランスとしてのキャリアを成功させるアイデア＆ヒント- を開催します, event 【4/2, 4/9リモート】クラスメソッドの会社説明会 〜フリーランスエンジニア/データ分析系案件特集〜 を開催します, event 【4/18（木）東京】gitlabで実現！初級者向けci/cd実践ハンズオン, event 【毎週開催】メソドロジック社共催！イチから始めるデータ活用！8週連続ウェビナー, event 【5/15（水）】classmethod showcase 事例で学ぶ、データ活用戦略の最新動向, event 【4/16（火）】snowflakeを触ってみよう！初めての方向けハンズオンセミナー, event 【4/10（水）リモート】クラスメソッドの会社説明会を開催します, event 【4/17（水）】初心者向け 問い合わせ対応業務をカンタンにする「zendesk for service」とは？, pytest-mockでモックを使ってみる.

kobayashi.m

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