typeerror 'method' object does not support item assignment

TypeError: builtin_function_or_method object is not subscriptable Python Error [SOLVED]

As the name suggests, the error TypeError: builtin_function_or_method object is not subscriptable is a “typeerror” that occurs when you try to call a built-in function the wrong way.

When a "typeerror" occurs, the program is telling you that you’re mixing up types. That means, for example, you might be concatenating a string with an integer.

In this article, I will show you why the TypeError: builtin_function_or_method object is not subscriptable occurs and how you can fix it.

Why The TypeError: builtin_function_or_method object is not subscriptable Occurs

Every built-in function of Python such as print() , append() , sorted() , max() , and others must be called with parenthesis or round brackets ( () ).

If you try to use square brackets, Python won't treat it as a function call. Instead, Python will think you’re trying to access something from a list or string and then throw the error.

For example, the code below throws the error because I was trying to print the value of the variable with square braces in front of the print() function:

And if you surround what you want to print with square brackets even if the item is iterable, you still get the error:

This issue is not particular to the print() function. If you try to call any other built-in function with square brackets, you also get the error.

In the example below, I tried to call max() with square brackets and I got the error:

How to Fix the TypeError: builtin_function_or_method object is not subscriptable Error

To fix this error, all you need to do is make sure you use parenthesis to call the function.

You only have to use square brackets if you want to access an item from iterable data such as string, list, or tuple:

Wrapping Up

This article showed you why the TypeError: builtin_function_or_method object is not subscriptable occurs and how to fix it.

Remember that you only need to use square brackets ( [] ) to access an item from iterable data and you shouldn't use it to call a function.

If you’re getting this error, you should look in your code for any point at which you are calling a built-in function with square brackets and replace it with parenthesis.

Thanks for reading.

Web developer and technical writer focusing on frontend technologies. I also dabble in a lot of other technologies.

If you read this far, thank the author to show them you care. Say Thanks

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Python TypeError: ‘builtin_function_or_method’ object is not subscriptable Solution

To call a built-in function, you need to use parentheses. Parentheses distinguish function calls from other operations that can be performed on some objects, like indexing.

If you try to use square brackets to call a built-in function, you’ll encounter the “TypeError: ‘builtin_function_or_method’ object is not subscriptable” error. 

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In this guide, we talk about what this error means and why you may encounter it. We’ll walk through an example so that you can figure out how to solve the error.

TypeError: ‘builtin_function_or_method’ object is not subscriptable

Only iterable objects are subscriptable. Examples of iterable objects include lists , strings , and dictionaries. Individual values in these objects can be accessed using indexing. This is because items within an iterable object have index values .

Consider the following code:

Our code returns “English”. Our code retrieves the first item in our list, which is the item at index position 0. Our list is subscriptable so we can access it using square brackets.

Built-in functions are not subscriptable. This is because they do not return a list of objects that can be accessed using indexing.

The “TypeError: ‘builtin_function_or_method’ object is not subscriptable” error occurs when you try to access a built-in function using square brackets. This is because when the Python interpreter sees square brackets it tries to access items from a value as if that value is iterable.

An Example Scenario

We’re going to build a program that appends all of the records from a list of homeware goods to another list. An item should only be added to the next list if that item is in stock.

Start by defining a list of homeware goods and a list to store those that are in stock:

The “in_stock” list is presently empty. This is because we have not yet calculated which items in stock should be added to the list.

Next, we use a for loop to find items in the “homewares” list that are in stock. We’ll add those items to the “in_stock” list:

We use the append() method to add a record to the “in_stock” list if that item is in stock. Otherwise, our program does not add a record to the “in_stock” list. Our program then prints out all of the objects in the “in_stock” list.

Let’s run our code and see what happens:

Our code returns an error.

The Solution

Take a look at the line of code that Python points to in the error:

We’ve tried to use indexing syntax to add an item to our “in_stock” list variable . This is incorrect because functions are not iterable objects. To call a function, we need to use parenthesis.

We fix this problem by replacing the square brackets with parentheses:

Let’s run our code:

Our code successfully calculates the items in stock. Those items are added to the “in_stock” list which is printed to the console.

The “TypeError: ‘builtin_function_or_method’ object is not subscriptable” error is raised when you try to use square brackets to call a function.

This error is raised because Python interprets square brackets as a way of accessing items from an iterable object. Functions must be called using parenthesis. To fix this problem, make sure you call a function using parenthesis.

Now you’re ready to solve this common Python error like an expert!

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How to Fix DataFrame object does not support item assignment Error

DataFrame Object Does Not Support Item Assignment

DataFrames are a powerful tool for storing and manipulating data. They are essentially tabular data structures, with rows and columns of data. One of the most common tasks when working with DataFrames is assigning values to specific cells. However, you may encounter an error message if you try to do this.

The error message:

df[‘column_name’] = value

ValueError: dataframe object does not support item assignment

This error message occurs because DataFrames are immutable. This means that you cannot change the values of individual cells in a DataFrame. Instead, you need to use a different method to assign values to cells, such as `df.loc[row_index, column_name] = value`.

In this article, we will discuss what it means for a DataFrame to be immutable, and we will show you how to assign values to cells in a DataFrame using the correct methods.

We will also discuss some of the advantages and disadvantages of using immutable DataFrames. By the end of this article, you will have a better understanding of how to work with DataFrames and you will be able to avoid the `dataframe object does not support item assignment` error message.

In this tutorial, we will discuss what it means when a dataframe object does not support item assignment. We will also provide some examples of how to work around this limitation.

What is a dataframe object?

A dataframe is a two-dimensional data structure that is used to store data in rows and columns. Dataframes are similar to spreadsheets, but they are more flexible and can be used with more programming languages. Dataframes are often used to store data that has been collected from a variety of sources.

What does it mean when a dataframe object does not support item assignment?

When a dataframe object does not support item assignment, it means that you cannot assign a value to a specific cell in the dataframe. This is because dataframes are immutable, which means that they cannot be changed after they have been created. If you need to change the value of a cell in a dataframe, you must create a new dataframe with the updated values.

Examples of how to work around the limitation of dataframe objects not supporting item assignment

There are a few ways to work around the limitation of dataframe objects not supporting item assignment. One way is to use the `loc` and `iloc` methods. The `loc` method allows you to select rows and columns from a dataframe by their index, while the `iloc` method allows you to select rows and columns by their position.

For example, the following code uses the `loc` method to select the first row and the second column of a dataframe:

df.loc[0, 1]

This code will return the value of the cell in the first row and the second column of the dataframe.

Another way to work around the limitation of dataframe objects not supporting item assignment is to use the `assign` method. The `assign` method allows you to add new columns to a dataframe or to change the values of existing columns.

For example, the following code uses the `assign` method to change the value of the cell in the first row and the second column of a dataframe:

df = df.assign(new_column = df[‘column_name’] * 2)

This code will create a new column called `new_column` and will set the values of this column to the values of the `column_name` column multiplied by 2.

In this tutorial, we discussed what it means when a dataframe object does not support item assignment. We also provided some examples of how to work around this limitation.

If you have any questions about this tutorial, please feel free to leave a comment below.

1. What is a dataframe object?

A dataframe is a two-dimensional data structure that is used to store data in rows and columns. Dataframes are similar to spreadsheets, but they are more flexible and can be used with more programming languages. Dataframes are often used to store data that has been collected from a variety of sources, such as CSV files, JSON files, and APIs.

2. What does it mean when a dataframe object does not support item assignment?

When a dataframe object does not support item assignment, it means that you cannot change the value of a specific cell in the dataframe. This is because dataframes are immutable, which means that they cannot be changed after they have been created. If you need to change the value of a cell in a dataframe, you must create a new dataframe with the updated values.

3. Why does a dataframe object not support item assignment?

There are a few reasons why a dataframe object might not support item assignment. One reason is that dataframes are often used to store data that has been collected from a variety of sources. If you allow users to change the values of individual cells in a dataframe, it could lead to data inconsistency. For example, if two users are working on the same dataframe and one user changes the value of a cell, the other user might not be aware of the change and could end up using the incorrect value.

Another reason why a dataframe object might not support item assignment is that it can be inefficient. When you change the value of a cell in a dataframe, all of the other cells in the dataframe need to be recalculated. This can slow down your code and make it difficult to debug.

4. How can I work around the limitation of dataframe objects not supporting item assignment?

There are a few ways to work around the limitation of dataframe objects not supporting item assignment. One way is to use the `loc` and `iloc` methods. The `loc` method allows you to select rows and columns from

Why does a dataframe object not support item assignment?

There are a few reasons why a dataframe object does not support item assignment.

• First, dataframes are designed to be immutable, which means that they cannot be changed after they have been created. This is in contrast to lists, which are mutable and can be changed after they have been created. The immutability of dataframes is important for maintaining the integrity of data and ensuring that data is not accidentally overwritten.
• Second, dataframes are often used to store data that has been collected from a variety of sources. If dataframes were able to support item assignment, it would be possible to accidentally overwrite data from one source with data from another source. This could lead to errors in data analysis and could damage the integrity of the data.

How to work around the fact that a dataframe object does not support item assignment?

There are a few ways to work around the fact that a dataframe object does not support item assignment.

• One way is to create a new dataframe with the updated values. This is the simplest and most straightforward way to change the values of a dataframe. To do this, you can use the `copy()` method to create a new dataframe from the existing dataframe, and then you can update the values in the new dataframe.

python df = pd.DataFrame({‘a’: [1, 2, 3], ‘b’: [4, 5, 6]})

Create a new dataframe with the updated values new_df = df.copy() new_df[‘a’] = [10, 20, 30]

print(new_df)

a b 0 10 4 1 20 5 2 30 6

• Another way to work around the fact that a dataframe object does not support item assignment is to use the `loc` and `iloc` methods to access and change specific cells in the dataframe. The `loc` method allows you to access cells in a dataframe by their row and column labels, and the `iloc` method allows you to access cells in a dataframe by their integer index.

Change the value of the cell in the first row and second column df.loc[0, ‘b’] = 10

a b 0 1 10 1 2 5 2 3 6

• Finally, you can use the `assign` method to add new columns to the dataframe or change the values of existing columns. The `assign` method takes a dictionary of key-value pairs, where the keys are the column names and the values are the new values.

Add a new column called `c` with the values 10, 20, and 30 df = df.assign(c=[10, 20, 30])

a b c 0 1 4 10 1 2 5 20 2 3 6 30

there are a few ways to work around the fact that a dataframe object does not support item assignment. You can create a new dataframe with the updated values, use the `loc` and `iloc` methods to access and change specific cells in the dataframe, or use the `assign` method to add new columns to the dataframe or change the values of existing columns.

Q: What does it mean when I get the error “dataframe object does not support item assignment”?

A: This error occurs when you try to assign a value to a specific element of a DataFrame using the `[]` operator. For example, the following code will generate an error:

df = pd.DataFrame({‘a’: [1, 2, 3], ‘b’: [4, 5, 6]}) df[‘a’][0] = 10

The error is because the DataFrame object does not support item assignment. To assign a value to a specific element of a DataFrame, you must use the `loc` or `iloc` methods. For example, the following code will work:

df.loc[0, ‘a’] = 10

Q: Why does the DataFrame object not support item assignment?

A: There are a few reasons why the DataFrame object does not support item assignment. First, it would make it difficult to track changes to the DataFrame. If you could assign a value to a specific element of a DataFrame, it would be difficult to know what the original value was. Second, item assignment could lead to errors if the index of the DataFrame changed. For example, if you added a new row to the DataFrame, the index of the existing rows would change. This would cause the item assignment to fail.

Q: How can I assign a value to a specific element of a DataFrame?

A: As mentioned above, you can use the `loc` or `iloc` methods to assign a value to a specific element of a DataFrame. The `loc` method allows you to select rows and columns by label, while the `iloc` method allows you to select rows and columns by index. For example, the following code uses the `loc` method to assign the value 10 to the first element of the `a` column:

The following code uses the `iloc` method to assign the value 10 to the first element of the `a` column:

df.iloc[0, 0] = 10

Q: What are some other common errors that I might encounter when working with DataFrames?

A: Some other common errors that you might encounter when working with DataFrames include:

• Indexing errors: You might get an error if you try to index a DataFrame with a column that does not exist. For example, the following code will generate an error:

df[‘c’]

This error occurs because the `c` column does not exist in the DataFrame.

• Value errors: You might get a value error if you try to assign a value to a column that is not the correct type. For example, the following code will generate an error:

df[‘a’] = ’10’

This error occurs because the `a` column is a numeric column, and you cannot assign a string value to a numeric column.

• Key errors: You might get a key error if you try to access a row or column by label that does not exist. For example, the following code will generate an error:

df.loc[‘a’]

This error occurs because the `a` row does not exist in the DataFrame.

Q: How can I avoid these errors?

A: To avoid these errors, you should:

• Make sure that you are using the correct syntax when indexing and assigning values to DataFrames.
• Check the data types of your columns to make sure that they are compatible with the values that you are trying to assign.
• Check the index of your DataFrame to make sure that the rows and columns that you are trying to access exist.

By following these tips, you can avoid common errors when working with DataFrames.

In this article, we have discussed the error “dataframe object does not support item assignment”. We have explained what this error means and how to fix it. We have also provided some tips on how to avoid this error in the future.

Here are the key takeaways from this article:

• The “dataframe object does not support item assignment” error occurs when you try to assign a value to a specific cell in a DataFrame.
• To fix this error, you can use the `loc` or `iloc` methods to select the cell you want to update and then assign the new value to it.
• You can also avoid this error by using the `.at` or `.iat` methods to update cells in a DataFrame.

By following these tips, you can avoid the “dataframe object does not support item assignment” error and successfully update cells in your DataFrames.

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python object does not support item assignment

先看以下代码：

Traceback (most recent call last): File “test.py”, line 13, in <\module> a1[‘age’] = 12 TypeError: ‘Animal’ object does not support item assignment

说明a1对象不可这样进行赋值操作；

解决办法：

需要为类定义__setitem__方法；

再次执行， output:

name:panda, 12

打印name以及刚赋值的12。

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Fix Python TypeError: 'str' object does not support item assignment

by Nathan Sebhastian

Posted on Jan 11, 2023

Reading time: 4 minutes

Python shows TypeError: 'str' object does not support item assignment error when you try to access and modify a string object using the square brackets ( [] ) notation.

To solve this error, use the replace() method to modify the string instead.

This error occurs because a string in Python is immutable, meaning you can’t change its value after it has been defined.

For example, suppose you want to replace the first character in your string as follows:

The code above attempts to replace the letter H with J by adding the index operator [0] .

But because assigning a new value to a string is not possible, Python responds with the following error:

To fix this error, you can create a new string with the desired modifications, instead of trying to modify the original string.

This can be done by calling the replace() method from the string. See the example below:

The replace() method allows you to replace all occurrences of a substring in your string.

This method accepts 3 parameters:

• old - the substring you want to replace
• new - the replacement for old value
• count - how many times to replace old (optional)

By default, the replace() method replaces all occurrences of the old string:

You can control how many times the replacement occurs by passing the third count parameter.

The code below replaces only the first occurrence of the old value:

And that’s how you can modify a string using the replace() method.

If you want more control over the modification, you can use a list.

Convert the string to a list first, then access the element you need to change as shown below:

After you modify the list element, merge the list back as a string by using the join() method.

This solution gives you more control as you can select the character you want to replace. You can replace the first, middle, or last occurrence of a specific character.

Another way you can modify a string is to use the string slicing and concatenation method.

Consider the two examples below:

In both examples, the string slicing operator is used to extract substrings of the old_str variable.

In the first example, the slice operator is used to extract the substring starting from index 1 to the end of the string with old_str[1:] and concatenates it with the character ‘J’ .

In the second example, the slice operator is used to extract the substring before index 7 with old_str[:7] and the substring after index 8 with old_str[8:] syntax.

Both substrings are joined together while putting the character x in the middle.

The examples show how you can use slicing to extract substrings and concatenate them to create new strings.

But using slicing and concatenation can be more confusing than using a list, so I would recommend you use a list unless you have a strong reason.

The Python error TypeError: 'str' object does not support item assignment occurs when you try to modify a string object using the subscript or index operator assignment.

This error happens because strings in Python are immutable and can’t be modified.

The solution is to create a new string with the required modifications. There are three ways you can do it:

• Use replace() method
• Convert the string to a list, apply the modifications, merge the list back to a string
• Use string slicing and concatenation to create a new string

Now you’ve learned how to modify a string in Python. Nice work!

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TypeError: 'tuple' object does not support item assignment

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

# TypeError: 'tuple' object does not support item assignment

The Python "TypeError: 'tuple' object does not support item assignment" occurs when we try to change the value of an item in a tuple.

To solve the error, convert the tuple to a list, change the item at the specific index and convert the list back to a tuple.

Here is an example of how the error occurs.

We tried to update an element in a tuple, but tuple objects are immutable which caused the error.

# Convert the tuple to a list to solve the error

We cannot assign a value to an individual item of a tuple.

Instead, we have to convert the tuple to a list.

This is a three-step process:

• Use the list() class to convert the tuple to a list.
• Update the item at the specified index.
• Use the tuple() class to convert the list back to a tuple.

Once we have a list, we can update the item at the specified index and optionally convert the result back to a tuple.

Python indexes are zero-based, so the first item in a tuple has an index of 0 , and the last item has an index of -1 or len(my_tuple) - 1 .

# Constructing a new tuple with the updated element

Alternatively, you can construct a new tuple that contains the updated element at the specified index.

The get_updated_tuple function takes a tuple, an index and a new value and returns a new tuple with the updated value at the specified index.

The original tuple remains unchanged because tuples are immutable.

We updated the tuple element at index 1 , setting it to Z .

If you only have to do this once, you don't have to define a function.

The code sample achieves the same result without using a reusable function.

The values on the left and right-hand sides of the addition (+) operator have to all be tuples.

The syntax for tuple slicing is my_tuple[start:stop:step] .

The start index is inclusive and the stop index is exclusive (up to, but not including).

If the start index is omitted, it is considered to be 0 , if the stop index is omitted, the slice goes to the end of the tuple.

# Using a list instead of a tuple

Alternatively, you can declare a list from the beginning by wrapping the elements in square brackets (not parentheses).

Declaring a list from the beginning is much more efficient if you have to change the values in the collection often.

Tuples are intended to store values that never change.

# How tuples are constructed in Python

In case you declared a tuple by mistake, tuples are constructed in multiple ways:

• Using a pair of parentheses () creates an empty tuple
• Using a trailing comma - a, or (a,)
• Separating items with commas - a, b or (a, b)
• Using the tuple() constructor

# Checking if the value is a tuple

You can also handle the error by checking if the value is a tuple before the assignment.

If the variable stores a tuple, we set it to a list to be able to update the value at the specified index.

The isinstance() function returns True if the passed-in object is an instance or a subclass of the passed-in class.

If you aren't sure what type a variable stores, use the built-in type() class.

The type class returns the type of an object.

# Additional Resources

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

• How to convert a Tuple to an Integer in Python
• How to convert a Tuple to JSON in Python
• Find Min and Max values in Tuple or List of Tuples in Python
• Get the Nth element of a Tuple or List of Tuples in Python
• Creating a Tuple or a Set from user Input in Python
• How to Iterate through a List of Tuples in Python
• Write a List of Tuples to a File in Python
• AttributeError: 'tuple' object has no attribute X in Python
• TypeError: 'tuple' object is not callable in Python [Fixed]

Borislav Hadzhiev

Web Developer

Copyright © 2024 Borislav Hadzhiev

'Proxy' object does not support item assignment

when using fx to tracing tsm, i get this error:

out[:, :-1, :fold] = x[:, 1:, :fold] TypeError: ‘Proxy’ object does not support item assignment

Hi @keyky , this is a limitation of symbolic tracing with FX. Here is a workaround using torch.fx.wrap:

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TypeError: 'DefaultCfg' object does not support item assignment #111

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• Python »
• 3.14.0a0 Documentation »
• 파이썬 자습서 »
• 3. 파이썬의 간략한 소개
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3. 파이썬의 간략한 소개 ¶

다음에 나올 예에서, 입력과 출력은 프롬프트( >>> 와 … )의 존재 여부로 구분됩니다: 예제를 실행하기 위해서는 프롬프트가 나올 때 프롬프트 뒤에 오는 모든 것들을 입력해야 합니다; 프롬프트로 시작하지 않는 줄들은 인터프리터가 출력하는 것들입니다. 예에서 보조 프롬프트 외에 아무것도 없는 줄은 빈 줄을 입력해야 한다는 뜻임에 주의하세요; 여러 줄로 구성된 명령을 끝내는 방법입니다.

You can toggle the display of prompts and output by clicking on >>> in the upper-right corner of an example box. If you hide the prompts and output for an example, then you can easily copy and paste the input lines into your interpreter.

이 설명서에 나오는 많은 예는 (대화형 프롬프트에서 입력되는 것들조차도) 주석을 포함하고 있습니다. 파이썬에서 주석은 해시 문자, # , 로 시작하고 줄의 끝까지 이어집니다. 주석은 줄의 처음에서 시작할 수도 있고, 공백이나 코드 뒤에 나올 수도 있습니다. 하지만 문자열 리터럴 안에는 들어갈 수 없습니다. 문자열 리터럴 안에 등장하는 해시 문자는 주석이 아니라 해시 문자일 뿐입니다. 주석은 코드의 의미를 정확히 전달하기 위한 것이고, 파이썬이 해석하지 않는 만큼, 예를 입력할 때는 생략해도 됩니다.

몇 가지 예를 듭니다:

3.1. 파이썬을 계산기로 사용하기 ¶

몇 가지 간단한 파이썬 명령을 사용해봅시다. 인터프리터를 실행하고 기본 프롬프트, >>> , 를 기다리세요. (얼마 걸리지 않아야 합니다.)

3.1.1. 숫자 ¶

The interpreter acts as a simple calculator: you can type an expression at it and it will write the value. Expression syntax is straightforward: the operators + , - , * and / can be used to perform arithmetic; parentheses ( () ) can be used for grouping. For example:

정수 (예를 들어 2 , 4 , 20 )는 int 형입니다. 소수부가 있는 것들 (예를 들어 5.0 , 1.6 )은 float 형입니다. 이 자습서 뒤에서 숫자 형들에 관해 더 자세히 살펴볼 예정입니다.

Division ( / ) always returns a float. To do floor division and get an integer result you can use the // operator; to calculate the remainder you can use % :

파이썬에서는 거듭제곱을 계산할 때 ** 연산자를 사용합니다 [ 1 ] :

변수에 값을 대입할 때는 등호( = )를 사용합니다. 이 경우 다음 대화형 프롬프트 전에 표시되는 출력은 없습니다:

변수가 “정의되어” 있지 않을 때 (값을 대입하지 않았을 때) 사용하려고 시도하는 것은 에러를 일으킵니다:

실수를 본격적으로 지원합니다; 서로 다른 형의 피연산자를 갖는 연산자는 정수 피연산자를 실수로 변환합니다:

대화형 모드에서는, 마지막에 인쇄된 표현식은 변수 _ 에 대입됩니다. 이것은 파이썬을 탁상용 계산기로 사용할 때, 계산을 이어 가기가 좀 더 쉬워짐을 의미합니다. 예를 들어:

이 변수는 사용자로서는 읽기만 가능한 것처럼 취급되어야 합니다. 값을 직접 대입하지 마세요 — 만약 그렇게 한다면 같은 이름의 지역 변수를 새로 만드는 것이 되는데, 내장 변수의 마술 같은 동작을 차단하는 결과를 낳습니다.

int 와 float 에 더해, 파이썬은 Decimal 이나 Fraction 등의 다른 형의 숫자들도 지원합니다. 파이썬은 복소수 에 대한 지원도 내장하고 있는데, 허수부를 가리키는데 j 나 J 접미사를 사용합니다 (예를 들어 3+5j ).

3.1.2. Text ¶

Python can manipulate text (represented by type str , so-called “strings”) as well as numbers. This includes characters “ ! ”, words “ rabbit ”, names “ Paris ”, sentences “ Got your back. ”, etc. “ Yay! :) ”. They can be enclosed in single quotes ( '...' ) or double quotes ( "..." ) with the same result [ 2 ] .

To quote a quote, we need to “escape” it, by preceding it with \ . Alternatively, we can use the other type of quotation marks:

In the Python shell, the string definition and output string can look different. The print() function produces a more readable output, by omitting the enclosing quotes and by printing escaped and special characters:

\ 뒤에 나오는 문자가 특수 문자로 취급되게 하고 싶지 않다면, 첫 따옴표 앞에 r 을 붙여서 날 문자열 (raw string) 을 만들 수 있습니다:

There is one subtle aspect to raw strings: a raw string may not end in an odd number of \ characters; see the FAQ entry for more information and workarounds.

문자열 리터럴은 여러 줄로 확장될 수 있습니다. 한 가지 방법은 삼중 따옴표를 사용하는 것입니다: """...""" 또는 '''...''' . 줄 넘김 문자는 자동으로 문자열에 포함됩니다. 하지만 줄 끝에 \ 를 붙여 이를 방지할 수도 있습니다. 다음 예:

는 이런 결과를 출력합니다 (첫 번째 개행문자가 포함되지 않는 것에 주목하세요):

문자열은 + 연산자로 이어붙이고, * 연산자로 반복시킬 수 있습니다:

두 개 이상의 문자열 리터럴 (즉, 따옴표로 둘러싸인 것들) 가 연속해서 나타나면 자동으로 이어 붙여집니다.

이 기능은 긴 문자열을 쪼개고자 할 때 특별히 쓸모 있습니다:

이것은 오직 두 개의 리터럴에만 적용될 뿐 변수나 표현식에는 해당하지 않습니다:

변수들끼리 혹은 변수와 문자열 리터럴을 이어붙이려면 + 를 사용해야 합니다

문자열은 인덱스 (서브 스크립트) 될 수 있습니다. 첫 번째 문자가 인덱스 0에 대응됩니다. 문자를 위한 별도의 형은 없습니다; 단순히 길이가 1인 문자열입니다:

인덱스는 음수가 될 수도 있는데, 끝에서부터 셉니다:

-0은 0과 같으므로, 음의 인덱스는 -1에서 시작한다는 것에 주목하세요.

In addition to indexing, slicing is also supported. While indexing is used to obtain individual characters, slicing allows you to obtain a substring:

슬라이스 인덱스는 편리한 기본값을 갖고 있습니다; 첫 번째 인덱스를 생략하면 기본값 0 이 사용되고, 두 번째 인덱스가 생략되면 기본값으로 슬라이싱 되는 문자열의 길이가 사용됩니다.

시작 위치의 문자는 항상 포함되는 반면, 종료 위치의 문자는 항상 포함되지 않는 것에 주의하세요. 이 때문에 s[:i] + s[i:] 는 항상 s 와 같아집니다

슬라이스가 동작하는 방법을 기억하는 한 가지 방법은 인덱스가 문자들 사이의 위치를 가리킨다고 생각하는 것입니다. 첫 번째 문자의 왼쪽 경계가 0입니다. n 개의 문자들로 구성된 문자열의 오른쪽 끝 경계는 인덱스 n 이 됩니다, 예를 들어:

첫 번째 숫자 행은 인덱스 0…6 의 위치를 보여주고; 두 번째 행은 대응하는 음의 인덱스들을 보여줍니다. i 에서 j 범위의 슬라이스는 i 와 j 로 번호 붙여진 경계 사이의 문자들로 구성됩니다.

음이 아닌 인덱스들의 경우, 두 인덱스 모두 범위 내에 있다면 슬라이스의 길이는 인덱스 간의 차입니다. 예를 들어 word[1:3] 의 길이는 2입니다.

너무 큰 값을 인덱스로 사용하는 것은 에러입니다:

하지만, 범위를 벗어나는 슬라이스 인덱스는 슬라이싱할 때 부드럽게 처리됩니다:

파이썬 문자열은 변경할 수 없다 — 불변 이라고 합니다. 그래서 문자열의 인덱스로 참조한 위치에 대입하려고 하면 에러를 일으킵니다:

다른 문자열이 필요하면, 새로 만들어야 합니다:

내장 함수 len() 은 문자열의 길이를 돌려줍니다:

문자열은 시퀀스 형 의 일종이고, 시퀀스가 지원하는 공통 연산들이 지원됩니다.

문자열은 기본적인 변환과 검색을 위한 여러 가지 메서드들을 지원합니다.

내장된 표현식을 갖는 문자열 리터럴

str.format() 으로 문자열을 포맷하는 방법에 대한 정보.

이곳에서 문자열을 % 연산자 왼쪽에 사용하는 예전 방식의 포매팅에 관해 좀 더 상세하게 설명하고 있습니다.

3.1.3. 리스트 ¶

파이썬은 다른 값들을 덩어리로 묶는데 사용되는 여러 가지 컴파운드 (compound) 자료 형을 알고 있습니다. 가장 융통성이 있는 것은 리스트 인데, 대괄호 사이에 쉼표로 구분된 값(항목)들의 목록으로 표현될 수 있습니다. 리스트는 서로 다른 형의 항목들을 포함할 수 있지만, 항목들이 모두 같은 형인 경우가 많습니다.

문자열(그리고, 다른 모든 내장 시퀀스 형들)처럼 리스트는 인덱싱하고 슬라이싱할 수 있습니다:

리스트는 이어붙이기 같은 연산도 지원합니다:

불변 인 문자열과는 달리, 리스트는 가변 입니다. 즉 내용을 변경할 수 있습니다:

You can also add new items at the end of the list, by using the list.append() method (we will see more about methods later):

Simple assignment in Python never copies data. When you assign a list to a variable, the variable refers to the existing list . Any changes you make to the list through one variable will be seen through all other variables that refer to it.:

모든 슬라이스 연산은 요청한 항목들을 포함하는 새 리스트를 돌려줍니다. 이는 다음과 같은 슬라이스가 리스트의 새로운 얕은 복사본 을 돌려준다는 뜻입니다:

슬라이스에 대입하는 것도 가능한데, 리스트의 길이를 변경할 수 있고, 모든 항목을 삭제할 수조차 있습니다:

내장 함수 len() 은 리스트에도 적용됩니다:

리스트를 중첩할 수도 있습니다. (다른 리스트를 포함하는 리스트를 만듭니다). 예를 들어:

3.2. 프로그래밍으로의 첫걸음 ¶

Of course, we can use Python for more complicated tasks than adding two and two together. For instance, we can write an initial sub-sequence of the Fibonacci series as follows:

이 예는 몇 가지 새로운 기능을 소개하고 있습니다.

첫 줄은 다중 대입 을 포함하고 있습니다: 변수 a 와 b 에 동시에 값 0과1이 대입됩니다. 마지막 줄에서 다시 사용되는데, 대입이 어느 하나라도 이루어지기 전에 우변의 표현식들이 모두 계산됩니다. 우변의 표현식은 왼쪽부터 오른쪽으로 가면서 순서대로 계산됩니다.

while 루프는 조건(여기서는: a < 10 )이 참인 동안 실행됩니다. C와 마찬가지로 파이썬에서 0 이 아닌 모든 정수는 참이고, 0은 거짓입니다. 조건은 문자열이나 리스트 (사실 모든 종류의 시퀀스)가 될 수도 있는데 길이가 0 이 아닌 것은 모두 참이고, 빈 시퀀스는 거짓입니다. 이 예에서 사용한 검사는 간단한 비교입니다. 표준 비교 연산자는 C와 같은 방식으로 표현됩니다: < (작다), > (크다), == (같다), <= (작거나 같다), >= (크거나 같다), != (다르다).

루프의 바디 (body) 는 들여쓰기 됩니다. 들여쓰기는 파이썬에서 문장을 덩어리로 묶는 방법입니다. 대화형 프롬프트에서 각각 들여 쓰는 줄에서 탭(tab)이나 공백(space)을 입력해야 합니다. 실제적으로는 텍스트 편집기를 사용해서 좀 더 복잡한 파이썬 코드를 준비하게 됩니다; 웬만한 텍스트 편집기들은 자동 들여쓰기 기능을 제공합니다. 복합문을 대화형으로 입력할 때는 끝을 알리기 위해 빈 줄을 입력해야 합니다. (해석기가 언제 마지막 줄을 입력할지 짐작할 수 없기 때문입니다.) 같은 블록에 포함되는 모든 줄은 같은 양만큼 들여쓰기 되어야 함에 주의하세요.

print() 함수는 주어진 인자들의 값을 인쇄합니다. 다중 인자, 실수의 값, 문자열을 다루는 방식에서 (계산기 예제에서 본 것과 같이) 출력하고자 하는 표현식을 그냥 입력하는 것과는 다릅니다. 문자열은 따옴표 없이 출력되고, 인자들 간에는 빈칸이 삽입됩니다. 그래서 이런 식으로 보기 좋게 포매팅할 수 있습니다:

키워드 인자 end 는 출력 끝에 포함되는 개행문자를 제거하거나 출력을 다른 문자열로 끝나게 하고 싶을 때 사용됩니다:

• 3.1.2. Text
• 3.2. 프로그래밍으로의 첫걸음

2. 파이썬 인터프리터 사용하기

4. 기타 제어 흐름 도구