os.path — Common pathname manipulations¶
Source code: Lib/posixpath.py (for POSIX) and Lib/ntpath.py (for Windows).
This module implements some useful functions on pathnames. To read or write files see open() , and for accessing the filesystem see the os module. The path parameters can be passed as strings, or bytes, or any object implementing the os.PathLike protocol.
Unlike a Unix shell, Python does not do any automatic path expansions. Functions such as expanduser() and expandvars() can be invoked explicitly when an application desires shell-like path expansion. (See also the glob module.)
The pathlib module offers high-level path objects.
All of these functions accept either only bytes or only string objects as their parameters. The result is an object of the same type, if a path or file name is returned.
Since different operating systems have different path name conventions, there are several versions of this module in the standard library. The os.path module is always the path module suitable for the operating system Python is running on, and therefore usable for local paths. However, you can also import and use the individual modules if you want to manipulate a path that is always in one of the different formats. They all have the same interface:
posixpath for UNIX-style paths
ntpath for Windows paths
Changed in version 3.8: exists() , lexists() , isdir() , isfile() , islink() , and ismount() now return False instead of raising an exception for paths that contain characters or bytes unrepresentable at the OS level.
Return a normalized absolutized version of the pathname path. On most platforms, this is equivalent to calling the function normpath() as follows: normpath(join(os.getcwd(), path)) .
Changed in version 3.6: Accepts a path-like object .
Return the base name of pathname path. This is the second element of the pair returned by passing path to the function split() . Note that the result of this function is different from the Unix basename program; where basename for ‘/foo/bar/’ returns ‘bar’ , the basename() function returns an empty string ( » ).
Changed in version 3.6: Accepts a path-like object .
Return the longest common sub-path of each pathname in the sequence paths. Raise ValueError if paths contain both absolute and relative pathnames, the paths are on the different drives or if paths is empty. Unlike commonprefix() , this returns a valid path.
New in version 3.5.
Changed in version 3.6: Accepts a sequence of path-like objects .
Return the longest path prefix (taken character-by-character) that is a prefix of all paths in list. If list is empty, return the empty string ( » ).
This function may return invalid paths because it works a character at a time. To obtain a valid path, see commonpath() .
Changed in version 3.6: Accepts a path-like object .
Return the directory name of pathname path. This is the first element of the pair returned by passing path to the function split() .
Changed in version 3.6: Accepts a path-like object .
Return True if path refers to an existing path or an open file descriptor. Returns False for broken symbolic links. On some platforms, this function may return False if permission is not granted to execute os.stat() on the requested file, even if the path physically exists.
Changed in version 3.3: path can now be an integer: True is returned if it is an open file descriptor, False otherwise.
Changed in version 3.6: Accepts a path-like object .
Return True if path refers to an existing path. Returns True for broken symbolic links. Equivalent to exists() on platforms lacking os.lstat() .
Changed in version 3.6: Accepts a path-like object .
On Unix and Windows, return the argument with an initial component of
user replaced by that user’s home directory.
On Unix, an initial
is replaced by the environment variable HOME if it is set; otherwise the current user’s home directory is looked up in the password directory through the built-in module pwd . An initial
user is looked up directly in the password directory.
On Windows, USERPROFILE will be used if set, otherwise a combination of HOMEPATH and HOMEDRIVE will be used. An initial
user is handled by checking that the last directory component of the current user’s home directory matches USERNAME , and replacing it if so.
If the expansion fails or if the path does not begin with a tilde, the path is returned unchanged.
Changed in version 3.6: Accepts a path-like object .
Changed in version 3.8: No longer uses HOME on Windows.
Return the argument with environment variables expanded. Substrings of the form $name or $
On Windows, %name% expansions are supported in addition to $name and $
Changed in version 3.6: Accepts a path-like object .
Return the time of last access of path. The return value is a floating point number giving the number of seconds since the epoch (see the time module). Raise OSError if the file does not exist or is inaccessible.
os.path. getmtime ( path ) ¶
Return the time of last modification of path. The return value is a floating point number giving the number of seconds since the epoch (see the time module). Raise OSError if the file does not exist or is inaccessible.
Changed in version 3.6: Accepts a path-like object .
Return the system’s ctime which, on some systems (like Unix) is the time of the last metadata change, and, on others (like Windows), is the creation time for path. The return value is a number giving the number of seconds since the epoch (see the time module). Raise OSError if the file does not exist or is inaccessible.
Changed in version 3.6: Accepts a path-like object .
Return the size, in bytes, of path. Raise OSError if the file does not exist or is inaccessible.
Changed in version 3.6: Accepts a path-like object .
Return True if path is an absolute pathname. On Unix, that means it begins with a slash, on Windows that it begins with a (back)slash after chopping off a potential drive letter.
Changed in version 3.6: Accepts a path-like object .
Return True if path is an existing regular file. This follows symbolic links, so both islink() and isfile() can be true for the same path.
Changed in version 3.6: Accepts a path-like object .
Return True if path is an existing directory. This follows symbolic links, so both islink() and isdir() can be true for the same path.
Changed in version 3.6: Accepts a path-like object .
Return True if path refers to an existing directory entry that is a symbolic link. Always False if symbolic links are not supported by the Python runtime.
Changed in version 3.6: Accepts a path-like object .
Return True if pathname path is a mount point: a point in a file system where a different file system has been mounted. On POSIX, the function checks whether path’s parent, path /.. , is on a different device than path, or whether path /.. and path point to the same i-node on the same device — this should detect mount points for all Unix and POSIX variants. It is not able to reliably detect bind mounts on the same filesystem. On Windows, a drive letter root and a share UNC are always mount points, and for any other path GetVolumePathName is called to see if it is different from the input path.
New in version 3.4: Support for detecting non-root mount points on Windows.
Changed in version 3.6: Accepts a path-like object .
Join one or more path segments intelligently. The return value is the concatenation of path and all members of *paths, with exactly one directory separator following each non-empty part, except the last. That is, the result will only end in a separator if the last part is either empty or ends in a separator. If a segment is an absolute path (which on Windows requires both a drive and a root), then all previous segments are ignored and joining continues from the absolute path segment.
On Windows, the drive is not reset when a rooted path segment (e.g., r’\foo’ ) is encountered. If a segment is on a different drive or is an absolute path, all previous segments are ignored and the drive is reset. Note that since there is a current directory for each drive, os.path.join("c:", "foo") represents a path relative to the current directory on drive C: ( c:foo ), not c:\foo .
Changed in version 3.6: Accepts a path-like object for path and paths.
Normalize the case of a pathname. On Windows, convert all characters in the pathname to lowercase, and also convert forward slashes to backward slashes. On other operating systems, return the path unchanged.
Changed in version 3.6: Accepts a path-like object .
Normalize a pathname by collapsing redundant separators and up-level references so that A//B , A/B/ , A/./B and A/foo/../B all become A/B . This string manipulation may change the meaning of a path that contains symbolic links. On Windows, it converts forward slashes to backward slashes. To normalize case, use normcase() .
On POSIX systems, in accordance with IEEE Std 1003.1 2013 Edition; 4.13 Pathname Resolution, if a pathname begins with exactly two slashes, the first component following the leading characters may be interpreted in an implementation-defined manner, although more than two leading characters shall be treated as a single character.
Changed in version 3.6: Accepts a path-like object .
Return the canonical path of the specified filename, eliminating any symbolic links encountered in the path (if they are supported by the operating system).
If a path doesn’t exist or a symlink loop is encountered, and strict is True , OSError is raised. If strict is False , the path is resolved as far as possible and any remainder is appended without checking whether it exists.
This function emulates the operating system’s procedure for making a path canonical, which differs slightly between Windows and UNIX with respect to how links and subsequent path components interact.
Operating system APIs make paths canonical as needed, so it’s not normally necessary to call this function.
Changed in version 3.6: Accepts a path-like object .
Changed in version 3.8: Symbolic links and junctions are now resolved on Windows.
Changed in version 3.10: The strict parameter was added.
Return a relative filepath to path either from the current directory or from an optional start directory. This is a path computation: the filesystem is not accessed to confirm the existence or nature of path or start. On Windows, ValueError is raised when path and start are on different drives.
Changed in version 3.6: Accepts a path-like object .
Return True if both pathname arguments refer to the same file or directory. This is determined by the device number and i-node number and raises an exception if an os.stat() call on either pathname fails.
Changed in version 3.2: Added Windows support.
Changed in version 3.4: Windows now uses the same implementation as all other platforms.
Changed in version 3.6: Accepts a path-like object .
Return True if the file descriptors fp1 and fp2 refer to the same file.
Changed in version 3.2: Added Windows support.
Changed in version 3.6: Accepts a path-like object .
Return True if the stat tuples stat1 and stat2 refer to the same file. These structures may have been returned by os.fstat() , os.lstat() , or os.stat() . This function implements the underlying comparison used by samefile() and sameopenfile() .
Changed in version 3.4: Added Windows support.
Changed in version 3.6: Accepts a path-like object .
Split the pathname path into a pair, (head, tail) where tail is the last pathname component and head is everything leading up to that. The tail part will never contain a slash; if path ends in a slash, tail will be empty. If there is no slash in path, head will be empty. If path is empty, both head and tail are empty. Trailing slashes are stripped from head unless it is the root (one or more slashes only). In all cases, join(head, tail) returns a path to the same location as path (but the strings may differ). Also see the functions dirname() and basename() .
Changed in version 3.6: Accepts a path-like object .
Split the pathname path into a pair (drive, tail) where drive is either a mount point or the empty string. On systems which do not use drive specifications, drive will always be the empty string. In all cases, drive + tail will be the same as path.
On Windows, splits a pathname into drive/UNC sharepoint and relative path.
If the path contains a drive letter, drive will contain everything up to and including the colon:
If the path contains a UNC path, drive will contain the host name and share, up to but not including the fourth separator:
Changed in version 3.6: Accepts a path-like object .
Split the pathname path into a pair (root, ext) such that root + ext == path , and the extension, ext, is empty or begins with a period and contains at most one period.
If the path contains no extension, ext will be » :
If the path contains an extension, then ext will be set to this extension, including the leading period. Note that previous periods will be ignored:
Leading periods of the last component of the path are considered to be part of the root:
Changed in version 3.6: Accepts a path-like object .
True if arbitrary Unicode strings can be used as file names (within limitations imposed by the file system).
How do I get the path of the current executed file in Python?
Is there a universal approach in Python, to find out the path to the file that is currently executing?
Failing approaches
path = os.path.abspath(os.path.dirname(sys.argv[0]))
This does not work if you are running from another Python script in another directory, for example by using execfile in 2.x.
path = os.path.abspath(os.path.dirname(__file__))
I found that this doesn’t work in the following cases:
- py2exe doesn’t have a __file__ attribute, although there is a workaround
- When the code is run from IDLE using execute() , in which case there is no __file__ attribute
- On Mac OS X v10.6 (Snow Leopard), I get NameError: global name ‘__file__’ is not defined
Test case
Directory tree
Content of a.py
Content of subdir/b.py
Output of python a.py (on Windows)
Related (but these answers are incomplete)
13 Answers 13
First, you need to import from inspect and os
Next, wherever you want to find the source file from you just use
You can’t directly determine the location of the main script being executed. After all, sometimes the script didn’t come from a file at all. For example, it could come from the interactive interpreter or dynamically generated code stored only in memory.
However, you can reliably determine the location of a module, since modules are always loaded from a file. If you create a module with the following code and put it in the same directory as your main script, then the main script can import the module and use that to locate itself.
If you have several main scripts in different directories, you may need more than one copy of module_locator.
Of course, if your main script is loaded by some other tool that doesn’t let you import modules that are co-located with your script, then you’re out of luck. In cases like that, the information you’re after simply doesn’t exist anywhere in your program. Your best bet would be to file a bug with the authors of the tool.
This solution is robust even in executables:
I was running into a similar problem, and I think this might solve the problem:
It works for regular scripts and in IDLE. All I can say is try it out for others!
My typical usage:
Now I use _modpath_ instead of _file_.
You have simply called:
abspath() gives you the absolute path of sys.argv[0] (the filename your code is in) and dirname() returns the directory path without the filename.
The short answer is that there is no guaranteed way to get the information you want, however there are heuristics that work almost always in practice. You might look at How do I find the location of the executable in C?. It discusses the problem from a C point of view, but the proposed solutions are easily transcribed into Python.
See my answer to the question Importing modules from parent folder for related information, including why my answer doesn’t use the unreliable __file__ variable. This simple solution should be cross-compatible with different operating systems as the modules os and inspect come as part of Python.
First, you need to import parts of the inspect and os modules.
Next, use the following line anywhere else it’s needed in your Python code:
How it works:
From the built-in module os (description below), the abspath tool is imported.
OS routines for Mac, NT, or Posix depending on what system we’re on.
Then getsourcefile (description below) is imported from the built-in module inspect .
Get useful information from live Python objects.
- abspath(path) returns the absolute/full version of a file path
- getsourcefile(lambda:0) somehow gets the internal source file of the lambda function object, so returns ‘<pyshell#nn>’ in the Python shell or returns the file path of the Python code currently being executed.
Using abspath on the result of getsourcefile(lambda:0) should make sure that the file path generated is the full file path of the Python file.
This explained solution was originally based on code from the answer at How do I get the path of the current executed file in Python?.
This should do the trick in a cross-platform way (so long as you’re not using the interpreter or something):
sys.path[0] is the directory that your calling script is in (the first place it looks for modules to be used by that script). We can take the name of the file itself off the end of sys.argv[0] (which is what I did with os.path.basename ). os.path.join just sticks them together in a cross-platform way. os.path.realpath just makes sure if we get any symbolic links with different names than the script itself that we still get the real name of the script.
I don’t have a Mac; so, I haven’t tested this on one. Please let me know if it works, as it seems it should. I tested this in Linux (Xubuntu) with Python 3.4. Note that many solutions for this problem don’t work on Macs (since I’ve heard that __file__ is not present on Macs).
Note that if your script is a symbolic link, it will give you the path of the file it links to (and not the path of the symbolic link).
Путь, имя и расширение файла
Достаточно часто возникают ситуации, когда у нас есть полное имя файла, а требуется узнать его расширение. Или добавить нужное расширение, когда не известно, ввел его пользователь или нет. Иногда у нас есть относительный путь до файла, а требуется узнать абсолютный. Про основные методы работы с именем файла и будет эта статья.
Абсолютный путь к файлу
Для того чтобы узнать в Python абсолютный путь к файлу, потребуется воспользоваться библиотекой os. Её подключаем с помощью команды import os. В классе path есть метод abspath. Вот пример использования.
Так же можно воспользоваться и стандартной библиотекой pathlib. Она вошла в состав основных библиотек, начиная с версии Python 3.4. До этого надо было ее инсталлировать с помощью команды pip install pathlib. Она предназначена для работы с путями файловой системы в разных ОС и отлично подойдет для решения данной задачи.
Имя файла
Чтобы узнать имя файла из полной строки с путем, воспользуемся методом basename модуля os.
Здесь перед строкой вставил r, чтобы подавить возможное возникновение служебных символов. Например, в данном случае если не указать r, то \f считалось бы символом перевода страницы.
Без расширения
Теперь разберемся, как в Python узнать имя файла без расширения. Воспользуемся методом splittext. В этот раз для примера возьмем файл с двойным расширением, чтобы проверить, как будут в этой ситуации работать стандартны функции.
Видно, что последнее расширение архиватора gz было отброшено, в то время как расширение несжатого архива tar осталось в имени.
Если же нам нужно только имя, то можно отбросить все символы полученной строки, которые идут после первой точки. Символ точки тоже отбросим.
Дополним предыдущий пример следующим кодом:
Расширение файла
В Python получить расширение файла можно аналогичным образом с помощью той же функции splitext. Она возвращает кортеж. Первый элемент кортежа имя, а второй – расширение. В данном случае нам нужен второй элемент. Индекс второго элемента равен единице, так как отсчет их идет от нуля.
Аналогично можно воспользоваться библиотекой pathlib. Воспользуемся методом suffix.
Но в нашем случае два расширения. Их можно узнать с помощью функции suffixes. Она возвращает список, элементами которого и будут расширения. Ниже приведен пример получения списка расширений.
Как задать путь к файлу в Python?
Для решения задач, связанных с редактированием или чтением файла, необходимо сообщить интерпретатору Python имя нужного нам файла, а также адрес, по которому этот файл располагается. Существуют разные способы указания пути к файлу в Python: от самого простого, до самого правильного. Давайте выясним, чем эти варианты отличаются и почему простой вариант не годится на роль лучшего!
Самый простой вариант — не самый верный!
Внимание! У этого способа обнаружен недостаток!
Самый простой вариант задания пути выглядит как последовательность директорий, в которых находится файл, с именем самого файла, разделенные знаками слеша:
Пример относительного пути:
Где вместо «Files» и «info.txt» Вы напишите названия ваших директорий и имя вашего файла соответственно.
Пример абсолютного пути:
Где вместо «C:\Python\pythonw.exe\Files\info.txt», «home/my_comp/Files/» и «info.txt» Вы напишите названия ваших директорий и имя вашего файла соответственно.
Этот вариант рабочий, однако, один существенный недостаток лишил его внимания разработчиков. Проблема заключается в том, что заданные таким способом пути адаптированы только к одному виду операционной системы: к Линукс, либо к Windows, так как в Windows используются обратные слеши «\», а в Линукс — обычные «/». Из-за этого скрипт, показывавший отличные результаты в Windows, начнет жаловаться на отсутствие файлов по прописанному пути в Linux, и наоборот. А с абсолютным путем вообще все сложно: никакого диска «C:» в Линуксе нет. Скрипт опять будет ругаться! Что же делать? Правильно указать путь к файлу!
Указываем путь к файлу правильно!
Внимание! Годный вариант!
Python — умный змей, поэтому в его арсенале, начиная с 3.4 версии появился модуль pathlib, который позволяет вытворять самые приятные вещи с путями к файлу, стоит только импортировать его класс Path:
Кстати, если у вас не установлен модуль pathlib, это легко исправить с помощью команды:
Задаем относительный путь с помощью Path!
После того, как класс импортирован, мы получаем власть над слешами! Теперь вопрос о прямых и обратных слешах в разных операционных системах ложится на плечи Path. Используя Path, вы можете корректно задать относительный путь, который будет работать в разных системах.
Например, в случае расположения файлов, как на представленном изображении, относительный путь, определяемый в скрипте «main_script.py», сформируется автоматически из перечисленных в скобках составных частей. Pathlib инициализирует новый объект класса Path, содержимым которого станет сформированный для Вашей системы относительный путь (в Windows части пути будут разделены обратными слешами, в Linux — обычными):
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Задаем абсолютный путь с помощью Path
- cwd() — возвращает путь к рабочей директории
- home() — возвращает путь к домашней директории
Полученную строку, содержащую путь к рабочей или домашней директории, объединим с недостающими участками пути при инициализации объекта класса Path :
Пример 1: с использованием функции cwd():
В данном случае путь к директории имеет вид: dir_path = «/home/my_comp/python», а полный путь к файлу «docs.txt» будет иметь вид: «/home/my_comp/python/files/info/docs.txt».
Представленный выше код можно оптимизировать и записать в одну строку:
Пример2: с использованием функции home():
В данном случае путь к директории имеет вид: dir_path = «/home/my_comp», а полный путь к файлу ‘docs.txt’ будет иметь вид: «/home/my_comp/files/info/docs.txt».
Сократим представленный выше код:
Подведем итог: начиная с версии Python 3.4, для задания пути к файлу рекомендуется использовать модуль pathlib с классом Path. Определить путь к рабочей директории можно с помощью функции cwd(), а путь к домашней директории подскажет функция home().