Python | os.path.dirname () method

dirname | Python Methods and Functions

os.path.dirname() in Python is used to get the name of a directory at a given path.

Syntax: os.path.dirname (path)

Parameter:
path : A path-like object representing a file system path.

Return Type: This method returns a string value which represents the directory name from the specified path.

Code: using method os.path.dirname ( )

# Python program to explain the os.path.dirname () method

 
# import of the os.path module

import os.path

 
# Path

path = '/ home / User / Documents '

  
# Get directory name
# from the specified path

dirname = os.path.dirname (path)

 
# Print directory name

print (dirname)

 

 
# Path

path = '/ home / User / Documents / file.txt'

 
# Get directory name
 # from the specified path

dirname = os .path.dirname (path)

 
# Print directory name

print (dirname)

 

  
# Path

path = 'file.txt'

 
# Get the directory name
# from the specified path

dirname = os.path.dirname (path)

 
# Print directory name

print (dirname)

  
# In the above path
# does not contain
# directory so
# Will not print anything

Exit :

 / home / User / home / User / Documents 

Link: https://docs.python.org/3/library/os.path.html





Python | os.path.dirname () method: StackOverflow Questions

os.path.dirname(__file__) returns empty

I want to get the path of the current directory under which a .py file is executed.

For example a simple file D: est.py with code:

import os

print os.getcwd()
print os.path.basename(__file__)
print os.path.abspath(__file__)
print os.path.dirname(__file__)

It is weird that the output is:

D:
test.py
D:	est.py
EMPTY

I am expecting the same results from the getcwd() and path.dirname().

Given os.path.abspath = os.path.dirname + os.path.basename, why

os.path.dirname(__file__)

returns empty?

What is the difference between os.path.basename() and os.path.dirname()?

What is the difference between os.path.basename() and os.path.dirname()?

I already searched for answers and read some links, but didn"t understand. Can anyone give a simple explanation?

Answer #1

@SpeedCoder5 "s comment deserves to be an answer;

Specifically, you can specify a dynamic working directory; (i.e. whichever directory where the currently-open Python file is located), using "cwd": "${fileDirname}"

If you"re using the Python: Current File (Integrated Terminal) option when you run Python, your launch.json file might look like mine, below.

{
    "version": "0.2.0",
    "configurations": [
    {
            "name": "Python: Current File (Integrated Terminal)",
            "type": "python",
            "request": "launch",
            "program": "${file}",
            "console": "integratedTerminal",
            "cwd": "${fileDirname}"
    }, 

    //... other settings, but I modified the "Current File" setting above ...
}

Remember the launch.json file controls the run/debug settings of your Visual Studio code project; my launch.json file was auto-generated by VS Code, in the directory of my current "Open Project". I just edited the file manually to add "cwd": "${fileDirname}" as shown above.

Remember the launch.json file may be specific to your project, or specific to your directory, so confirm you"re editing the correct launch.json (see comment)

If you don"t have a launch.json file, try this:

To create a launch.json file, open your project folder in VS Code (File > Open Folder) and then select the Configure gear icon on the Debug view top bar.

Answer #2

It depends on how you want to launch your script.

If you want to launch your UnitTest from the command line in a classic way, that is:

python tests/core_test.py

Then, since in this case "components" and "tests" are siblings folders, you can import the relative module either using the insert or the append method of the sys.path module. Something like:

import sys
from os import path
sys.path.append( path.dirname( path.dirname( path.abspath(__file__) ) ) )
from components.core import GameLoopEvents

Otherwise, you can launch your script with the "-m" argument (note that in this case, we are talking about a package, and thus you must not give the ".py" extension), that is:

python -m pkg.tests.core_test

In such a case, you can simply use the relative import as you were doing:

from ..components.core import GameLoopEvents

You can finally mix the two approaches, so that your script will work no matter how it is called. For example:

if __name__ == "__main__":
    if __package__ is None:
        import sys
        from os import path
        sys.path.append( path.dirname( path.dirname( path.abspath(__file__) ) ) )
        from components.core import GameLoopEvents
    else:
        from ..components.core import GameLoopEvents

Answer #3

You can use dirname:

os.path.dirname(path)

Return the directory name of pathname path. This is the first element of the pair returned by passing path to the function split().

And given the full path, then you can split normally to get the last portion of the path. For example, by using basename:

os.path.basename(path)

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 ("").


All together:

>>> import os
>>> path=os.path.dirname("C:/folder1/folder2/filename.xml")
>>> path
"C:/folder1/folder2"
>>> os.path.basename(path)
"folder2"

Answer #4

tl;dr

Call the is_path_exists_or_creatable() function defined below.

Strictly Python 3. That"s just how we roll.

A Tale of Two Questions

The question of "How do I test pathname validity and, for valid pathnames, the existence or writability of those paths?" is clearly two separate questions. Both are interesting, and neither have received a genuinely satisfactory answer here... or, well, anywhere that I could grep.

vikki"s answer probably hews the closest, but has the remarkable disadvantages of:

  • Needlessly opening (...and then failing to reliably close) file handles.
  • Needlessly writing (...and then failing to reliable close or delete) 0-byte files.
  • Ignoring OS-specific errors differentiating between non-ignorable invalid pathnames and ignorable filesystem issues. Unsurprisingly, this is critical under Windows. (See below.)
  • Ignoring race conditions resulting from external processes concurrently (re)moving parent directories of the pathname to be tested. (See below.)
  • Ignoring connection timeouts resulting from this pathname residing on stale, slow, or otherwise temporarily inaccessible filesystems. This could expose public-facing services to potential DoS-driven attacks. (See below.)

We"re gonna fix all that.

Question #0: What"s Pathname Validity Again?

Before hurling our fragile meat suits into the python-riddled moshpits of pain, we should probably define what we mean by "pathname validity." What defines validity, exactly?

By "pathname validity," we mean the syntactic correctness of a pathname with respect to the root filesystem of the current system – regardless of whether that path or parent directories thereof physically exist. A pathname is syntactically correct under this definition if it complies with all syntactic requirements of the root filesystem.

By "root filesystem," we mean:

  • On POSIX-compatible systems, the filesystem mounted to the root directory (/).
  • On Windows, the filesystem mounted to %HOMEDRIVE%, the colon-suffixed drive letter containing the current Windows installation (typically but not necessarily C:).

The meaning of "syntactic correctness," in turn, depends on the type of root filesystem. For ext4 (and most but not all POSIX-compatible) filesystems, a pathname is syntactically correct if and only if that pathname:

  • Contains no null bytes (i.e., x00 in Python). This is a hard requirement for all POSIX-compatible filesystems.
  • Contains no path components longer than 255 bytes (e.g., "a"*256 in Python). A path component is a longest substring of a pathname containing no / character (e.g., bergtatt, ind, i, and fjeldkamrene in the pathname /bergtatt/ind/i/fjeldkamrene).

Syntactic correctness. Root filesystem. That"s it.

Question #1: How Now Shall We Do Pathname Validity?

Validating pathnames in Python is surprisingly non-intuitive. I"m in firm agreement with Fake Name here: the official os.path package should provide an out-of-the-box solution for this. For unknown (and probably uncompelling) reasons, it doesn"t. Fortunately, unrolling your own ad-hoc solution isn"t that gut-wrenching...

O.K., it actually is. It"s hairy; it"s nasty; it probably chortles as it burbles and giggles as it glows. But what you gonna do? Nuthin".

We"ll soon descend into the radioactive abyss of low-level code. But first, let"s talk high-level shop. The standard os.stat() and os.lstat() functions raise the following exceptions when passed invalid pathnames:

  • For pathnames residing in non-existing directories, instances of FileNotFoundError.
  • For pathnames residing in existing directories:
    • Under Windows, instances of WindowsError whose winerror attribute is 123 (i.e., ERROR_INVALID_NAME).
    • Under all other OSes:
    • For pathnames containing null bytes (i.e., "x00"), instances of TypeError.
    • For pathnames containing path components longer than 255 bytes, instances of OSError whose errcode attribute is:
      • Under SunOS and the *BSD family of OSes, errno.ERANGE. (This appears to be an OS-level bug, otherwise referred to as "selective interpretation" of the POSIX standard.)
      • Under all other OSes, errno.ENAMETOOLONG.

Crucially, this implies that only pathnames residing in existing directories are validatable. The os.stat() and os.lstat() functions raise generic FileNotFoundError exceptions when passed pathnames residing in non-existing directories, regardless of whether those pathnames are invalid or not. Directory existence takes precedence over pathname invalidity.

Does this mean that pathnames residing in non-existing directories are not validatable? Yes – unless we modify those pathnames to reside in existing directories. Is that even safely feasible, however? Shouldn"t modifying a pathname prevent us from validating the original pathname?

To answer this question, recall from above that syntactically correct pathnames on the ext4 filesystem contain no path components (A) containing null bytes or (B) over 255 bytes in length. Hence, an ext4 pathname is valid if and only if all path components in that pathname are valid. This is true of most real-world filesystems of interest.

Does that pedantic insight actually help us? Yes. It reduces the larger problem of validating the full pathname in one fell swoop to the smaller problem of only validating all path components in that pathname. Any arbitrary pathname is validatable (regardless of whether that pathname resides in an existing directory or not) in a cross-platform manner by following the following algorithm:

  1. Split that pathname into path components (e.g., the pathname /troldskog/faren/vild into the list ["", "troldskog", "faren", "vild"]).
  2. For each such component:
    1. Join the pathname of a directory guaranteed to exist with that component into a new temporary pathname (e.g., /troldskog) .
    2. Pass that pathname to os.stat() or os.lstat(). If that pathname and hence that component is invalid, this call is guaranteed to raise an exception exposing the type of invalidity rather than a generic FileNotFoundError exception. Why? Because that pathname resides in an existing directory. (Circular logic is circular.)

Is there a directory guaranteed to exist? Yes, but typically only one: the topmost directory of the root filesystem (as defined above).

Passing pathnames residing in any other directory (and hence not guaranteed to exist) to os.stat() or os.lstat() invites race conditions, even if that directory was previously tested to exist. Why? Because external processes cannot be prevented from concurrently removing that directory after that test has been performed but before that pathname is passed to os.stat() or os.lstat(). Unleash the dogs of mind-fellating insanity!

There exists a substantial side benefit to the above approach as well: security. (Isn"t that nice?) Specifically:

Front-facing applications validating arbitrary pathnames from untrusted sources by simply passing such pathnames to os.stat() or os.lstat() are susceptible to Denial of Service (DoS) attacks and other black-hat shenanigans. Malicious users may attempt to repeatedly validate pathnames residing on filesystems known to be stale or otherwise slow (e.g., NFS Samba shares); in that case, blindly statting incoming pathnames is liable to either eventually fail with connection timeouts or consume more time and resources than your feeble capacity to withstand unemployment.

The above approach obviates this by only validating the path components of a pathname against the root directory of the root filesystem. (If even that"s stale, slow, or inaccessible, you"ve got larger problems than pathname validation.)

Lost? Great. Let"s begin. (Python 3 assumed. See "What Is Fragile Hope for 300, leycec?")

import errno, os

# Sadly, Python fails to provide the following magic number for us.
ERROR_INVALID_NAME = 123
"""
Windows-specific error code indicating an invalid pathname.

See Also
----------
https://docs.microsoft.com/en-us/windows/win32/debug/system-error-codes--0-499-
    Official listing of all such codes.
"""

def is_pathname_valid(pathname: str) -> bool:
    """
    `True` if the passed pathname is a valid pathname for the current OS;
    `False` otherwise.
    """
    # If this pathname is either not a string or is but is empty, this pathname
    # is invalid.
    try:
        if not isinstance(pathname, str) or not pathname:
            return False

        # Strip this pathname"s Windows-specific drive specifier (e.g., `C:`)
        # if any. Since Windows prohibits path components from containing `:`
        # characters, failing to strip this `:`-suffixed prefix would
        # erroneously invalidate all valid absolute Windows pathnames.
        _, pathname = os.path.splitdrive(pathname)

        # Directory guaranteed to exist. If the current OS is Windows, this is
        # the drive to which Windows was installed (e.g., the "%HOMEDRIVE%"
        # environment variable); else, the typical root directory.
        root_dirname = os.environ.get("HOMEDRIVE", "C:") 
            if sys.platform == "win32" else os.path.sep
        assert os.path.isdir(root_dirname)   # ...Murphy and her ironclad Law

        # Append a path separator to this directory if needed.
        root_dirname = root_dirname.rstrip(os.path.sep) + os.path.sep

        # Test whether each path component split from this pathname is valid or
        # not, ignoring non-existent and non-readable path components.
        for pathname_part in pathname.split(os.path.sep):
            try:
                os.lstat(root_dirname + pathname_part)
            # If an OS-specific exception is raised, its error code
            # indicates whether this pathname is valid or not. Unless this
            # is the case, this exception implies an ignorable kernel or
            # filesystem complaint (e.g., path not found or inaccessible).
            #
            # Only the following exceptions indicate invalid pathnames:
            #
            # * Instances of the Windows-specific "WindowsError" class
            #   defining the "winerror" attribute whose value is
            #   "ERROR_INVALID_NAME". Under Windows, "winerror" is more
            #   fine-grained and hence useful than the generic "errno"
            #   attribute. When a too-long pathname is passed, for example,
            #   "errno" is "ENOENT" (i.e., no such file or directory) rather
            #   than "ENAMETOOLONG" (i.e., file name too long).
            # * Instances of the cross-platform "OSError" class defining the
            #   generic "errno" attribute whose value is either:
            #   * Under most POSIX-compatible OSes, "ENAMETOOLONG".
            #   * Under some edge-case OSes (e.g., SunOS, *BSD), "ERANGE".
            except OSError as exc:
                if hasattr(exc, "winerror"):
                    if exc.winerror == ERROR_INVALID_NAME:
                        return False
                elif exc.errno in {errno.ENAMETOOLONG, errno.ERANGE}:
                    return False
    # If a "TypeError" exception was raised, it almost certainly has the
    # error message "embedded NUL character" indicating an invalid pathname.
    except TypeError as exc:
        return False
    # If no exception was raised, all path components and hence this
    # pathname itself are valid. (Praise be to the curmudgeonly python.)
    else:
        return True
    # If any other exception was raised, this is an unrelated fatal issue
    # (e.g., a bug). Permit this exception to unwind the call stack.
    #
    # Did we mention this should be shipped with Python already?

Done. Don"t squint at that code. (It bites.)

Question #2: Possibly Invalid Pathname Existence or Creatability, Eh?

Testing the existence or creatability of possibly invalid pathnames is, given the above solution, mostly trivial. The little key here is to call the previously defined function before testing the passed path:

def is_path_creatable(pathname: str) -> bool:
    """
    `True` if the current user has sufficient permissions to create the passed
    pathname; `False` otherwise.
    """
    # Parent directory of the passed path. If empty, we substitute the current
    # working directory (CWD) instead.
    dirname = os.path.dirname(pathname) or os.getcwd()
    return os.access(dirname, os.W_OK)

def is_path_exists_or_creatable(pathname: str) -> bool:
    """
    `True` if the passed pathname is a valid pathname for the current OS _and_
    either currently exists or is hypothetically creatable; `False` otherwise.

    This function is guaranteed to _never_ raise exceptions.
    """
    try:
        # To prevent "os" module calls from raising undesirable exceptions on
        # invalid pathnames, is_pathname_valid() is explicitly called first.
        return is_pathname_valid(pathname) and (
            os.path.exists(pathname) or is_path_creatable(pathname))
    # Report failure on non-fatal filesystem complaints (e.g., connection
    # timeouts, permissions issues) implying this path to be inaccessible. All
    # other exceptions are unrelated fatal issues and should not be caught here.
    except OSError:
        return False

Done and done. Except not quite.

Question #3: Possibly Invalid Pathname Existence or Writability on Windows

There exists a caveat. Of course there does.

As the official os.access() documentation admits:

Note: I/O operations may fail even when os.access() indicates that they would succeed, particularly for operations on network filesystems which may have permissions semantics beyond the usual POSIX permission-bit model.

To no one"s surprise, Windows is the usual suspect here. Thanks to extensive use of Access Control Lists (ACL) on NTFS filesystems, the simplistic POSIX permission-bit model maps poorly to the underlying Windows reality. While this (arguably) isn"t Python"s fault, it might nonetheless be of concern for Windows-compatible applications.

If this is you, a more robust alternative is wanted. If the passed path does not exist, we instead attempt to create a temporary file guaranteed to be immediately deleted in the parent directory of that path – a more portable (if expensive) test of creatability:

import os, tempfile

def is_path_sibling_creatable(pathname: str) -> bool:
    """
    `True` if the current user has sufficient permissions to create **siblings**
    (i.e., arbitrary files in the parent directory) of the passed pathname;
    `False` otherwise.
    """
    # Parent directory of the passed path. If empty, we substitute the current
    # working directory (CWD) instead.
    dirname = os.path.dirname(pathname) or os.getcwd()

    try:
        # For safety, explicitly close and hence delete this temporary file
        # immediately after creating it in the passed path"s parent directory.
        with tempfile.TemporaryFile(dir=dirname): pass
        return True
    # While the exact type of exception raised by the above function depends on
    # the current version of the Python interpreter, all such types subclass the
    # following exception superclass.
    except EnvironmentError:
        return False

def is_path_exists_or_creatable_portable(pathname: str) -> bool:
    """
    `True` if the passed pathname is a valid pathname on the current OS _and_
    either currently exists or is hypothetically creatable in a cross-platform
    manner optimized for POSIX-unfriendly filesystems; `False` otherwise.

    This function is guaranteed to _never_ raise exceptions.
    """
    try:
        # To prevent "os" module calls from raising undesirable exceptions on
        # invalid pathnames, is_pathname_valid() is explicitly called first.
        return is_pathname_valid(pathname) and (
            os.path.exists(pathname) or is_path_sibling_creatable(pathname))
    # Report failure on non-fatal filesystem complaints (e.g., connection
    # timeouts, permissions issues) implying this path to be inaccessible. All
    # other exceptions are unrelated fatal issues and should not be caught here.
    except OSError:
        return False

Note, however, that even this may not be enough.

Thanks to User Access Control (UAC), the ever-inimicable Windows Vista and all subsequent iterations thereof blatantly lie about permissions pertaining to system directories. When non-Administrator users attempt to create files in either the canonical C:Windows or C:Windowssystem32 directories, UAC superficially permits the user to do so while actually isolating all created files into a "Virtual Store" in that user"s profile. (Who could have possibly imagined that deceiving users would have harmful long-term consequences?)

This is crazy. This is Windows.

Prove It

Dare we? It"s time to test-drive the above tests.

Since NULL is the only character prohibited in pathnames on UNIX-oriented filesystems, let"s leverage that to demonstrate the cold, hard truth – ignoring non-ignorable Windows shenanigans, which frankly bore and anger me in equal measure:

>>> print(""foo.bar" valid? " + str(is_pathname_valid("foo.bar")))
"foo.bar" valid? True
>>> print("Null byte valid? " + str(is_pathname_valid("x00")))
Null byte valid? False
>>> print("Long path valid? " + str(is_pathname_valid("a" * 256)))
Long path valid? False
>>> print(""/dev" exists or creatable? " + str(is_path_exists_or_creatable("/dev")))
"/dev" exists or creatable? True
>>> print(""/dev/foo.bar" exists or creatable? " + str(is_path_exists_or_creatable("/dev/foo.bar")))
"/dev/foo.bar" exists or creatable? False
>>> print("Null byte exists or creatable? " + str(is_path_exists_or_creatable("x00")))
Null byte exists or creatable? False

Beyond sanity. Beyond pain. You will find Python portability concerns.

Answer #5

Put this inside your package"s __init__.py file:

# For relative imports to work in Python 3.6
import os, sys; sys.path.append(os.path.dirname(os.path.realpath(__file__)))

Assuming your package is like this:

├── project
│   ├── package
│   │   ├── __init__.py
│   │   ├── module1.py
│   │   └── module2.py
│   └── setup.py

Now use regular imports in you package, like:

# in module2.py
from module1 import class1

This works in both python 2 and 3.

Answer #6

Much nicer than the above, because you don"t need several os.path.join() and you will get the full path directly (if you wish), you can do this in Python 3.5 and above.

subfolders = [ f.path for f in os.scandir(folder) if f.is_dir() ]

This will give the complete path to the subdirectory. If you only want the name of the subdirectory use f.name instead of f.path

https://docs.python.org/3/library/os.html#os.scandir


Slightly OT: In case you need all subfolder recursively and/or all files recursively, have a look at this function, that is faster than os.walk & glob and will return a list of all subfolders as well as all files inside those (sub-)subfolders: https://stackoverflow.com/a/59803793/2441026

In case you want only all subfolders recursively:

def fast_scandir(dirname):
    subfolders= [f.path for f in os.scandir(dirname) if f.is_dir()]
    for dirname in list(subfolders):
        subfolders.extend(fast_scandir(dirname))
    return subfolders

Returns a list of all subfolders with their full paths. This again is faster than os.walk and a lot faster than glob.


An analysis of all functions

tl;dr:
- If you want to get all immediate subdirectories for a folder use os.scandir.
- If you want to get all subdirectories, even nested ones, use os.walk or - slightly faster - the fast_scandir function above.
- Never use os.walk for only top-level subdirectories, as it can be hundreds(!) of times slower than os.scandir.

  • If you run the code below, make sure to run it once so that your OS will have accessed the folder, discard the results and run the test, otherwise results will be screwed.
  • You might want to mix up the function calls, but I tested it, and it did not really matter.
  • All examples will give the full path to the folder. The pathlib example as a (Windows)Path object.
  • The first element of os.walk will be the base folder. So you will not get only subdirectories. You can use fu.pop(0) to remove it.
  • None of the results will use natural sorting. This means results will be sorted like this: 1, 10, 2. To get natural sorting (1, 2, 10), please have a look at https://stackoverflow.com/a/48030307/2441026


Results:

os.scandir      took   1 ms. Found dirs: 439
os.walk         took 463 ms. Found dirs: 441 -> it found the nested one + base folder.
glob.glob       took  20 ms. Found dirs: 439
pathlib.iterdir took  18 ms. Found dirs: 439
os.listdir      took  18 ms. Found dirs: 439

Tested with W7x64, Python 3.8.1.

# -*- coding: utf-8 -*-
# Python 3


import time
import os
from glob import glob
from pathlib import Path


directory = r"<insert_folder>"
RUNS = 1


def run_os_walk():
    a = time.time_ns()
    for i in range(RUNS):
        fu = [x[0] for x in os.walk(directory)]
    print(f"os.walk			took {(time.time_ns() - a) / 1000 / 1000 / RUNS:.0f} ms. Found dirs: {len(fu)}")


def run_glob():
    a = time.time_ns()
    for i in range(RUNS):
        fu = glob(directory + "/*/")
    print(f"glob.glob		took {(time.time_ns() - a) / 1000 / 1000 / RUNS:.0f} ms. Found dirs: {len(fu)}")


def run_pathlib_iterdir():
    a = time.time_ns()
    for i in range(RUNS):
        dirname = Path(directory)
        fu = [f for f in dirname.iterdir() if f.is_dir()]
    print(f"pathlib.iterdir	took {(time.time_ns() - a) / 1000 / 1000 / RUNS:.0f} ms. Found dirs: {len(fu)}")


def run_os_listdir():
    a = time.time_ns()
    for i in range(RUNS):
        dirname = Path(directory)
        fu = [os.path.join(directory, o) for o in os.listdir(directory) if os.path.isdir(os.path.join(directory, o))]
    print(f"os.listdir		took {(time.time_ns() - a) / 1000 / 1000 / RUNS:.0f} ms. Found dirs: {len(fu)}")


def run_os_scandir():
    a = time.time_ns()
    for i in range(RUNS):
        fu = [f.path for f in os.scandir(directory) if f.is_dir()]
    print(f"os.scandir		took {(time.time_ns() - a) / 1000 / 1000 / RUNS:.0f} ms.	Found dirs: {len(fu)}")


if __name__ == "__main__":
    run_os_scandir()
    run_os_walk()
    run_glob()
    run_pathlib_iterdir()
    run_os_listdir()

Answer #7

In Python 3, TemporaryDirectory in the tempfile module can be used.

This is straight from the examples:

import tempfile
with tempfile.TemporaryDirectory() as tmpdirname:
     print("created temporary directory", tmpdirname)
# directory and contents have been removed

If you would like to keep the directory a bit longer, you could do something like this:

import tempfile

temp_dir = tempfile.TemporaryDirectory()
print(temp_dir.name)
# use temp_dir, and when done:
temp_dir.cleanup()

The documentation also says that "On completion of the context or destruction of the temporary directory object the newly created temporary directory and all its contents are removed from the filesystem." So at the end of the program, for example, Python will clean up the directory if it wasn"t explicitly removed. Python"s unittest may complain of ResourceWarning: Implicitly cleaning up <TemporaryDirectory... if you rely on this, though.

Answer #8

The Pathlib method (Python 3.4+)

from pathlib import Path
Path("C:Program Files").parent
# Returns a Pathlib object

The traditional method

import os.path
os.path.dirname("C:Program Files")
# Returns a string


Which method should I use?

Use the traditional method if:

  • You are worried about existing code generating errors if it were to use a Pathlib object. (Since Pathlib objects cannot be concatenated with strings.)

  • Your Python version is less than 3.4.

  • You need a string, and you received a string. Say for example you have a string representing a filepath, and you want to get the parent directory so you can put it in a JSON string. It would be kind of silly to convert to a Pathlib object and back again for that.

If none of the above apply, use Pathlib.



What is Pathlib?

If you don"t know what Pathlib is, the Pathlib module is a terrific module that makes working with files even easier for you. Most if not all of the built in Python modules that work with files will accept both Pathlib objects and strings. I"ve highlighted below a couple of examples from the Pathlib documentation that showcase some of the neat things you can do with Pathlib.

Navigating inside a directory tree:

>>> p = Path("/etc")
>>> q = p / "init.d" / "reboot"
>>> q
PosixPath("/etc/init.d/reboot")
>>> q.resolve()
PosixPath("/etc/rc.d/init.d/halt")

Querying path properties:

>>> q.exists()
True
>>> q.is_dir()
False

Answer #9

I had the exact same issue. If you are using a relative path os.path.dirname(path) will only return the relative path. os.path.realpath does the trick:

>>> import os
>>> f = open("file.txt")
>>> os.path.realpath(f.name)

Answer #10

Starting from Python 3.5, pathlib.Path.mkdir has an exist_ok flag:

from pathlib import Path
path = Path("/my/directory/filename.txt")
path.parent.mkdir(parents=True, exist_ok=True) 
# path.parent ~ os.path.dirname(path)

This recursively creates the directory and does not raise an exception if the directory already exists.

(just as os.makedirs got an exist_ok flag starting from python 3.2 e.g os.makedirs(path, exist_ok=True))


Note: when i posted this answer none of the other answers mentioned exist_ok...

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