How do I create a directory, and any missing parent directories?

Question

How do I create a directory at a given path, and also create any missing parent directories along that path? For example, the Bash command mkdir -p /path/to/nested/directory does this.

Metcalfe

In general you might need to account for the case where there's no directory in the filename. On my machine dirname('foo.txt') gives '', which doesn't exist and causes makedirs() to fail. – Argali

if the path exists one has not only to check if it is a directory and not a regular file or another object (many answers check this) it is also necessary to check if it is writable (I did not find an answer that checked this) – Southeastwardly

In case you came here to create parent directories of file path string p, here is my code snippet: os.makedirs(p[:p.rindex(os.path.sep)], exist_ok=True) – Cowage

Answer 1

On Python ≥ 3.5, use pathlib.Path.mkdir:

from pathlib import Path
Path("/my/directory").mkdir(parents=True, exist_ok=True)

For older versions of Python, I see two answers with good qualities, each with a small flaw, so I will give my take on it:

Try os.path.exists, and consider os.makedirs for the creation.

import os
if not os.path.exists(directory):
    os.makedirs(directory)

As noted in comments and elsewhere, there's a race condition – if the directory is created between the os.path.exists and the os.makedirs calls, the os.makedirs will fail with an OSError. Unfortunately, blanket-catching OSError and continuing is not foolproof, as it will ignore a failure to create the directory due to other factors, such as insufficient permissions, full disk, etc.

One option would be to trap the OSError and examine the embedded error code (see Is there a cross-platform way of getting information from Python’s OSError):

import os, errno

try:
    os.makedirs(directory)
except OSError as e:
    if e.errno != errno.EEXIST:
        raise

Alternatively, there could be a second os.path.exists, but suppose another created the directory after the first check, then removed it before the second one – we could still be fooled.

Depending on the application, the danger of concurrent operations may be more or less than the danger posed by other factors such as file permissions. The developer would have to know more about the particular application being developed and its expected environment before choosing an implementation.

Modern versions of Python improve this code quite a bit, both by exposing FileExistsError (in 3.3+)...

try:
    os.makedirs("path/to/directory")
except FileExistsError:
    # directory already exists
    pass

...and by allowing a keyword argument to os.makedirs called exist_ok (in 3.2+).

os.makedirs("path/to/directory", exist_ok=True)  # succeeds even if directory exists.

Ceric

The race condition is a good point, but the approach in stackoverflow.com/questions/273192/#273208, will mask a failure to create the directory. Don't feel bad for voting down - you don't like the answer. It's what votes are for. – Ceric

Remember that os.path.exists() isn't free. If the normal case is that the directory will be there, then the case where it isn't should be handled as an exception. In other words, try to open and write to your file, catch the OSError exception and, based on errno, do your makedir() and re-try or re-raise. This creates duplication of code unless you wrap the writing in a local method. – Undertake

os.path.exists also returns True for a file. I have posted an answer to address this. – Tezel

As commenters to other answers here have noted, the exists_ok parameter to os.makedirs() can be used to cover how prior existence of the path is handled, since Python 3.2. – Southward

os.mkdirs() can create unintended folders if a path separator is accidentally left out, the current folder is not as expected, a path element contains the path separator. If you use os.mkdir() these bugs will raise an exception, alerting you to their existence. – Capacitate

If you want to create the directory in your current working directory leave out the first slash Path("my_directory").mkdir(parents=True, exist_ok=True). – Later

Answer 2

Python 3.5+:

import pathlib
pathlib.Path('/my/directory').mkdir(parents=True, exist_ok=True) 

pathlib.Path.mkdir as used above recursively creates the directory and does not raise an exception if the directory already exists. If you don't need or want the parents to be created, skip the parents argument.

Python 3.2+:

Using pathlib:

If you can, install the current pathlib backport named pathlib2. Do not install the older unmaintained backport named pathlib. Next, refer to the Python 3.5+ section above and use it the same.

If using Python 3.4, even though it comes with pathlib, it is missing the useful exist_ok option. The backport is intended to offer a newer and superior implementation of mkdir which includes this missing option.

Using os:

import os
os.makedirs(path, exist_ok=True)

os.makedirs as used above recursively creates the directory and does not raise an exception if the directory already exists. It has the optional exist_ok argument only if using Python 3.2+, with a default value of False. This argument does not exist in Python 2.x up to 2.7. As such, there is no need for manual exception handling as with Python 2.7.

Python 2.7+:

Using pathlib:

If you can, install the current pathlib backport named pathlib2. Do not install the older unmaintained backport named pathlib. Next, refer to the Python 3.5+ section above and use it the same.

Using os:

import os
try: 
    os.makedirs(path)
except OSError:
    if not os.path.isdir(path):
        raise

While a naive solution may first use os.path.isdir followed by os.makedirs, the solution above reverses the order of the two operations. In doing so, it prevents a common race condition having to do with a duplicated attempt at creating the directory, and also disambiguates files from directories.

Note that capturing the exception and using errno is of limited usefulness because OSError: [Errno 17] File exists, i.e. errno.EEXIST, is raised for both files and directories. It is more reliable simply to check if the directory exists.

Alternative:

mkpath creates the nested directory, and does nothing if the directory already exists. This works in both Python 2 and 3. Note however that distutils has been deprecated, and is scheduled for removal in Python 3.12.

import distutils.dir_util
distutils.dir_util.mkpath(path)

Per Bug 10948, a severe limitation of this alternative is that it works only once per python process for a given path. In other words, if you use it to create a directory, then delete the directory from inside or outside Python, then use mkpath again to recreate the same directory, mkpath will simply silently use its invalid cached info of having previously created the directory, and will not actually make the directory again. In contrast, os.makedirs doesn't rely on any such cache. This limitation may be okay for some applications.

---

With regard to the directory's mode, please refer to the documentation if you care about it.

Tezel

This answer covers pretty much every special case as far as I can tell. I plan on wrapping this in a "if not os.path.isdir()" though since I expect the directory to exist almost every time and I can avoid the exception that way. – Sharl

@CharlesL. An exception is probably cheaper than the disk IO of the check, if your reason is performance. – Thorma

@Thorma but makedirs does additional stat, umask, lstat when only checking to throw OSError. – Kippy

@CharesL Anytime there's a directory separator, os.makedirs() starts off with a stat() so a pre-check with os.path.isdir() is just extra effort. – Malcom

This is the wrong answer, as it introduces a potential FS race cond. See answer from Aaron Hall. – Merwin

as @sleepycal has said, this suffers from a similar race condition as the accepted answer. If between raising the error and checking os.path.isdir someone else deletes the folder, you will raise the wrong, outdated, and confusing error that folder exists. – Borage

The concern about race conditions noted by @sleepycal is addressed in the comments on Heikki Toivonen's answer for many (most) use cases, including mine. If you're worried that someone else is going to remove the directory, you're still going to need to check that before using it later on. Creating multiple places where you check for what you'll need later just adds complications. – Shrink

On Windows, what about permissions errors when parent directory is owned by Administrator, or else a special directory e.g. C:\Program Files or ...(x86) or Documents and Settings, and hence creating directory fails? Does this code propagate the exact permissions error? – Lineup

@Lineup It doesn't seem so, but at least on Linux the available permissions can then be checked. I don't know about Windows. – Tezel

Note that NotADirectoryError: [Errno 20] Not a directory: ... is raised by pathlib.Path('/my/directory').mkdir(parents=True, exist_ok=True) in cases where a file exists matching the proposed name of a directory in the path, e.g., if my is a file. – Fanatic

Answer 3

Using try except and the right error code from errno module gets rid of the race condition and is cross-platform:

import os
import errno

def make_sure_path_exists(path):
    try:
        os.makedirs(path)
    except OSError as exception:
        if exception.errno != errno.EEXIST:
            raise

In other words, we try to create the directories, but if they already exist we ignore the error. On the other hand, any other error gets reported. For example, if you create dir 'a' beforehand and remove all permissions from it, you will get an OSError raised with errno.EACCES (Permission denied, error 13).

Flameout

What's the common reason against this? ie: why do people prefer the if statement to the try/except? Is it that try/except is more dangerous if you miss the error? – Sturm

The accepted answer is actually dangerous because it has a race-condition. It is simpler, though, so if you are unaware of the race-condition, or think it won't apply to you, that would be your obvious first pick. – Flameout

Raising the exception only when exception.errno != errno.EEXIST will unintentionally ignore the case when path exists but is a non-directory object such as a file. The exception should ideally be raised if the path is a non-directory object. – Tezel

Note that the above code is equivalent to os.makedirs(path,exist_ok=True) – Pollaiuolo

@Pollaiuolo The exist_ok parameter was introduced in Python 3.2. It is not present in Python 2.x. I will incorporate it into my answer. – Tezel

@HeikkiToivonen Technically speaking, if another program is modifying the directories and files at the same time your program is, your entire program is one giant race condition. What's to stop another program from just deleting this directory after the code creates it and before you actually put files in it? – Thorma

In Python 3.3 and above, you can use FileExistsError. – Prototype

Downvoted. The source code of mkdirs actually warns against checking for EEXIST: github.com/python/cpython/blob/… – Nosography

@Clément You're looking at Python 3.x source code, the answer given above applies to Python 2.7: github.com/python/cpython/blob/2.7/Lib/os.py#L153 – Ima

@jpmc26: It's not that there are no race conditions; it's that your program doesn't contain two different places where you might detect what (due to races) is really the same condition. – Wootan

@DavisHerring What you say isn't exactly true. An existence check occurs any time you access the directory or its contents. So if you create the directory and then open a file in it, you're still vulnerable to the same race condition. The answer is good; it's just still not safe if another program is modifying stuff, which is what Heikki (not the answer) asserted. – Thorma

@jpmc26: The races are materially worse the other way: with if not path.exists(...) a concurrent mkdir can give you an exception (despite the check). More broadly put, the check doesn't protect you from having to deal with any race condition, so it's strictly inferior to include it. – Wootan

@jpmc26: It prevents the mkdir race (in that you have two clear cases and obvious code paths to deal with them). Contention doesn't have to be a "battle": it's easily useful to have several programs create different parts of a directory. – Wootan

Answer 4

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...

Sophi

one note: it will be '[Errno 17] File exists' error if foldername and filename are the same in a location, even an exist_ok flag – Compeer

Answer 5

I would personally recommend that you use os.path.isdir() to test instead of os.path.exists().

>>> os.path.exists('/tmp/dirname')
True
>>> os.path.exists('/tmp/dirname/filename.etc')
True
>>> os.path.isdir('/tmp/dirname/filename.etc')
False
>>> os.path.isdir('/tmp/fakedirname')
False

If you have:

>>> directory = raw_input(":: ")

And a foolish user input:

:: /tmp/dirname/filename.etc

... You're going to end up with a directory named filename.etc when you pass that argument to os.makedirs() if you test with os.path.exists().

Fulmar

Answer 6

Check os.makedirs: (It makes sure the complete path exists.)
To handle the fact the directory might exist, catch OSError. (If exist_ok is False (the default), an OSError is raised if the target directory already exists.)

import os
try:
    os.makedirs('./path/to/somewhere')
except OSError:
    pass

Ddene

with the try/except, you will mask errors in directory creation, in the case when the directory didn't exist but for some reason you can't make it – Ceric

Answer 7

Try the os.path.exists function

if not os.path.exists(dir):
    os.mkdir(dir)

Unsought

This doesn't create parent directories. From the os.mkdir docs: "If a parent directory in the path does not exist, FileNotFoundError is raised." – Luana

Answer 8

Insights on the specifics of this situation

You give a particular file at a certain path and you pull the directory from the file path. Then after making sure you have the directory, you attempt to open a file for reading. To comment on this code:

filename = "/my/directory/filename.txt"
dir = os.path.dirname(filename)

We want to avoid overwriting the builtin function, dir. Also, filepath or perhaps fullfilepath is probably a better semantic name than filename so this would be better written:

import os
filepath = '/my/directory/filename.txt'
directory = os.path.dirname(filepath)

Your end goal is to open this file, you initially state, for writing, but you're essentially approaching this goal (based on your code) like this, which opens the file for reading:

if not os.path.exists(directory):
    os.makedirs(directory)
f = file(filename)

Assuming opening for reading

Why would you make a directory for a file that you expect to be there and be able to read?

Just attempt to open the file.

with open(filepath) as my_file:
    do_stuff(my_file)

If the directory or file isn't there, you'll get an IOError with an associated error number: errno.ENOENT will point to the correct error number regardless of your platform. You can catch it if you want, for example:

import errno
try:
    with open(filepath) as my_file:
        do_stuff(my_file)
except IOError as error:
    if error.errno == errno.ENOENT:
        print 'ignoring error because directory or file is not there'
    else:
        raise

Assuming we're opening for writing

This is probably what you're wanting.

In this case, we probably aren't facing any race conditions. So just do as you were, but note that for writing, you need to open with the w mode (or a to append). It's also a Python best practice to use the context manager for opening files.

import os
if not os.path.exists(directory):
    os.makedirs(directory)
with open(filepath, 'w') as my_file:
    do_stuff(my_file)

However, say we have several Python processes that attempt to put all their data into the same directory. Then we may have contention over creation of the directory. In that case it's best to wrap the makedirs call in a try-except block.

import os
import errno
if not os.path.exists(directory):
    try:
        os.makedirs(directory)
    except OSError as error:
        if error.errno != errno.EEXIST:
            raise
with open(filepath, 'w') as my_file:
    do_stuff(my_file)

Melancholy

Answer 9

I have put the following down. It's not totally foolproof though.

import os

dirname = 'create/me'

try:
    os.makedirs(dirname)
except OSError:
    if os.path.exists(dirname):
        # We are nearly safe
        pass
    else:
        # There was an error on creation, so make sure we know about it
        raise

Now as I say, this is not really foolproof, because we have the possiblity of failing to create the directory, and another process creating it during that period.

Profiteer

Answer 10

Check if a directory exists and create it if necessary?

The direct answer to this is, assuming a simple situation where you don't expect other users or processes to be messing with your directory:

if not os.path.exists(d):
    os.makedirs(d)

or if making the directory is subject to race conditions (i.e. if after checking the path exists, something else may have already made it) do this:

import errno
try:
    os.makedirs(d)
except OSError as exception:
    if exception.errno != errno.EEXIST:
        raise

But perhaps an even better approach is to sidestep the resource contention issue, by using temporary directories via tempfile:

import tempfile

d = tempfile.mkdtemp()

Here's the essentials from the online doc:

mkdtemp(suffix='', prefix='tmp', dir=None)
    User-callable function to create and return a unique temporary
    directory.  The return value is the pathname of the directory.

    The directory is readable, writable, and searchable only by the
    creating user.

    Caller is responsible for deleting the directory when done with it.

New in Python 3.5: pathlib.Path with exist_ok

There's a new Path object (as of 3.4) with lots of methods one would want to use with paths - one of which is mkdir.

(For context, I'm tracking my weekly rep with a script. Here's the relevant parts of code from the script that allow me to avoid hitting Stack Overflow more than once a day for the same data.)

First the relevant imports:

from pathlib import Path
import tempfile

We don't have to deal with os.path.join now - just join path parts with a /:

directory = Path(tempfile.gettempdir()) / 'sodata'

Then I idempotently ensure the directory exists - the exist_ok argument shows up in Python 3.5:

directory.mkdir(exist_ok=True)

Here's the relevant part of the documentation:

If exist_ok is true, FileExistsError exceptions will be ignored (same behavior as the POSIX mkdir -p command), but only if the last path component is not an existing non-directory file.

Here's a little more of the script - in my case, I'm not subject to a race condition, I only have one process that expects the directory (or contained files) to be there, and I don't have anything trying to remove the directory.

todays_file = directory / str(datetime.datetime.utcnow().date())
if todays_file.exists():
    logger.info("todays_file exists: " + str(todays_file))
    df = pd.read_json(str(todays_file))

Path objects have to be coerced to str before other APIs that expect str paths can use them.

Perhaps Pandas should be updated to accept instances of the abstract base class, os.PathLike.

Melancholy

Answer 11

fastest safest way to do it is: it will create if not exists and skip if exists:

from pathlib import Path
Path("path/with/childs/.../").mkdir(parents=True, exist_ok=True)

Goodfornothing

Answer 12

import os
directory = "./out_dir/subdir1/subdir2"
if not os.path.exists(directory):
    os.makedirs(directory)

Jargonize

Answer 13

In Python 3.4 you can also use the brand new pathlib module:

from pathlib import Path
path = Path("/my/directory/filename.txt")
try:
    if not path.parent.exists():
        path.parent.mkdir(parents=True)
except OSError:
    # handle error; you can also catch specific errors like
    # FileExistsError and so on.

Conservatism

Answer 14

For a one-liner solution, you can use IPython.utils.path.ensure_dir_exists():

from IPython.utils.path import ensure_dir_exists
ensure_dir_exists(dir)

From the documentation: Ensure that a directory exists. If it doesn’t exist, try to create it and protect against a race condition if another process is doing the same.

IPython is an extension package, not part of the standard library.

Girvin

Answer 15

In Python3, os.makedirs supports setting exist_ok. The default setting is False, which means an OSError will be raised if the target directory already exists. By setting exist_ok to True, OSError (directory exists) will be ignored and the directory will not be created.

os.makedirs(path,exist_ok=True)

In Python2, os.makedirs doesn't support setting exist_ok. You can use the approach in heikki-toivonen's answer:

import os
import errno

def make_sure_path_exists(path):
    try:
        os.makedirs(path)
    except OSError as exception:
        if exception.errno != errno.EEXIST:
            raise

Hogwash

Answer 16

The relevant Python documentation suggests the use of the EAFP coding style (Easier to Ask for Forgiveness than Permission). This means that the code

try:
    os.makedirs(path)
except OSError as exception:
    if exception.errno != errno.EEXIST:
        raise
    else:
        print "\nBE CAREFUL! Directory %s already exists." % path

is better than the alternative

if not os.path.exists(path):
    os.makedirs(path)
else:
    print "\nBE CAREFUL! Directory %s already exists." % path

The documentation suggests this exactly because of the race condition discussed in this question. In addition, as others mention here, there is a performance advantage in querying once instead of twice the OS. Finally, the argument placed forward, potentially, in favour of the second code in some cases --when the developer knows the environment the application is running-- can only be advocated in the special case that the program has set up a private environment for itself (and other instances of the same program).

Even in that case, this is a bad practice and can lead to long useless debugging. For example, the fact we set the permissions for a directory should not leave us with the impression permissions are set appropriately for our purposes. A parent directory could be mounted with other permissions. In general, a program should always work correctly and the programmer should not expect one specific environment.

Napoleon

Answer 17

I found this Q/A after I was puzzled by some of the failures and errors I was getting while working with directories in Python. I am working in Python 3 (v.3.5 in an Anaconda virtual environment on an Arch Linux x86_64 system).

Consider this directory structure:

└── output/         ## dir
   ├── corpus       ## file
   ├── corpus2/     ## dir
   └── subdir/      ## dir

Here are my experiments/notes, which provides clarification:

# ----------------------------------------------------------------------------
# [1] https://mcmap.net/q/36061/-how-do-i-create-a-directory-and-any-missing-parent-directories

import pathlib

""" Notes:
        1.  Include a trailing slash at the end of the directory path
            ("Method 1," below).
        2.  If a subdirectory in your intended path matches an existing file
            with same name, you will get the following error:
            "NotADirectoryError: [Errno 20] Not a directory:" ...
"""
# Uncomment and try each of these "out_dir" paths, singly:

# ----------------------------------------------------------------------------
# METHOD 1:
# Re-running does not overwrite existing directories and files; no errors.

# out_dir = 'output/corpus3'                ## no error but no dir created (missing tailing /)
# out_dir = 'output/corpus3/'               ## works
# out_dir = 'output/corpus3/doc1'           ## no error but no dir created (missing tailing /)
# out_dir = 'output/corpus3/doc1/'          ## works
# out_dir = 'output/corpus3/doc1/doc.txt'   ## no error but no file created (os.makedirs creates dir, not files!  ;-)
# out_dir = 'output/corpus2/tfidf/'         ## fails with "Errno 20" (existing file named "corpus2")
# out_dir = 'output/corpus3/tfidf/'         ## works
# out_dir = 'output/corpus3/a/b/c/d/'       ## works

# [2] https://docs.python.org/3/library/os.html#os.makedirs

# Uncomment these to run "Method 1":

#directory = os.path.dirname(out_dir)
#os.makedirs(directory, mode=0o777, exist_ok=True)

# ----------------------------------------------------------------------------
# METHOD 2:
# Re-running does not overwrite existing directories and files; no errors.

# out_dir = 'output/corpus3'                ## works
# out_dir = 'output/corpus3/'               ## works
# out_dir = 'output/corpus3/doc1'           ## works
# out_dir = 'output/corpus3/doc1/'          ## works
# out_dir = 'output/corpus3/doc1/doc.txt'   ## no error but creates a .../doc.txt./ dir
# out_dir = 'output/corpus2/tfidf/'         ## fails with "Errno 20" (existing file named "corpus2")
# out_dir = 'output/corpus3/tfidf/'         ## works
# out_dir = 'output/corpus3/a/b/c/d/'       ## works

# Uncomment these to run "Method 2":

#import os, errno
#try:
#       os.makedirs(out_dir)
#except OSError as e:
#       if e.errno != errno.EEXIST:
#               raise
# ----------------------------------------------------------------------------

Conclusion: in my opinion, "Method 2" is more robust.

[1] How can I safely create a nested directory?

[2] https://docs.python.org/3/library/os.html#os.makedirs

Tapia

The reason your method 1 doesn't work is the call to os.dirname. Without it, method 1 works as expected. – Necrotomy

Answer 18

You can use mkpath

# Create a directory and any missing ancestor directories. 
# If the directory already exists, do nothing.

from distutils.dir_util import mkpath
mkpath("test")    

Note that it will create the ancestor directories as well.

It works for Python 2 and 3.

Poetaster

Answer 19

You can pass the exist_ok=True parameter to the os.makedirs() function to suppress the error in case the directory already exists:

import os

# Create directory /path/to/nested/directory if it doesn't already exist
os.makedirs('/path/to/nested/directory', exist_ok=True)

Raspy

Very clean/concise solution that does not require if statements! – Tonina

Answer 20

In case you're writing a file to a variable path, you can use this on the file's path to make sure that the parent directories are created.

from pathlib import Path

path_to_file = Path("zero/or/more/directories/file.ext")
parent_directory_of_file = path_to_file.parent
parent_directory_of_file.mkdir(parents=True, exist_ok=True)

Works even if path_to_file is file.ext (zero directories deep).

See pathlib.PurePath.parent and pathlib.Path.mkdir.

Driggers

Answer 21

Why not use subprocess module if running on a machine that supports command mkdir with -p option ? Works on python 2.7 and python 3.6

from subprocess import call
call(['mkdir', '-p', 'path1/path2/path3'])

Should do the trick on most systems.

In situations where portability doesn't matter (ex, using docker) the solution is a clean 2 lines. You also don't have to add logic to check if directories exist or not. Finally, it is safe to re-run without any side effects

If you need error handling:

from subprocess import check_call
try:
    check_call(['mkdir', '-p', 'path1/path2/path3'])
except:
    handle...

Spokesman

Answer 22

You have to set the full path before creating the directory:

import os,sys,inspect
import pathlib

currentdir = os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe())))
your_folder = currentdir + "/" + "your_folder"

if not os.path.exists(your_folder):
   pathlib.Path(your_folder).mkdir(parents=True, exist_ok=True)

This works for me and hopefully, it will works for you as well

Elanaeland

Answer 23

I saw Heikki Toivonen and A-B-B's answers and thought of this variation.

import os
import errno

def make_sure_path_exists(path):
    try:
        os.makedirs(path)
    except OSError as exception:
        if exception.errno != errno.EEXIST or not os.path.isdir(path):
            raise

Exaggerative

Answer 24

I use os.path.exists(), here is a Python 3 script that can be used to check if a directory exists, create one if it does not exist, and delete it if it does exist (if desired).

It prompts users for input of the directory and can be easily modified.

Diorio

If you created this script, please include it in the answer here. Links can become invalid if the linked page changes. – Luana

Answer 25

Use this command check and create dir

 if not os.path.isdir(test_img_dir):
     os.mkdir(test_img_dir)

Ironlike

This doesn't create parent directories. From the os.mkdir docs: "If a parent directory in the path does not exist, FileNotFoundError is raised." – Luana

Answer 26

Call the function create_dir() at the entry point of your program/project.

import os

def create_dir(directory):
    if not os.path.exists(directory):
        print('Creating Directory '+directory)
        os.makedirs(directory)

create_dir('Project directory')

Friedrick

Answer 27

This may not exactly answer the question. But I guess your real intention is to create a file and its parent directories, given its content all in 1 command.

You can do that with fastcore extension to pathlib: path.mk_write(data)

from fastcore.utils import Path
Path('/dir/to/file.txt').mk_write('Hello World')

See more in fastcore documentation

Stalinist